# Magnetic Force

Claimed by Janel Ennin Fall 2023

## The Main Idea

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.

By applying forces to charged particles depending on both their charges and the properties of the electric field itself, an electric field acts as a mediator in interactions between charged particles. A charged particle encounters a force because of the existence of an electric field while it is in one. This force is closely correlated with the particle's charge and the strength of the electric field. The equation \( \mathbf{F} = q \cdot \mathbf{E} \) mathematically expresses this relationship, with \( \mathbf{F} \) representing the force, \( q \) representing the particle's charge, and \( \mathbf{E} \) representing the electric field vector.

The sign of the particle's charge determines the force's direction. The force acts in the direction of the electric field if the charge is positive. In contrast, the force acts in the opposite direction of the electric field when the charge is negative. This differentiation emphasizes that the force and the electric field are both vectors.

Crucially, the existence of the electric field is independent of the mobility or stationary state of the source charge that generates it. This notion follows from the fact that an electric field is created when there are changes in the distribution of charge, whether such changes are brought about by the motion of charges or by other factors.

Magnetic forces are on moving particles, not stationary particles which means that the calculation of magnetic force **MUST** relate to the particle's velocity (we see this quantitatively with the Biot-Savart Law).

If the source charge is moving, it also generate a magnetic field; so not only is velocity involved in calculation of the magnetic force on a moving particle, or collection of moving particles (as we see in a rod or a wire), but this phenomenal relationship includes magnetic field as well.

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. "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." 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. *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.*

**The Main Idea - Aurora Borealis Edition**

The Aurora Borealis or more commonly called, 'The Northern Lights' 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.

The excited molecules and atoms, having absorbed the energy imparted by the electrons, undergo a subsequent relaxation process. As they return to lower energy levels, a radiant emission of light ensues — the hallmark Northern Lights spectacle. The distinctive hues and vibrant colors characterizing the display are contingent upon the specific molecules and atoms involved in these collisions. The variance in colors is a testament to the diversity of atmospheric constituents participating in this cosmic ballet. This celestial dance, orchestrated by the interaction between charged particles and the Earth's magnetic field, not only enchants observers but also serves as a captivating reminder of the intricate interplay between astrophysical phenomena and the Earth's atmospheric composition.

## A Mathematical Model

Suppose we have a moving particle. It has a charge given by *q*. It has a velocity given by [math]\displaystyle{ {\vec{v}} }[/math]. It is also in the presence of a magnetic field given by [math]\displaystyle{ {\vec{B}} }[/math]. The force that this particle will experience is given by the following:

**(1)** [math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

Therefore, for a particle at rest ([math]\displaystyle{ {\vec{v} = \vec{0}} }[/math]), the particle will experience a force given by [math]\displaystyle{ {\vec{F} = \vec{0}} }[/math].

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).

Note that the above equation **(1)** 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:

**(2)** [math]\displaystyle{ {|\vec{F}| = q|\vec{v}||\vec{B}|sin(\theta)} }[/math]

In equation **(2)**, the angle [math]\displaystyle{ {\theta} }[/math] represents the angle spanning the velocity vector and the magnetic field vector at some given position that the particle is at, and equation **(2)** 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'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.

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.

Recall... **(1)** [math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

Thus, for some section of charge [math]\displaystyle{ {\Delta q} }[/math]... **(3)** [math]\displaystyle{ {\Delta\vec{F} = \Delta q (\vec{v}\times\vec{B})} }[/math]

... hence, for n charged particles, A cross sectional area, and sectional length [math]\displaystyle{ {\Delta L} }[/math], we have... **(4)** [math]\displaystyle{ {\Delta\vec{F} = n A \Delta L (\vec{v}\times\vec{B})} }[/math]

... and now, by re-arranging the terms to collectively represent some current I, we have... **(5)** [math]\displaystyle{ {\Delta\vec{F} = I (\vec{\Delta L}\times\vec{B})} }[/math]

Equation **(5)** 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!

Recall that a moving charged particle generates a magnetic field [math]\displaystyle{ {\vec{B}} }[/math] given by the following:

**(6)** [math]\displaystyle{ {\vec{B} = \frac {\mu_0} {4\pi} \frac {q\vec{v}\times\hat{r}} {r^2}} }[/math]

Equation **(6)** involves the vector [math]\displaystyle{ {\hat{r}} }[/math] 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's important to remember that a charge won'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.

For our purposes we're going to focus on two things: 1- The circular orbit of the electrons in the Earth's magnetic field 2- The helical orbit of the electrons in the Earth's magnetic field. The combination of these two phenomenons contribute to the creation of the Northern Lights.

Let'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 **(1)**. However, if there is a magnetic field that is large enough so that the electron cannot escape (i.e. Earth's magnetic field) then the charged particle will continue to move in a circular path in the x-z plane.

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.

## A Computational Model

The following Glowscript model displays a moving particle'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.

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'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 force always point in the positive z direction?

No, the direction of the magnetic force consistently changes since the direction of the particle'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'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.

[Magnetic Force on a Moving Particle Perpendicular to the Magnetic Field]

However, consider the case where the initial direction of the electron'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's path, and thus the overall path of the electron resembles that of a helix.

[Magnetic Force on a Moving Particle not Directly Perpendicular to the Magnetic Field]

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'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's position and extend and append the trail with the particle's current location to display the path.

## Examples

We can now consider several example problems related to this topic.

### Simple

**Question:**

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?

**Solution:**

This situation involves a simple case of the velocity vector and the magnetic field vector appropriately combining to generate a force on our given particle. We have...

[math]\displaystyle{ {\vec{v} = \lt 4 \times 10^5,0,0\gt m/s} }[/math]

[math]\displaystyle{ {\vec{B} = \lt 0,0,0.2\gt T} }[/math]

[math]\displaystyle{ q = {1.6 \times 10^{-19} C} }[/math]

Therefore:

[math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

[math]\displaystyle{ {\vec{F} = (1.6 \times 10^{-19}) \lt 4 \times 10^5,0,0\gt \times \lt 0,0,0.2\gt } }[/math]

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 (middle finger) in the positive z-direction, then the resulting force vector (direction of your thumb) must be in the negative y-direction.

Thus...

[math]\displaystyle{ {\vec{F} = (1.6 \times 10^{-19})(4 \times 10^5 * 0.2) = \lt 0, -1.28 \times 10^{-14}, 0\gt N} }[/math]

### Middling

**Question:**

Suppose we have a situation where a positively charged particle ([math]\displaystyle{ {+ q} }[/math]) of mass *m* is in a region where a magnetic field ([math]\displaystyle{ {\vec{B}} }[/math]) is applied. It travels at a velocity ([math]\displaystyle{ {\vec{v}} }[/math]). 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 in which this particle travels in terms of values given?

**Solution:**

This problem may appear complicated, but it's not as hard as it seems.

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:

First... the magnetic force on the particle is given by the following:

[math]\displaystyle{ {\vec{F} = q\vec{v}\times\vec{B}} }[/math]

Because [math]\displaystyle{ {\vec{v}} }[/math] and [math]\displaystyle{ {\vec{B}} }[/math] are effectively perpendicular, the two vectors can be effectively combined in the following way:

[math]\displaystyle{ {|\vec{F}| = q|\vec{v}| |\vec{B}|} }[/math]

The force [math]\displaystyle{ {\vec{F}} }[/math] is constantly inward to generate a circular motion based path of the particle.

Recall that for circular motion with a constant inward force, the force is given by:

[math]\displaystyle{ {|\vec{F}| = m \frac{|\vec{v}|^2} {r}} }[/math]

Thus, we can set the forces equal to each other:

[math]\displaystyle{ {|\vec{F}| = q|\vec{v}| |\vec{B}| = m \frac{|\vec{v}|^2} {r}} }[/math]

Therefore, the radius r of the circular path can be defined in terms of the given variables in the problem...

[math]\displaystyle{ {r = \frac {m v^2} {q v B} = \frac {m v} {q B}} }[/math]

### Difficult

**Problem:**

Suppose we have a negatively charged particle at the origin of a standard xyz coordinate system. A current loop of radius [math]\displaystyle{ {R_1} }[/math] exists to the left of the origin a distance [math]\displaystyle{ {d_1} }[/math] which maintains a current [math]\displaystyle{ {I_1} }[/math]. Another current loop of radius [math]\displaystyle{ {R_2} }[/math] exists to the right of the origin a distance [math]\displaystyle{ {d_2} }[/math], and it maintains a current [math]\displaystyle{ {I_2} }[/math]. The particle itself moves upward on the positive z-axis with a velocity [math]\displaystyle{ {\vec{v}} }[/math].

Assume the following:

[math]\displaystyle{ {I_1 = I_2} }[/math]

[math]\displaystyle{ {R_1 = 0.5R_2} }[/math]

[math]\displaystyle{ {d_1 = 3d_2} }[/math]

[math]\displaystyle{ {d_1, d_2 \gt \gt R_1, R_2} }[/math]

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).

What is the net force exerted on the particle at this exact position? Determine an expression in terms of any of the variables staed above.

**Solution:**

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.

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.

*For loop 1:*

[math]\displaystyle{ {B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_1^{2}} {d_1^3}} }[/math] (This approximation can be used because of the fourth assumption made above, where [math]\displaystyle{ {d_1} }[/math] is considerably larger than [math]\displaystyle{ {R_1} }[/math])

*For loop 2:*

[math]\displaystyle{ {B_2 = \frac {\mu_0} {4\pi} \frac {2 I_2 \pi R_2^{2}} {d_2^3}} }[/math] (This approximation can be used because of the fourth assumption made above, where [math]\displaystyle{ {d_2} }[/math] is considerably larger than [math]\displaystyle{ {d_2} }[/math])

Now we combine the appropriate values for radius and distance in terms of [math]\displaystyle{ {R_2} }[/math] and [math]\displaystyle{ {d_2} }[/math], 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.

[math]\displaystyle{ {B_1 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi 0.5 R_2^{2}} {3 d_2^3}} }[/math]

[math]\displaystyle{ {B_2 = \frac {\mu_0} {4\pi} \frac {2 I_1 \pi R_2^{2}} {d_2^3}} }[/math]

[math]\displaystyle{ {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}} }[/math]

[math]\displaystyle{ {B_{net} = \frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3}} }[/math]

Now let'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... *but wait!* 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:

[math]\displaystyle{ {|\vec{F_{net}}| = e |\vec{v}| |\vec{B_{net}}|} }[/math]

We can now involve our determined magnetic field that was generated by the two current carrying rings.

[math]\displaystyle{ {|\vec{F_{net}}| = e |\vec{v}| (\frac {7} {3} \frac {\mu_0} {4\pi} \frac {I_1 \pi R_2^{2}} {d_2^3})} }[/math]

The above represents the magnitude of the force. Its direction is (as stated previously) in the positive y-direction (+y).

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.

### Magnetic Forces in Wires

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:

[math]\displaystyle{ |\vec{F_{mag}}| = qnAv_{drift}(L\times\vec{B}) = I(L\times\vec B) = ILBsin\ominus }[/math]

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.

#### Simple

A wire is laying in the xy plane, with I, conventional current, flowing to the right. B, the magnetic field on the wire, is at a 45 degree angle to the wire, and pointing down. I = 0.6 A, B = 0.005 T. What is the magnetic force on the wire?

[math]\displaystyle{ |\vec F_{mag}| = ILBsin(45) }[/math]

[math]\displaystyle{ |\vec F_{mag}| = (0.6)(0.005)(sin(45)) }[/math]

[math]\displaystyle{ |\vec F_{mag}| = 0.002 }[/math] N into the page

#### Middling

A horizontal bar is falling at a constant velocity v. B, the magnetic field, points into the page. What is the the magnitude and direction of current in the bar?

[math]\displaystyle{ |\vec F_{grav}| = mg }[/math]

[math]\displaystyle{ |\vec F_{grav}| = \vec F_{mag} }[/math] because there is no gravitational acceleration, the net force must equal zero.

[math]\displaystyle{ mg = I(L\times\vec B) }[/math]

[math]\displaystyle{ I = \frac{mg}{LB} }[/math]

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.

I, the conventional current, flows to the left.

## Application (i.e. What Does This Have To Do With Anything?)

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.

As a chemist (*The Astrochemist*, 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.

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!

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.

## History

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.

Additionally, the topic of magnetic force can't be explored without magnetic fields. Although magnetic fields had been known for a long time, the direct connection between electricity and magnetism wasn'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.

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 act on moving charges.

Félix Savart and Jean-Baptiste Biot, discovered the phenomenon that supports the Biot- Savart law in 1820.

Hendrik Lorentz provided the actual "Lorentz Force Law" of which the component above (F = qv x B) is a main feature. This was published in 1865 in the Netherlands.

In 1907, a Norwegian physicist determined that electrons and positive ions follow the magnetic field lines of the earth towards the polar regions.

In 1973, two US scientists, Al Zmuda and Jim Williamson mapped the magnetic field lines of the Earth with help from a US Navy navigational satellite.

## See also

### Further reading

Argonne National Laboratory information regarding the Advanced Photon Source

National Oceanic and Atmospheric Administration's explanation of the Northern Lights

Secrets of the Polar Aurora - NASA

National Geographic - Heavenly Lights

### External links

Footage from space of Aurora Borealis

Magnetic force fields generated in copper (with more advanced and complex applications)

## References

1. Chabay, R.W; Sherwood, B.A.; *Matter and Interactions*. **2015**. *4*. 805-812.