Physics Book - User contributions [en]

Magnetic Field of a Disk

2018-02-24T20:00:56Z

Astevens37: /* A Mathematical Model */

'''Claimed by Andy Stevens (Spring 2018)'''

==The Main Idea==
Through this page, you will understand how to solve for the magnetic field produced by a moving charged, circular disk.

First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object allowed them to produce unique electric fields.

In order to figure out this magnetic field, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by this disk.

===A Mathematical Model===
Say we have a flat circular disk with radius <math>R</math> and carrying a uniform surface charge density <math>\sigma</math>. It rotates with an angular velocity <math>\omega</math> about the z-axis.

[[File:magfielddisk1.png|400px|center]]

To find the magnetic field <math>B</math> at any point <math>z</math> along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width <math>dr</math> and with radius <math>r</math>, we will have a total area of

<math>dA = 2\pi r dr</math>

And therefore have a total charge of

<math>dQ = 2\pi \sigma rdr</math>

This loop is rotating at an angular velocity <math>\omega</math>, which means the charge <math>dQ</math> on our loop makes a full rotation around the axis every period,

<math> T = \frac{2\pi}{\omega} </math> seconds.

Since a charge <math>dQ</math> makes a full rotation every <math>T</math> seconds, our infinitesimally thing ring is a current loop:

<math>dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{2\pi / \omega} = \sigma\omega rdr </math>

For this single loop, equivalent to a current <math>dI</math>, we can calculate the field a distance <math>z</math> above the axis. Using this proof from the Magnetic Field of a Loop page, we can get

<math> dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}</math>

Now to find the total field for all of the concentric rings, we must integrate from <math>r = 0</math> to <math>r = R</math>:

<math> B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}</math>

<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}</math>

<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]</math>

The absolute value around the z is important because we want the field to be the same on either side of the disk, whether in the positive z or negative z direction.

==Examples==
===Simple===
===Middling===
===Difficult===

==See also==
[http://www.physicsbook.gatech.edu/Magnetic_Field_of_a_Long_Straight_Wire Magnetic Field of a Long Straight Wire]

[http://www.physicsbook.gatech.edu/Magnetic_Field_of_a_Loop Magnetic Field of a Loop]

Magnetic Field of a Disk

2018-02-24T20:00:42Z

Astevens37: /* The Main Idea */

'''Claimed by Andy Stevens (Spring 2018)'''

==The Main Idea==
Through this page, you will understand how to solve for the magnetic field produced by a moving charged, circular disk.

First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object allowed them to produce unique electric fields.

In order to figure out this magnetic field, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by this disk.

===A Mathematical Model===
Say we have a flat circular disk with radius <math>R</math> and carrying a uniform surface charge density <math>\sigma</math>. It rotates with an angular velocity <math>\omega</math> about the z-axis.

[[File:magfielddisk1.png|200px|center]]

To find the magnetic field <math>B</math> at any point <math>z</math> along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width <math>dr</math> and with radius <math>r</math>, we will have a total area of

<math>dA = 2\pi r dr</math>

And therefore have a total charge of

<math>dQ = 2\pi \sigma rdr</math>

This loop is rotating at an angular velocity <math>\omega</math>, which means the charge <math>dQ</math> on our loop makes a full rotation around the axis every period,

<math> T = \frac{2\pi}{\omega} </math> seconds.

Since a charge <math>dQ</math> makes a full rotation every <math>T</math> seconds, our infinitesimally thing ring is a current loop:

<math>dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{2\pi / \omega} = \sigma\omega rdr </math>

For this single loop, equivalent to a current <math>dI</math>, we can calculate the field a distance <math>z</math> above the axis. Using this proof from the Magnetic Field of a Loop page, we can get

<math> dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}</math>

Now to find the total field for all of the concentric rings, we must integrate from <math>r = 0</math> to <math>r = R</math>:

<math> B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}</math>

<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}</math>

<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]</math>

The absolute value around the z is important because we want the field to be the same on either side of the disk, whether in the positive z or negative z direction.

==Examples==
===Simple===
===Middling===
===Difficult===

==See also==
[http://www.physicsbook.gatech.edu/Magnetic_Field_of_a_Long_Straight_Wire Magnetic Field of a Long Straight Wire]

[http://www.physicsbook.gatech.edu/Magnetic_Field_of_a_Loop Magnetic Field of a Loop]

Magnetic Field of a Disk

2018-02-24T19:58:53Z

Astevens37: /* The Main Idea */

'''Claimed by Andy Stevens (Spring 2018)'''

==The Main Idea==
Through this page, you will understand how to solve for the magnetic field produced by a moving charged, circular disk.

First, let us start with the basics. We know that moving charges spread out over the surface of an object will produce a magnetic field. This is similar to the concept of how charges spread out over an object allowed them to produce unique electric fields.

In order to figure out this magnetic field, we will start from the fundamental principles that we have learned already with regards to how magnetic fields are produced. We will then build on that and include the geometry of the object in question, in this a circular disk, in order to solve for the magnetic field produced by this disk.

===A Mathematical Model===
Say we have a flat circular disk with radius <math>R</math> and carrying a uniform surface charge density <math>\sigma</math>. It rotates with an angular velocity <math>\omega</math> about the z-axis.

[[File:magfielddisk1.png]]

To find the magnetic field <math>B</math> at any point <math>z</math> along the rotation axis, we can consider the disk to be a collection of concentric current loops. Breaking the disk into a series of loops with infinitesimal width <math>dr</math> and with radius <math>r</math>, we will have a total area of

<math>dA = 2\pi r dr</math>

And therefore have a total charge of

<math>dQ = 2\pi \sigma rdr</math>

This loop is rotating at an angular velocity <math>\omega</math>, which means the charge <math>dQ</math> on our loop makes a full rotation around the axis every period,

<math> T = \frac{2\pi}{\omega} </math> seconds.

Since a charge <math>dQ</math> makes a full rotation every <math>T</math> seconds, our infinitesimally thing ring is a current loop:

<math>dI = \frac{dQ}{T} = \frac{2\pi \sigma rdr}{2\pi / \omega} = \sigma\omega rdr </math>

For this single loop, equivalent to a current <math>dI</math>, we can calculate the field a distance <math>z</math> above the axis. Using this proof from the Magnetic Field of a Loop page, we can get

<math> dB = \frac{\mu_0 r^2 dI}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}}</math>

Now to find the total field for all of the concentric rings, we must integrate from <math>r = 0</math> to <math>r = R</math>:

<math> B = \int_{0}^{R} \frac{\mu_0 r^3 \sigma \omega dr}{2(r^2+z^2)^{3/2}} = \frac{\mu_0 \sigma\omega}{2} \int_{0}^{R} \frac{r^3 dr}{(r^2+z^2)^{3/2}}</math>

<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+r^2}{\sqrt{r^2+z^2}}\right]_{0}^{R}</math>

<math> B = \frac{\mu_0 \sigma\omega}{2} \left[\frac{2z^2+R^2}{\sqrt{R^2+z^2}} - 2|z| \right]</math>

The absolute value around the z is important because we want the field to be the same on either side of the disk, whether in the positive z or negative z direction.

==Examples==
===Simple===
===Middling===
===Difficult===

==See also==
[http://www.physicsbook.gatech.edu/Magnetic_Field_of_a_Long_Straight_Wire Magnetic Field of a Long Straight Wire]

[http://www.physicsbook.gatech.edu/Magnetic_Field_of_a_Loop Magnetic Field of a Loop]

Length and Stiffness of an Interatomic Bond

2017-04-10T03:44:07Z

Astevens37: /* Example 1 */

'''Claimed by Andy Stevens Spring 2017'''

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid objects is made up of a bunch of tiny balls (atoms) that are held together by springs (chemical bonds). The relaxed legth of the little spring between two atoms (the interatomic bond) is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

We can determine the stiffness of an interatomic bond by considering it as a spring. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms. The diameter of an atom is the space-filling model of a solid. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond and the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, considered as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<div align ="center">

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

</div>

''b. What is the diameter of a single zinc atom?''

<div align ="center">

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

</div>

''c. How many zinc atoms make up one side of the penny?''

<div align ="center">

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

</div>

''d. Calculate the interatomic spring stiffness for zinc.''

<div align ="center">

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

</div>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

'''Step 2'''
Cross-Sectional Area of Wire

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

'''Step 3'''
Cross-Sectional Area of One Atom

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

'''Step 4'''
Number of Chains

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

'''Step 5'''
Number of Bonds in One Chain

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

'''Step 6'''
Stiffness of One Interatomic Spring

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Length and Stiffness of an Interatomic Bond

2017-04-10T03:43:35Z

Astevens37:

'''Claimed by Andy Stevens Spring 2017'''

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid objects is made up of a bunch of tiny balls (atoms) that are held together by springs (chemical bonds). The relaxed legth of the little spring between two atoms (the interatomic bond) is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

We can determine the stiffness of an interatomic bond by considering it as a spring. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms. The diameter of an atom is the space-filling model of a solid. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond and the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, considered as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<div align ="center">

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

</div>

''b. What is the diameter of a single zinc atom?''

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

''c. How many zinc atoms make up one side of the penny?''

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

''d. Calculate the interatomic spring stiffness for zinc.''

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

'''Step 2'''
Cross-Sectional Area of Wire

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

'''Step 3'''
Cross-Sectional Area of One Atom

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

'''Step 4'''
Number of Chains

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

'''Step 5'''
Number of Bonds in One Chain

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

'''Step 6'''
Stiffness of One Interatomic Spring

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Length and Stiffness of an Interatomic Bond

2017-04-10T03:42:49Z

Astevens37: /* See also */

'''Claimed by Andy Stevens Spring 2017'''

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid objects is made up of a bunch of tiny balls (atoms) that are held together by springs (chemical bonds). The relaxed legth of the little spring between two atoms (the interatomic bond) is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

We can determine the stiffness of an interatomic bond by considering it as a spring. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms. The diameter of an atom is the space-filling model of a solid. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond and the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, considered as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

''b. What is the diameter of a single zinc atom?''

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

''c. How many zinc atoms make up one side of the penny?''

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

''d. Calculate the interatomic spring stiffness for zinc.''

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

'''Step 2'''
Cross-Sectional Area of Wire

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

'''Step 3'''
Cross-Sectional Area of One Atom

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

'''Step 4'''
Number of Chains

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

'''Step 5'''
Number of Bonds in One Chain

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

'''Step 6'''
Stiffness of One Interatomic Spring

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's Modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn

Length and Stiffness of an Interatomic Bond

2017-04-10T03:42:32Z

Astevens37: /* Connectedness */

'''Claimed by Andy Stevens Spring 2017'''

This topic covers find the length and stiffness of an Interatomic Bond.

==The Main Idea==

A solid objects is made up of a bunch of tiny balls (atoms) that are held together by springs (chemical bonds). The relaxed legth of the little spring between two atoms (the interatomic bond) is just the distance from the center of one atom to the center of the other atom, <math>d</math>. For our object, the distance is just twice the radius of one of the atoms since the electron cloud of the atom fills in the extra space. If the atoms are in a cubic arrangement that means that the volume of each atom would be <math>d × d × d</math> and would have a cross-sectional area of <math>d × d</math>.

[[File:ballcube.jpg|center]]

Since the interatomic bonds are modeled as springs, they have a stiffness, <math>k_{si}</math>, that relates the interatomic force to the stretch of the interatomic bonds. We use <math>s</math> for the microscopic stretch.

We can determine the stiffness of an interatomic bond by considering it as a spring. In order to determine stiffness, we must determine the length of an interatomic bond in a particular material. For different materials, bond lengths will vary slightly depending on the size of the atoms. The length of one interatomic bond is defined as the center-to-center distance between two adjacent atoms. The diameter of an atom is the space-filling model of a solid. To find the radius, we divide the diameter in half. If we can calculate the length of the interatomic bond and the diameter of a single atom, we can use this data to find the stiffness of the interatomic bond, considered as a spring.

==Length of an Interatomic Bond==

The length of an interatomic bond is defined as the center-to-center distance between adjacent atoms. This is the same as the diameter of an atom (including the full electron cloud).

[[File:electroncloudstuff.jpg|center]]

We can calculate atomic diameters for crystals of particular elements by using the measured density of the material in kilograms per cubic meter and Avogadro's number (the number of atoms in one mole of the material), 6.02 X 10^23 atoms per 1 mol.

The mass of one atom can be determined using the mass of one mole and dividing it by <math>6.022 x 10^{23} </math> atoms (Avogadro's number)

==The Stiffness of an Interatomic Bond==

It is difficult to measure the stiffness of an interatomic bond directly, so instead we can analyze data from macroscopic experiments to determine this quantity. We will consider the stiffness of an interatomic bond as a spring.

The equation for stiffness is:
<math>|F| = k_{si}|s|</math>

===Springs in Series===

Springs in series refers to when springs are linked end-to-end.

Two identical springs linked end to end stretch twice as much as one spring when the same force is applied. The combined spring therefore is only half as stiff as the individual springs.

[[File:springsinseries.jpg]]

===Springs in Parallel===

Springs in parallel is when springs are linked side-by-side.

We can think of the two springs as a single, wider spring. Two springs side by side are effectively twice as stuff as a single spring.

[[File:springsinparallel.jpg]]

===Cross-Sectional Area===

The cross-sectional area of an object is the area of a flat surface made by slicing through the object.

For example, the cross-sectional area of a cylinder is the area of a cicle and the cross-sectional area of a rectangular solid is the area of a rectangle.

[[File:crosssectionalarea.jpg]]

===Formulas to know===

Number of bonds in one "chain" of the solid:
<math>n_{bonds}={\frac{L}{d_{atomic}}}</math>

Spring Constant for one chain in the overall material:
<math>k_{chain}({\frac{k_{atomic}}{n_{bonds}}})</math>

Number of "chains" in the solid:
<math>n_{chains}={\frac{A_{wire}}{A_{atom}}}={\frac{A_{wire}}{{(d_{atomic})}^2}}</math>

Spring constant for the overall material:
<math>k_{wire}=k_{chain}×n_{chains} = k_{atomic} ({\frac{n_{chains}}{n_{bonds}}})</math>
<math>k_{wire}=k_{atomic}( {\frac{A_{wire}}{L_{wire}}})({\frac{1}{d_{atomic}}})</math>

==Examples==

===Example 1===

The US Penny is actually made of zinc. A typical penny has a diameter of 1.905 cm and an average thickness fo 1.228 mm. The density of zinc is 7140 kg/m^3 and its atomic weight is 65.4 amu = 65.4 g/mol. Young's modulus of zinc is 1.022 e11 N/m^2.

''a. Determine the mass of a typical penny.''

<math align = 'center'>ρ = {\frac{m}{v}}</math>

<math>m = ρ v = ρ(hπr^2) = ρ(hπ({\frac{d}{2}})^2)</math>

<math>m = (7140)(1.228e-3)(π)({\frac{1.905e-2}{2}}^2)</math>

<math>m = 0.002499\ kg = 2.499\ g</math>

''b. What is the diameter of a single zinc atom?''

<math> n_{atoms} = (2.499\ g)({\frac{1\ mol}{65.4\ g}})({\frac{6.022e23\ atoms}{1\ mol}})=2.3e22\ atoms</math>
<math>m_{atom} = {\frac{0.002499\ kg}{2.3e22\ atoms}} = 1.09e-25\ kg*m^3 </math>

<math>ρ = {\frac{m_{atom}}{v_{atom}}}</math>

<math> v_{atom} = {\frac{m_{atom}}{ρ_{atom}}} = {\frac{1.09e-25\ kg*m^3}{7140\ kg}}=1.53e-29\ m^3</math>

<math> v_{atom} = (d_{atom})^3</math>

<math> d_{atom} = (v_{atom})^{1/3}=(1.53e-29\ m^3)^{1/3} = 2.48e-10\ m</math>

''c. How many zinc atoms make up one side of the penny?''

<math>{\frac{area\ of\ penny\ face}{cross-sectional\ area\ of\ one\ atom}}={\frac{π({\frac{d}{2}})^2}{π({\frac{d_{atom}}{2}})^2}}={\frac{(1.095e-2)^2}{(2.48e-10)^2}} = 1.95e15\ atoms</math>

''d. Calculate the interatomic spring stiffness for zinc.''

<math>y = {\frac{k_{si}}{d_{atom}}}</math>

<math>k_{si} = y*d_{atom} = (1.077e11)(2.48e-10) = 26.7096\ {\frac{N}{m}}</math>

===Example 2===

A copper wire is 2m long. The wire has a square cross section. Each side of the wire is 1mm in width. Making sure the wire is straight, you hang a 10 kg mass on the end of the wire. The wrie is now 1.67 mm longer. Determine the stiffness of one interatomic bond in copper.

'''Step 1'''
Spring Stiffness of Entire Wire

<math>Δp_y = 0 = (k_{s,\ wire}*s-mg)Δt</math>

<math>k_{s, wire} = {\frac{mg}{s}} = {\frac{(10\ kg)(9.8\ N/kg)}{(1.67e-3\ m)}}= 5.87e4\ N/m</math>

'''Step 2'''
Cross-Sectional Area of Wire

<math>A_{wire} = (1e-3\ m)^2 = 1.6e-6\ m^2</math>

'''Step 3'''
Cross-Sectional Area of One Atom

<math>A_{atom} = (2.28e-10\ m)^2 = 5.20e-20\ m^2 </math>

'''Step 4'''
Number of Chains

<math>n_{chains} = {\frac{A_{wire}}{A_{atom}}} = {\frac{1e-6\ m^2}{5.2e-20\ m^2}} = 1.92e13</math>

'''Step 5'''
Number of Bonds in One Chain

<math>n_{bonds} = {\frac{L_{wire}}{d}} = {\frac{2\ m}{2.28e-10\ m}} = 8.77e9</math>

'''Step 6'''
Stiffness of One Interatomic Spring

<math>k_{s, wire} = {\frac{(k_{si})(n_{chains})}{n_{bonds}}}</math>

<math>k_{si} = {\frac{(k_{s,wire})(n_{bonds})}{n_{chains}}}={\frac{(5.87e4)(8.77e9)}{1.92e13}} = 26.8\ N/m</math>

==Connectedness==

#How is this topic connected to something that you are interested in?
Materials are all made up of atoms and their connections. Its interesting to see how we can apply macroscopic observations to a microscopic topic.

#How is it connected to your major?
Biomedical Engineers must always chose the right materials when creating devices for humans. Depending on their use, a material might not need to be very durable, or it may need to be the opposite, and be extremely durable.

#Is there an interesting industrial application?
Bonds are found in every material and object, so how these different bonds are connected create different objects and its really cool.

== See also ==

To further your reading and increase your knowledge, check out [[Young's Modulus]]. Like density and interatomic spring stiffness, Young's modulus is a property of a particular material and is independent of the share or size of a particular object made of that material.

==References==

Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 4

http://www.webassign.net/question_assets/ncsucalcphysmechl3/lab_10_1/manual.html

Created by Emily Milburn