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		<title>Magnetic Force</title>
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		<updated>2017-11-28T21:04:51Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. Now the question is, would the direction of the magnetic field always point in the positive z direction?&lt;br /&gt;
&lt;br /&gt;
No, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative to the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpendicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Also, take note of the iterative calculations made in the code. Within the code, we must initalize values for the initial velocity and momentum, position, mass, and charge of the particle, and magnetic field present in the location of the electron. In the iterative calculations, we must update the value of the magnetic force, as it is constantly changing directions since the electron&#039;s velocity is also changing in direction. Similarly, a net force causes a change in momentum, so we must update the momentum and velocity of the particle by utilizing the momentum principle where the derivative of momentum with respect to time is equivalent to the net force acting upon the particle. Furthermore, we update the particle&#039;s position and extend and append the trail with the particle&#039;s current location to display the path. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Additionally, the topic of magnetic force can&#039;t be ignored without mentioning magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn&#039;t discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other. &lt;br /&gt;
&lt;br /&gt;
Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only affect moving charges. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29907</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29907"/>
		<updated>2017-11-28T21:04:10Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. Now the question is, would the direction of the magnetic field always point in the positive z direction?&lt;br /&gt;
&lt;br /&gt;
No, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative to the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpendicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
Also, take note of the iterative calculations made in the code. Within the code, we must initalize values for the initial velocity and momentum, position, mass, and charge of the particle, and magnetic field present in the location of the electron. In the iterative calculations, we must update the value of the magnetic force, as it is constantly changing directions since the electron&#039;s velocity is also changing in direction. Similarly, a net force causes a change in momentum, so we must update the momentum and velocity of the particle by utilizing the momentum principle where the derivative of momentum with respect to time is equivalent to the net force acting upon the particle. Furthermore, we update the particle&#039;s position and extend and append the trail with the particle&#039;s current location to display the path. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Additionally, the topic of magnetic force can&#039;t be ignored without mentioning magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn&#039;t discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other. &lt;br /&gt;
&lt;br /&gt;
Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only affect moving charges. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29906</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29906"/>
		<updated>2017-11-28T20:58:57Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. Now the question is, would the direction of the magnetic field always point in the positive z direction?&lt;br /&gt;
&lt;br /&gt;
No, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative to the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpendicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Additionally, the topic of magnetic force can&#039;t be ignored without mentioning magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn&#039;t discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other. &lt;br /&gt;
&lt;br /&gt;
Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only affect moving charges. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29891</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29891"/>
		<updated>2017-11-28T20:47:48Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. Now the question is, would the direction of the magnetic field always point in the positive z direction?&lt;br /&gt;
&lt;br /&gt;
No, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative to the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpendicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Additionally, the topic of magnetic force can&#039;t be ignored without mentioning magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn&#039;t discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other. &lt;br /&gt;
&lt;br /&gt;
Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only affect moving charges. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29883</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29883"/>
		<updated>2017-11-28T20:40:27Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. However, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative ot the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpindicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Additionally, the topic of magnetic force can&#039;t be ignored without mentioning magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn&#039;t discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other. &lt;br /&gt;
&lt;br /&gt;
Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only affect moving charges. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29882</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29882"/>
		<updated>2017-11-28T20:39:54Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. However, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative ot the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpindicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Additionally, the topic of magnetic force can&#039;t be ignored without mentioning magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn&#039;t discovered until the early 1800s by Hans Christian Oersted, who used compass needles. Experiments in the 1800s demonstrated that wires set adjacent together with currents in the same direction were attracted to each other, while those with opposing currents repelled each other. &lt;br /&gt;
&lt;br /&gt;
Consequently, similar experiements were conducted with a static charge placed next to a current carrying wire, where no force was acted upon the static charge. Additionally, another experiment was conducted with a conductor placed in between two current carrying wires. Therefore, scientists could later come to a conclusion that magnetic fields are caused by moving charges, and later scientists determined that any charged particle with a velocity can produce a magnetic field, and magnetic forces can only affect moving charges. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29881</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29881"/>
		<updated>2017-11-28T20:25:01Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation &#039;&#039;&#039;(2)&#039;&#039;&#039; gives the magnitude of the magnetic force. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. Similarly, if the velocity and magnetic field direction vectors are parallel to each other, and thus the angle spanning the two vectors is zero, then the value of theta is zero. Consequently, the magnitude of the magnetic force is zero. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. However, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
However, consider the case where the initial direction of the electron&#039;s velocity was not directly perpendicular to the direction of the magnetic field. Because the magnetic field is not completely perpendicular to the magnetic field, the velocity will have parallel and perpendicular components relative to the magnetic field. As a result, the parallel component of the velocity relative ot the magnetic field causes the electron to move upwards as demonstrated in the glowscript simulation below rather than a simple circle on the x-z plane. The perpindicular component of the velocity, however, still contributes to the overall circular motion of the electron&#039;s path, and thus the overall path of the electron resembles that of a helix.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/894615d7dc?showInstructions=true Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29873</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29873"/>
		<updated>2017-11-28T20:14:01Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. However, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant and no other outside forces suddenly begin to act on the particle. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29866</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29866"/>
		<updated>2017-11-28T20:06:07Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following Glowscript model displays a moving particle&#039;s path in the presence of a magnetic field. Initially, the particle moves in the negative x direction in the presence of a magneetic field that points in the positive y direction. Therefore, because there is a the particle is moving in some perpendicular component relative to the magnetic field, the particle, in this case an electron, experiences a magnetic force. &lt;br /&gt;
&lt;br /&gt;
Initially when the particle moves in the negative x direction, the magnetic force is in the positive z direction since the cross product of particle&#039;s velocity and magnetic field yields a direction in the negative z direction. Because the particle is an electron, however, the particle experiences a force in the positive z direction. However, the direction of the magnetic force consistently changes since the direction of the particle&#039;s velocity continuously changes, and the direction of the magnetic force is dependent on the direction of the velocity of the electron. In fact, because the magnetic force is always perpendicular to the particle&#039;s velocity, the magnetic force also acts as a centripetal force that allows the electron to travel in a continuous circle as long as the magnetic field stays constant. &lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29865</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29865"/>
		<updated>2017-11-28T19:58:19Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following [http://harker-honors-physics.wikispaces.com/Charge+in+Magnetic+Fields?responseToken=041c41bb1719edcec16e23fb69c51ad37 link] contains an excellent set of Glowscript-based visualizations involving a moving charge in the presence of a given magnetic field.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true Magnetic Force on a Moving Particle]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29864</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29864"/>
		<updated>2017-11-28T19:57:24Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following [http://harker-honors-physics.wikispaces.com/Charge+in+Magnetic+Fields?responseToken=041c41bb1719edcec16e23fb69c51ad37 link] contains an excellent set of Glowscript-based visualizations involving a moving charge in the presence of a given magnetic field.&lt;br /&gt;
&lt;br /&gt;
[[https://trinket.io/glowscript/060ed7ba46?start=result&amp;amp;showInstructions=true]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29863</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29863"/>
		<updated>2017-11-28T19:53:47Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following [http://harker-honors-physics.wikispaces.com/Charge+in+Magnetic+Fields?responseToken=041c41bb1719edcec16e23fb69c51ad37 link] contains an excellent set of Glowscript-based visualizations involving a moving charge in the presence of a given magnetic field.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;iframe src=&amp;quot;https://trinket.io/embed/glowscript/060ed7ba46&amp;quot; width=&amp;quot;100%&amp;quot; height=&amp;quot;356&amp;quot; frameborder=&amp;quot;0&amp;quot; marginwidth=&amp;quot;0&amp;quot; marginheight=&amp;quot;0&amp;quot; allowfullscreen&amp;gt;&amp;lt;/iframe&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29862</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29862"/>
		<updated>2017-11-28T19:51:09Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following [http://harker-honors-physics.wikispaces.com/Charge+in+Magnetic+Fields?responseToken=041c41bb1719edcec16e23fb69c51ad37 link] contains an excellent set of Glowscript-based visualizations involving a moving charge in the presence of a given magnetic field.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29861</id>
		<title>Magnetic Force</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Magnetic_Force&amp;diff=29861"/>
		<updated>2017-11-28T19:50:58Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[File:Aurora156.jpeg]]&lt;br /&gt;
Edited by Brandon Kang Fall 2017&lt;br /&gt;
Authored by TheAstroChemist (This page was claimed first by TheAstroChemist - check the page history)&lt;br /&gt;
KALEY PARCHINSKI 11/6/16&lt;br /&gt;
JULIA LEONARD Spring 2017&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
So far we have learned that an electric field can act on a charged particle, causing a force. This applied force is based on the magnitude of the electric field applied and the sign of the particle that the electric field is acting on. The electric field is generated regardless of whether the source charge (i.e. what was responsible for that electric field) is stationary or moving.&lt;br /&gt;
&lt;br /&gt;
If the source charge is moving, it can in fact also generate a magnetic field (we see this quantitatively with the Biot-Savart Law). So to be explicit, if a charge has a velocity, it can produce a magnetic field. This applies whether you&#039;re dealing with a single point charge or a charge distribution such as a uniformly charged rod or disk. &lt;br /&gt;
&lt;br /&gt;
Now, you might reasonably guess that because an electric field brings about a force on a charged particle, then so too a magnetic field should bring about a force on a particle. However, in order for a magnetic field to implement this force, the particle of interest (i.e. not the source charge but the actual charged particle of our system of study) must be moving. &amp;quot;If the charge is not moving, the magnetic field has no effect on it, whereas electric fields affect charges even if they are at rest.&amp;quot; These two forces (electric and magnetic) can be combined to be known as the Lorentz Force, but that will be covered in further detail later. &#039;&#039;For now, we shall only focus on the specifics of the magnetic force and ignore the effects of an electric field on our system of interest. We will combine the two in later sections.&#039;&#039;   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;The Main Idea - Aurora Borealis Edition&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The Aurora Borealis or more commonly called, &#039;The Northern Lights&#039; is caused by the acceleration of electrons when they collide with the upper atmosphere of the Earth. These electrons then follow the magnetic field of the Earth towards the polar regions (in our case, specifically the North Pole). Once there, the accelerated electrons collide with and transfer their energy to other molecules/atoms in the atmosphere. The molecules/atoms are then excited to higher energy levels, and when they settle back down to lower energy levels, they emit light -- THE NORTHERN LIGHTS!! Depending on what kind of molecules/atoms the electrons collide with, determines the color of the light emitted (we&#039;ve all done this admittedly, kinda fun experiment in chemistry).&lt;br /&gt;
&lt;br /&gt;
However, the part that we&#039;re interested in is when the electrons collide with the magnetic field of the earth. &lt;br /&gt;
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===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
Suppose we have a moving particle. It has a charge given by &#039;&#039;q&#039;&#039;. It has a velocity given by &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. It is also in the presence of a magnetic field given by &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;. The force that this particle will experience is given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, for a particle at rest (&amp;lt;math&amp;gt;{\vec{v} = \vec{0}}&amp;lt;/math&amp;gt;), the particle will experience a force given by &amp;lt;math&amp;gt;{\vec{F} = \vec{0}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
The SI units involved? Force is in newtons (N), magnetic field is in tesla (T), charge is in coloumbs (C), and velocity is in meters per second (m/s).&lt;br /&gt;
&lt;br /&gt;
Note that the above equation &#039;&#039;&#039;(1)&#039;&#039;&#039; denotes a cross product of the vectors of velocity and magnetic field. Therefore, the force that the moving charged particle will experience is perpendicular to the plane spanned by those two vectors. It could also be written in another way: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(2)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q|\vec{v}||\vec{B}|sin(\theta)}&amp;lt;/math&amp;gt;            &lt;br /&gt;
&lt;br /&gt;
In equation &#039;&#039;&#039;(2)&#039;&#039;&#039;, the angle &amp;lt;math&amp;gt;{\theta}&amp;lt;/math&amp;gt; represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at. Thus, if and only if the velocity and the magnetic field vectors are exactly perpendicular, then the force that the particle will experience is simply given by the multiplication of the velocity and magnetic field magnitudes with the charge of the particle of interest. It&#039;s important to remember that the true force involved is a vector and thus it needs to be treated appropriately in most, if not all cases you will encounter. &lt;br /&gt;
&lt;br /&gt;
What about the force that a moving charged distribution will experience? This is relevant in the case of something such as a current carrying wire. &lt;br /&gt;
&lt;br /&gt;
Recall... &#039;&#039;&#039;(1)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, for some section of charge &amp;lt;math&amp;gt;{\Delta q}&amp;lt;/math&amp;gt;... &#039;&#039;&#039;(3)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... hence, for n charged particles, A cross sectional area, and sectional length &amp;lt;math&amp;gt;{\Delta L}&amp;lt;/math&amp;gt;, we have...  &#039;&#039;&#039;(4)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
... and now, by re-arranging the terms to collectively represent some current I, we have... &#039;&#039;&#039;(5)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(5)&#039;&#039;&#039; can be readily applied to any given charge distribution, whereby the partial length is in the same direction as current. It can be integrated to find the total force if need be in a manner similar to how you computed this for previous charge distributions!&lt;br /&gt;
&lt;br /&gt;
Recall that a moving charged particle generates a magnetic field &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; given by the following: &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(6)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Equation &#039;&#039;&#039;(6)&#039;&#039;&#039; involves the vector &amp;lt;math&amp;gt;{\hat{r}}&amp;lt;/math&amp;gt; which is the unit vector for the (r) vector going from the initial source position to the observation location. This is your familiar Biot-Savart Law. It&#039;s important to remember that a charge won&#039;t enact a force on itself, but a moving charge in the presence of a magnetic field will undergo a force. Thus, some problems will require you to identify the magnetic field involved and then calculate the effects of that magnetic field on a given moving charge or charge distribution.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
For our purposes we&#039;re going to focus on two things: 1- The circular orbit of the electrons in the Earth&#039;s magnetic field 2- The helical orbit of the electrons in the Earth&#039;s magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights. &lt;br /&gt;
&lt;br /&gt;
Let&#039;s imagine that a charged particle moves in a straight trajectory with some velocity, v in the x-z plane. The charged particle then encounters a uniform magnetic field in the +y direction (perpendicular to the plane of trajectory). This magnetic field also only exists in a specified region. When the charged particle encounters this B field, a force is applied that causes the particle to deflect from its straight trajectory. As soon as the particle exits the specified region of B field, it will then continue in a straight trajectory. The applied force which causes the curve in the trajectory is given to us by equation &#039;&#039;&#039;(1)&#039;&#039;&#039;. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth&#039;s magnetic field) then the charged particle will continue to move in a circular path in the x-z plane. &lt;br /&gt;
&lt;br /&gt;
What if the applied B field is not perpendicular to the trajectory? The particle will then follow a helical path. Because the B field is not perpendicular to the velocity, the velocity will have two components (parallel and perpendicular). The parallel component of the velocity is responsible for the movement that occurs in the third dimension (in our case +y). The perpendicular velocity is still responsible for the circular motion of the charged particle. Together, both of these motions create a helix.  &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;(7)&#039;&#039;&#039; &amp;lt;math&amp;gt;{\vec{p} = |\vec{T}||\vec{vparallel}|sin(\theta)}&amp;lt;/math&amp;gt;    &lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
The following [http://harker-honors-physics.wikispaces.com/Charge+in+Magnetic+Fields?responseToken=041c41bb1719edcec16e23fb69c51ad37 link] contains an excellent set of Glowscript-based visualizations involving a moving charge in the presence of a given magnetic field.&lt;br /&gt;
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==Examples==&lt;br /&gt;
&lt;br /&gt;
We can now consider several example problems related to this topic. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
A proton of velocity (4E5 m/s, +x-direction) travels through a region of magnetic field (0.2 T, +z-direction). What is the force exerted on this particle? &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
This situation involves a simple case of the velocity vector and the magnetic field vector combining appropriately to generate a force on our given particle. We have... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{v} = &amp;lt;4 \times 10^5,0,0&amp;gt; m/s}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{B} = &amp;lt;0,0,0.2&amp;gt; T}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;q = {1.6 \times 10^{-19} C}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19}) &amp;lt;4 \times 10^5,0,0&amp;gt; \times &amp;lt;0,0,0.2&amp;gt;}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The right-hand rule indicates that because the velocity vector is in the positive x-direction (index-finger), and this is crossed with the magnetic field vector in the positive z-direction, then the resulting force vector must necessarily be in the positive y-direction. &lt;br /&gt;
&lt;br /&gt;
Thus...&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = &amp;lt;0, 1.28 \times 10^{-14}, 0&amp;gt; N}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Question:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a situation where a positively charged particle (&amp;lt;math&amp;gt;{+ q}&amp;lt;/math&amp;gt;) of mass &#039;&#039;m&#039;&#039; is in a region where a magnetic field (&amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt;) is applied. It travels at a velocity (&amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;). Assume that the velocity spans the xy-plane, and that the magnetic field is upward in the z-direction. What is the radius of the circular path that this particle travels in?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This may appear to be a rather complicated situation to elucidate, but if you think about the situation carefully, it&#039;s not as hard as it seems. &lt;br /&gt;
&lt;br /&gt;
Imagine the positively charged particle travels in the xy-plane. Its magnetic field vector is directly perpendicular to it, so the particle will follow a circular path, with a constant inward force. We can then apply both what we know about magnetic force and then subsequently what we know about circular motion:&lt;br /&gt;
&lt;br /&gt;
First... the magnetic force on the particle is given by the following:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{\vec{F} = q\vec{v}\times\vec{B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Because &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{\vec{B}}&amp;lt;/math&amp;gt; are effectively perpendicular, the two vectors can be effectively combined in the following way:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The force &amp;lt;math&amp;gt;{\vec{F}}&amp;lt;/math&amp;gt; is constantly inward to generate a circular motion based path of the particle. &lt;br /&gt;
&lt;br /&gt;
Recall that for circular motion with a constant inward force, the force is given by: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, we can set the forces equal to each other: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem... &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{r = \frac {m v^2} {q v B} = \frac {m v} {q B}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt; exists to the left of the origin a distance &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; which maintains a current &amp;lt;math&amp;gt;{I_1}&amp;lt;/math&amp;gt;. Another current loop of radius &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; exists to the right of the origin a distance &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, and it maintains a current &amp;lt;math&amp;gt;{I_2}&amp;lt;/math&amp;gt;. The particle itself moves upward on the positive z-axis with a velocity &amp;lt;math&amp;gt;{\vec{v}}&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
Assume the following: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I_1 = I_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
 &lt;br /&gt;
&amp;lt;math&amp;gt;{R_1 = 0.5R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1 = 3d_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{d_1, d_2 &amp;gt;&amp;gt; R_1, R_2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The conventional current direction (for both loops) is counter-clockwise looking face-on the current loops from the left. Additionally, the center of each loop is on the x-axis (left to right). &lt;br /&gt;
&lt;br /&gt;
What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the above state variables. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
We must carefully analyze this situation. It might help to draw a diagram representing the problem specifics, but we will describe the situation here purely in terms of description, as we have deliberately made the visualization not too challenging. &lt;br /&gt;
&lt;br /&gt;
Consider each loop separately and then accordingly calculate the involved magnetic field for each along. The magnetic field acts along the x-axis and to the left (-x direction) based upon the right-hand rule for loop current, and this is the direction for each loop. For now we will focus on the magnitudes and then rationalize the directions based on the right hand-rule, mainly because the directions are accordingly perpendicular to each other in terms of the velocity and magnetic field.   &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 1:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_1}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{R_1}&amp;lt;/math&amp;gt;) &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;For loop 2:&#039;&#039; &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt; (This approximation can be used because of the fourth assumption made above, where &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt; is considerably larger than &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;)&lt;br /&gt;
&lt;br /&gt;
Now we combine the appropriate values for radius and distance in terms of &amp;lt;math&amp;gt;{R_2}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;{d_2}&amp;lt;/math&amp;gt;, so that we can combine the two magnetic field expressions for each loop and add them together accordingly. We refer to the given constraints listed above.  &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = B_1 + B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3} + \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Now let&#039;s pause to think about where we are at so far... we have a net magnetic field where the particle is positioned at the origin with a given velocity at that instant. The net magnetic field is directed in the negative x-direction, while the velocity is directed upward on the z-axis. Right hand rule would dictate that the force would be towards the negative y-direction... &#039;&#039;but wait!&#039;&#039; The particle involved here is an electron! Every good physics student knows that an electron is negatively charged and they will therefore have to reverse the sign of direction in a right-hand rule case. So, the electron would experience a force in the positive y-direction. Therefore, we can say:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can now involve our determined magnetic field that was generated by the two current carrying rings. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y). &lt;br /&gt;
&lt;br /&gt;
This was an example of a situation where we had to determine the magnetic field due to the current-carrying wires and then use that information to determine the force on the electron. Even more difficult situations may involve the current varying direction or a variation of the assumptions we applied.&lt;br /&gt;
&lt;br /&gt;
==Magnetic Forces in Wires==&lt;br /&gt;
&lt;br /&gt;
Because a current carrying wire contains moving electrons, there is a magnetic force exerted on the wire as well that can be represented by the following equation: &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math display = &amp;quot;block&amp;quot;&amp;gt;|\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For these problems, Right Hand Rule still applies. Point index finger in the direction of I, middle finger in direction of B, and thumb will point in the direction of F. &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A wire is laying in the xy plane, with I, conventional current, flowing to the right. B,magnetic field is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the force?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = ILBsin(45)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = (0.6)(0.005)(sin(45))&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{mag}| = 0.002&amp;lt;/math&amp;gt; N into the page&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
A horizontal bar is falling at a constant velocity. B, the magnetic field, points into the page. What is the amount of current and in what direction?&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = mg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;|\vec F_{grav}| = \vec F_{mag}&amp;lt;/math&amp;gt;     because there is no gravitational acceleration, the net force must equal zero. &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;mg = I(L\times\vec B)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;I = \frac{mg}{LB}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To find the direction of I: the bar is falling in the -y direction, and the magnetic field points in the -z direction. In order for the net force to equal 0, the magnetic force must point in the opposite direction of gravity. Therefore, the magnetic force is in the +y direction. Using Right Hand Rule, your thumb in the +y direction for the magnetic force, your middle finger (B) points in the -z direction, and therefore, your index finger points in the -x direction.  &lt;br /&gt;
&lt;br /&gt;
I, the conventional current, flows to the left. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
==Application (i.e. What Does This Have To Do With Anything?)==&lt;br /&gt;
&lt;br /&gt;
[[File:Aps15.jpg]]&lt;br /&gt;
&lt;br /&gt;
This topic of magnetic force is highly relevant to many specific areas in physics, engineering, chemistry, and biology. It helps to introduce another possible agent of force as a result of a magnetic field in much the same way as electric field acts as an agent of force on a charged particle. &lt;br /&gt;
&lt;br /&gt;
As a chemist (&#039;&#039;The Astrochemist&#039;&#039;, in fact), I have had the extremely exciting opportunity to work at the x-ray synchrotron at Argonne National Laboratory near Chicago (called the Advance Photon Source) where strong magnetic fields are applied to generate an extremely large acceleration of electrons that can then generate x-ray radiation. The above photo showcases this facility, which is a massive building one kilometer in circumference. While the part involving radiation will be discussed in the future of this textbook, the very core fundamentals of accelerating charged particles in a circular orbit is very well defined by the idea of magnetic force. &lt;br /&gt;
&lt;br /&gt;
These particle accelerators are utilized all over the world (in a huge number of locations) to do a vast number of useful things such as investigating material properties (at Argonne National Laboratory) or at CERN in Switzerland where they are currently conducting extremely fascinating experiments aimed at understanding the mechanics and dynamics  of the early universe. None of this would be possible without the dynamics of magnetic force!&lt;br /&gt;
&lt;br /&gt;
The aurora borealis has intrigued humans for centuries, appearing in many mythologies and folklores. But besides being a central player in ancient stories or just an awe-inspiring site, the study of the aurora borealis and the surrounding reasons for its existence has led to a host of other applications that include military exploits. &lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
The fundamental history of the core basics surrounding magnetic force is somewhat brief. The extremely well known scottish physicist James Clerk Maxwell was the first scientist to publish an equation describing the force generated by a magnetic field in 1861. &lt;br /&gt;
&lt;br /&gt;
Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820. &lt;br /&gt;
&lt;br /&gt;
Hendrik Lorentz provided the actual &amp;quot;Lorentz Force Law&amp;quot; of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands. &lt;br /&gt;
&lt;br /&gt;
These were important steps in figuring out how just how a magnetic field could generate a force on a charged particle much in the same way that an electric field did. It was already known that an electric field would generate a force on a charged particle, but this was just another piece in the puzzle.&lt;br /&gt;
&lt;br /&gt;
In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions (why electrons and positive ions alike is beyond the scope of the course and my understanding of physics, sorry). &lt;br /&gt;
&lt;br /&gt;
In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with some help from a US Navy navigational satellite. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
[http://press.web.cern.ch/press-releases/2015/11/lhc-collides-ions-new-record-energy CERN news article regarding a new collision energy achieved by their main particle accelerator, the Large Hadron Collider]&lt;br /&gt;
&lt;br /&gt;
[https://www1.aps.anl.gov/About/Welcome Argonne National Laboratory information regarding the Advanced Photon Source]&lt;br /&gt;
&lt;br /&gt;
[http://www.swpc.noaa.gov/phenomena/aurora National Oceanic and Atmospheric Administration&#039;s explanation of the Northern Lights]&lt;br /&gt;
&lt;br /&gt;
[http://www-spof.gsfc.nasa.gov/Education/aurora.htm Secrets of the Polar Aurora - NASA]&lt;br /&gt;
&lt;br /&gt;
[http://science.nationalgeographic.com/science/space/universe/auroras-heavenly-lights/  National Geographic - Heavenly Lights]&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=dFT7-_s0jh0 A short, eight minute video that covers and reviews some basic ideas, particularly in regards to getting down the direction of magnetic force in a given situation]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=X4dXXnUMHbQ&amp;amp;t=21m26s Walter Lewin, a famous former MIT Physics lecturer, demonstrates and discusses an interesting example involving magnetic force... you might find much of this lecture very helpful]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=PeGs4Eec_lc An in depth lecture conducted by Walter Lewin regarding magnetic force, something that you might find useful in your studies]&lt;br /&gt;
&lt;br /&gt;
[https://www.youtube.com/watch?v=fVMgnmi2D1w Footage from space of Aurora Borealis]&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
1. Chabay, R.W; Sherwood, B.A.; &#039;&#039;Matter and Interactions&#039;&#039;. &#039;&#039;&#039;2015&#039;&#039;&#039;. &#039;&#039;4&#039;&#039;. 805-812.&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=28902</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=28902"/>
		<updated>2017-04-10T03:04:53Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by Brandon Kang for Spring 2017&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
Claimed by dansher3 (Daniel Ansher) for Fall 2016&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;. Because the net force is equal to zero, this means that the force due to tension is equal in magnitude and opposite in direction to the force due to gravity.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
In the next example, we will analyze the code to observe how the fan cart moves and gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration since the force applied is in the same direction as the momentum.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, which is defined only on the x axis, and the final values of the momentum, position, and velocity are printed. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=28053</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=28053"/>
		<updated>2017-04-09T18:44:42Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by Brandon Kang for Spring 2017&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
Claimed by dansher3 (Daniel Ansher) for Fall 2016&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
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===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
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&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
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The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
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===A Computational Model===&lt;br /&gt;
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Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
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Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
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[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
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====Breaking Down the Code====&lt;br /&gt;
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[[File:Img111.png]]&lt;br /&gt;
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[[File:Img2.png]]&lt;br /&gt;
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[[File:Img3.png]]&lt;br /&gt;
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[[File:Img4.png]]&lt;br /&gt;
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[[File:Img5.png]]&lt;br /&gt;
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In the next example, we will analyze the code to observe how the fan cart moves and gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration since the force applied is in the same direction as the momentum.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, which is defined only on the x axis, and the final values of the momentum, position, and velocity are printed. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt;\vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
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==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
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===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
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====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
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===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
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==History==&lt;br /&gt;
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Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
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== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
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[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
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[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
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===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
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[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
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==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
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[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
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[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=28052</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=28052"/>
		<updated>2017-04-09T18:44:06Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
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&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
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&#039;&#039;&#039;Claimed by Brandon Kang for Spring 2017&#039;&#039;&#039;&lt;br /&gt;
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Claimed by hyk96610&lt;br /&gt;
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Claimed by dansher3 (Daniel Ansher) for Fall 2016&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
In the next example, we will analyze the code to observe how the fan cart moves and gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration since the force applied is in the same direction as the momentum.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, which is defined only on the x axis, and the final values of the momentum, position, and velocity are printed. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27365</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27365"/>
		<updated>2017-04-04T21:26:20Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by Brandon Kang for Spring 2017&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
Claimed by dansher3 (Daniel Ansher) for Fall 2016&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
In the next example, we will analyze the code to observe how the fan cart moves an gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, which is defined only on the x axis, and the final values of the momentum, position, and velocity are printed. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27364</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27364"/>
		<updated>2017-04-04T21:26:10Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by Brandon Kang for Spring 2017&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by dansher3 (Daniel Ansher) for Fall 2016&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
In the next example, we will analyze the code to observe how the fan cart moves an gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, which is defined only on the x axis, and the final values of the momentum, position, and velocity are printed. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27363</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27363"/>
		<updated>2017-04-04T21:17:57Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
Claimed by Brandon Kang for Spring 2017&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by dansher3 (Daniel Ansher) for Fall 2016&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
In the next example, we will analyze the code to observe how the fan cart moves an gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, which is defined only on the x axis, and the final values of the momentum, position, and velocity are printed. &lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27362</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27362"/>
		<updated>2017-04-04T21:12:45Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
Claimed by Brandon Kang for Spring 2017&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by dansher3 (Daniel Ansher) for Fall 2016&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &amp;lt;math&amp;gt;\vec{v}_{i}&amp;lt;/math&amp;gt; is the initial velocity of the system. The difference between the two can be shortened to &amp;lt;math&amp;gt;\vec{{∆v}}&amp;lt;/math&amp;gt;, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;m&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
Below is another demonstration of the Momentum Principle that involves a fan cart traveling on a track with an applied force.&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/5e9d2cfad4 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
In the next example, we will analyze the code to observe how the fan cart moves an gains momentum. &lt;br /&gt;
&lt;br /&gt;
[[File:iteration.png]]&lt;br /&gt;
&lt;br /&gt;
In lines 5-10, we initialize the values of the mass, velocity, and momentum of the fan cart, which is crucial and necessary for the code to not error when it enters the iteration, as these values must be defined. Line 14 defines the &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt;, which should be a small value to avoid inaccurate measurements of the final position, momentum, and velocity. Allowing &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; to be a large number would lead to inaccurate estimations of the change in momentum, and the momentum principle utilizes the derivative of momentum with respect to time, thus smaller measurements of &amp;lt;math&amp;gt;{∆t}&amp;lt;/math&amp;gt; should be used. Lines 15-16  initialize the values of the constant, applied force. Although this force is constant, it will increase the momentum of the fan cart through each iteration.&lt;br /&gt;
&lt;br /&gt;
Entering the iteration, the momentum of the cart is updated with the Momentum Principle. We must keep track of initial and final values of velocity before and after the Momentum Principle is utilized, as we use the average velocity defined as the numerical average of the initial and final values of velocity due to a constant force acting on the system. Therefore, the final velocity is updated after the Momentum update. The iteration continues until the cart has reached the end of the track, and the final values of the momentum, position, and velocity are printed.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:Iteration.png&amp;diff=27361</id>
		<title>File:Iteration.png</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:Iteration.png&amp;diff=27361"/>
		<updated>2017-04-04T21:10:27Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
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		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=File:6377ed10eb45e8dc55a7ef3c2c4529bd.png&amp;diff=27360</id>
		<title>File:6377ed10eb45e8dc55a7ef3c2c4529bd.png</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=File:6377ed10eb45e8dc55a7ef3c2c4529bd.png&amp;diff=27360"/>
		<updated>2017-04-04T21:02:19Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27355</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27355"/>
		<updated>2017-04-04T20:45:49Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
Claimed by Brandon Kang for Spring 2017&lt;br /&gt;
&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by dansher3 (Daniel Ansher) for Fall 2016&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass and is consequently directly proportional to the magnitude of the mass and velocity of any particular object. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{p}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in &amp;lt;math&amp;gt;kg*{\frac{m}{s}}&amp;lt;/math&amp;gt; ) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum. Therefore, momentum is a vector quantity. &lt;br /&gt;
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&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. Because momentum is defined as a vector, in the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. Therefore, a free body diagram of the system and the forces acting on the system is a helpful way to determine the various forces acting on the system. For example, a free body diagram of a very simplistic model of a ball held by a string is demonstrated below in order to analyze the forces acting on the ball. &lt;br /&gt;
&lt;br /&gt;
[[File:Fbd.png]]&lt;br /&gt;
&lt;br /&gt;
Therefore, in the above example, the net force acting on the ball is 0. We can conclude this through the momentum principle such that initial and final momentum are equal due to the ball not moving. Therefore, &amp;lt;math&amp;gt;∆\vec{p}=0&amp;lt;/math&amp;gt;, and thus &amp;lt;math&amp;gt;\vec{F}_{net} = 0&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed. Overall, if a force is applied to a system over a greater duration, the change in momentum is greater, as the momentum principle, &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;, also demonstrates that the change in momentum is directly proportional to the duration of the applied force. &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &amp;lt;math&amp;gt;\vec{v}_{f}=\vec{v}_{i}+{\frac{\vec{F}_{net}}{m}}*{∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;&amp;lt;math&amp;gt;\vec{v}_{f}&amp;lt;/math&amp;gt;&#039;&#039;&#039; is the final velocity of the system, and &#039;&#039;&#039;Vi&#039;&#039;&#039; is the initial velocity of the system. The difference between the two can be shortened to deltaV, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Fnet&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;m&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;deltat&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
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[[File:Img3.png]]&lt;br /&gt;
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[[File:Img4.png]]&lt;br /&gt;
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[[File:Img5.png]]&lt;br /&gt;
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==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
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==Connectedness==&lt;br /&gt;
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===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
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===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
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====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
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== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
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[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
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		<updated>2017-04-04T20:34:59Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: Bkang70 uploaded a new version of &amp;amp;quot;File:Fbd.png&amp;amp;quot;&lt;/p&gt;
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		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27343</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27343"/>
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		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
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&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
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Claimed by Brandon Kang for Spring 2017&lt;br /&gt;
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Claimed by hyk96610&lt;br /&gt;
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&#039;&#039;&#039;Claimed by dansher3 (Daniel Ansher) for Fall 2016&#039;&#039;&#039;&lt;br /&gt;
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==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;p&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in kg*m/s) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;F&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. In the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;t&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed.&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &#039;&#039;&#039;Vf = Vi + (Fnet/m)*deltat&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Vf&#039;&#039;&#039; is the final velocity of the system, and &#039;&#039;&#039;Vi&#039;&#039;&#039; is the initial velocity of the system. The difference between the two can be shortened to deltaV, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Fnet&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;m&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;deltat&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
&lt;br /&gt;
[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
&lt;br /&gt;
[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
&lt;br /&gt;
[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
&lt;br /&gt;
[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27342</id>
		<title>Momentum Principle</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Momentum_Principle&amp;diff=27342"/>
		<updated>2017-04-04T20:15:55Z</updated>

		<summary type="html">&lt;p&gt;Bkang70: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;This page discusses the Momentum Principle and examples of how it is used.&lt;br /&gt;
&lt;br /&gt;
Claimed by Brandon Kang for Spring 2017&lt;br /&gt;
Claimed by hyk96610&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Claimed by dansher3 (Daniel Ansher) for Fall 2016&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is the first fundamental principle of mechanics and describes the relationship between the change in momentum of a system and the total amount of interaction (or total amount of force) within the surroundings. In terms of the system and surroundings, both can be set in any way necessary, where the system may just include a person or the entire Earth. The Momentum Principle can be used in nearly all situations, and it is typically advised to start a problem by first writing out the Momentum Principle and then branching out (by rearranging or substituting values) in order to solve a problem. Overall, the momentum principle assumes that the final momentum is dependent on the initial momentum, the force applied, and the duration in which the force is applied. &lt;br /&gt;
&lt;br /&gt;
Momentum, in general, is a property and by nature is consisted of velocity and mass. This in turn determines the required amount of force and time required to stop it. This section focuses on linear momentum as opposed to angular momentum (which is when an object is rotational in motion).&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle is defined as &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system}=  \vec{F}_{net}&amp;lt;/math&amp;gt; (or &amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;). As a result, the momentum principle can alternatively be written as &amp;lt;math&amp;gt;\vec{p}_{f}=\vec{p}_{i}+\vec{F}_{net}* {∆t}&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;p&#039;&#039;&#039; is the momentum of the system. In the equation, momentum (measured in kg*m/s) is expressed as the &amp;quot;change in momentum&amp;quot; (&amp;lt;math&amp;gt;∆\vec{p} = \vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt;), which includes both the magnitude and direction of the momentum.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;F&#039;&#039;&#039; is the net force from the surroundings. Force (measured in Newtons, or N) includes the interactions between system and the surroundings, like the gravitational force exerted by the Earth on us or the force that a compressed spring exerts on a mass. In the Momentum Principle, the force includes both the magnitude and direction. Also, it is important to note that the Momentum Principle calls for the &#039;&#039;net&#039;&#039; force, which is the sum of all the different forces from the surroundings, like adding both the force of gravity and the force of the spring together to calculate the net force. Because of this, it is even more crucial to pay attention to the direction of the forces as a positive or negative sign error could cause an error in the calculated net force. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;t&#039;&#039;&#039; is the time (measured in seconds, or s). Specifically, the Momentum Principle calls for the &amp;quot;change in time&amp;quot; (&amp;lt;math&amp;gt;∆\vec{t} = \vec{t}_{final} - \vec{t}_{initial}&amp;lt;/math&amp;gt;), or in other words, the duration of the interaction is needed.&lt;br /&gt;
&lt;br /&gt;
The Momentum Principle can be further manipulated to find the change in velocity of the system. This is represented by the formula: &#039;&#039;&#039;Vf = Vi + (Fnet/m)*deltat&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Vf&#039;&#039;&#039; is the final velocity of the system, and &#039;&#039;&#039;Vi&#039;&#039;&#039; is the initial velocity of the system. The difference between the two can be shortened to deltaV, and represents the change in velocity of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Fnet&#039;&#039;&#039; is the net force acting on the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;m&#039;&#039;&#039; is the mass of the system.&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;deltat&#039;&#039;&#039; is the change in time of the system in which it is acted on by the force. &lt;br /&gt;
&lt;br /&gt;
This manipulation of the momentum principle is very useful when it comes to updating the position of the system because it gives you the change in velocity over the observed time interval.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
Click on the link to see the Momentum Principle through VPython!&lt;br /&gt;
&lt;br /&gt;
Make sure to press &amp;quot;Run&amp;quot; to see the principle in action!&lt;br /&gt;
&lt;br /&gt;
[https://trinket.io/embed/glowscript/8271b15824 Teach hand-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
====Breaking Down the Code====&lt;br /&gt;
&lt;br /&gt;
[[File:Img111.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img2.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img3.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img4.png]]&lt;br /&gt;
&lt;br /&gt;
[[File:Img5.png]]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&lt;br /&gt;
Two external forces &amp;lt;40,-70,0&amp;gt;N and &amp;lt;20,10,0&amp;gt;N, act on a system. What is the net force acting on the system?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;60,-60,0&amp;gt;N&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The net force is just synonymous for the overall force acting on the system. In this case, this is the sum of the forces given in the problem above.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{F}_{net}&amp;lt;/math&amp;gt; = &amp;lt;40,-70,0)N + &amp;lt;20,10,0&amp;gt;N = &amp;lt;60,-60,0&amp;gt;N&lt;br /&gt;
&lt;br /&gt;
===Middling===&lt;br /&gt;
&lt;br /&gt;
A hockey puck is sliding along the ice with nearly constant momentum &amp;lt;10,0,5&amp;gt;kg*m/s when it is suddenly struck by a hockey stick with a force &amp;lt;0,0,2000&amp;gt;N that lasts for only 3 milliseconds (3e-3s). What is the new (vector) momentum of the puck?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;10,0,11&amp;gt;kg*m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
In this example, we are manipulating the momentum principle in order to calculate the new momentum at a given time. This is also called the update momentum formula.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial}&amp;lt;/math&amp;gt; = \vec{F}_{net}  * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; - &amp;lt;10,0,5&amp;gt;kg*m/s = &amp;lt;0,0,2000&amp;gt;N * (3e-3)s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final}&amp;lt;/math&amp;gt; = &amp;lt;10,0,11&amp;gt;kg*m/s&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
In outer space a rock of mass 5kg is acted on by a constant net force &amp;lt;29,-15,40&amp;gt;N during a 4s time interval. At the end of this time interval the rock has a velocity of &amp;lt;114,94,112&amp;gt;m/s. What is the rock&#039;s velocity at the beginning of the time interval?&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Answer: &amp;lt;90.8,106,80&amp;gt;m/s&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Explanation:&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
This example is similar to the first example. Only now, instead of manipulating variables to solve for momentum, we must manipulate the variables to solve for velocity, specifically initial velocity. Velocity can be connected to the momentum principle since momentum is just mass multiplied by velocity.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;∆\vec{p} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p}_{final} - \vec{p}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{p} = m * \vec{v}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;m\vec{v}_{final} - m\vec{v}_{initial} = \vec{F}_{net} * {∆t}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
(5kg * &amp;lt;114,94,112&amp;gt;m/s) - (5kg * &amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt;) = &amp;lt;29,-15,40&amp;gt;N * 4s&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\vec{v}_{initial}&amp;lt;/math&amp;gt; = &amp;lt;90.8,106,80&amp;gt;m/s&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
&lt;br /&gt;
===Why the Momentum Principle Connects to Your Interests===&lt;br /&gt;
&lt;br /&gt;
All over the world and at every point in time, interactions are continuously occurring, and I thought it was interesting to see how the Momentum Principle was the most fundamental principle that is used in starting to the analyze the different interactions. This fundamental idea allows us to examine how items collide, interact with each other, and most importantly, allow us to gain knowledge about how our everyday lives work as a whole.&lt;br /&gt;
&lt;br /&gt;
===How the Momentum Principle Connects to Your Major===&lt;br /&gt;
&lt;br /&gt;
====Computer Science====&lt;br /&gt;
&lt;br /&gt;
While the momentum principle is often related to the tangible physical world, there are applications or tools when building software applications that apply to computer science. For example, if I was a software engineer trying to create Roller Coaster Tycoon, I would need to virtually calculate the correct and verifiable momentums for the roller coasters to mimic real world movements. This idea can even be expanded with relation to robotics where computer scientists aim to make hardware movement mimic human movement in the real world. Again, while not directly related, the concept of momentum and this fundamental idea must be learned and incorporated in order to benefit the future of technologies, especially with regard to the computer science realm.&lt;br /&gt;
&lt;br /&gt;
===Interesting Industrial Applications===&lt;br /&gt;
&lt;br /&gt;
While the Momentum Principle is not directly connected to the applications, it is often used in the process (especially in the beginning) of industrial application. For example, when creating life saving airbags and seat belts for cars, the Momentum Principle is used. The final momentum of a car during an accident would be zero, or would stop, and the initial momentum would be based on the mass and velocity of the car. With the change in momentum fixed, the airbag and seat belt would focus on increasing the time taken for the body&#039;s momentum to reach zero (final momentum), which would consequently reduce the force of the collision and protect the body from getting as injured. With the Momentum Principle being applicable in so many areas of my life, I found the concept even more interesting.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Although the Momentum Principle is credited as Newton’s second law of motion, it is difficult to just credit Isaac Newton (1643AD – 1727AD) for the development of the principle. As the Momentum Principle is the quantitative and more in-depth representation of Newton’s first law of motion (“An object tends to be at rest or moves in a straight line and a constant speed except to the extent that it interact with other objects”), the development of the first law also serves an important role in the history of the Momentum Principle. Aristotle (384BC – 322BC) initially proposed that objects had the natural tendency to be at rest and that a push (or a force) was absolutely needed to keep the object moving. His proposal was challenged by Galileo (1564AD – 1642AD), who introduced the idea that objects had the natural tendency to travel in a straight line at constant speed unless something (or a force) was interacting with something. Likewise, Descartes (1596AD – 1650AD) also contributed as he proposed three laws of nature in his “Principle of Philosophy,” which actually outlined the later published Newton’s first law of motion. After studying Descartes, Newton adopted Descartes’ principles as his first law of motion, and alongside the famous story of Newton sitting under an apple, Newton was able to create the Momentum Principle, or his second law of motion.&lt;br /&gt;
&lt;br /&gt;
===Why is this principle so important to us?===&lt;br /&gt;
&lt;br /&gt;
While we have known about the momentum principle for quite a long time now, many (including students) still wonder why the momentum principle is so important and crucial to learn. The reality of the matter is that this fundamental principle applies to many real world applications and can explain movement in the world of science. Whether its examining collisions between objects or impact breaking/building, the principle is a practical and useful way for us to explain how our real physical world works.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
As the Momentum Principle is the first of three fundamental principles of mechanics, the next possible topics to examine would be the other fundamental principles, the [http://www.physicsbook.gatech.edu/The_Energy_Principle Energy Principle] and the [http://www.physicsbook.gatech.edu/The_Angular_Momentum_Principle Angular Momentum Principle]. Also, although the Momentum Principle is an extremely important concept that usually signals the start of a momentum related problem, the principle branches out into other momentum topics like [http://www.physicsbook.gatech.edu/Impulse_Momentum Impulse] and [http://www.physicsbook.gatech.edu/Iterative_Prediction Iterative Prediction], which are used to solve other types of problems.&lt;br /&gt;
&lt;br /&gt;
===Further Reading===&lt;br /&gt;
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[1] http://espace.library.uq.edu.au/view/UQ:273253/Pap_290f_postprint.pdf&lt;br /&gt;
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[2] http://authors.library.caltech.edu/2577/1/TOLpr30c.pdf&lt;br /&gt;
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===External links===&lt;br /&gt;
[1] https://www.khanacademy.org/science/physics/linear-momentum/momentum-tutorial/v/introduction-to-momentum&lt;br /&gt;
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[2] https://www.youtube.com/watch?v=ZvPrn3aBQG8&lt;br /&gt;
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==References==&lt;br /&gt;
[1] Harris, William. &amp;quot;How Netwon&#039;s Laws of Motion Works.&amp;quot; HowStuffWorks. HowStuffWorks.com, 29 July 2008. Web. 29 Nov. 2015. &amp;lt;http://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion.htm&amp;gt;&lt;br /&gt;
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[2] Sherwood, Bruce A. &amp;quot;2.1 The Momentum Principle.&amp;quot; Matter &amp;amp; Interactions. By Ruth W. Chabay. 4th ed. Vol. 1. N.p.: John Wiley &amp;amp; Sons, 2015. 45-50. Print. Modern Mechanics.&lt;br /&gt;
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[3] Fenton, Flavio. &amp;quot;Momentum and Second Newton&#039;s Law.&amp;quot; 26 Aug. 2015. Lecture.&lt;br /&gt;
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[[Category:Momentum]]&lt;/div&gt;</summary>
		<author><name>Bkang70</name></author>
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