Difference between revisions of "Introduction to Electrodynamics/Chapter 7/1"
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'''Ohm's Law''' is <math>\vec{J} = \sigma \vec{E}\;</math>, the current is proportional to the electric field. | '''Ohm's Law''' is <math>\vec{J} = \sigma \vec{E}\;</math>, the current is proportional to the electric field. | ||
| − | + | <br clear="all"/> | |
---- | ---- | ||
| + | {{YouTubeRight|rwohn6OXCUw}} | ||
'''Example 1''' | '''Example 1''' | ||
| + | A cylinder of constant cross-section has a potential put across the ends. How do the currents and potential relate? | ||
| + | |||
| + | First, we're going to assume that the electric field inside the cylinder is going to be constant. Example 3 explains why this is true. | ||
| + | |||
| + | Next, we relate current to the electric field: | ||
| + | |||
| + | :<math>\vec{J} = \sigma \vec{E}\;</math> | ||
| + | |||
| + | This is multiplied by the cross-sectional area to get the total current: | ||
| + | |||
| + | :<math>I = A J = \sigma A E\;</math> | ||
| + | |||
| + | The potential difference is equal to the electric field times the distance: | ||
| + | |||
| + | :<math>I = \sigma A V / L\;</math> | ||
| + | |||
| + | <br clear="all"/> | ||
---- | ---- | ||
| + | {{YouTubeRight|oj-9SQQWLKo}} | ||
'''Example 2''' | '''Example 2''' | ||
| + | We have two concentric cylinders of potential difference V. How much current flows between them over a distance L in terms of the conductivity and potential difference? | ||
| + | |||
| + | The field on the inside, thanks to Gauss's Law, is equal to the contained charge (linear charge density times the length) divided by the surface area. | ||
| + | |||
| + | :<math>\vec{E} = {1 \over \epsilon_0} {\lambda L \over 2 \pi r L} \hat r\;</math> | ||
| + | |||
| + | L's cancel, of course. | ||
| + | |||
| + | :<math>\vec{E} = {1 \over \epsilon_0} {\lambda \over 2 \pi r} \hat r\;</math> | ||
| + | |||
| + | The total current is: | ||
| + | |||
| + | :<math>I = J A = \sigma E A\;</math> | ||
| + | |||
| + | E A turns out to be the flux through a gaussian surface, so it equals the charge enclosed. | ||
| + | |||
| + | :<math>I = {\sigma \over \epsilon_0} \lambda L</math> | ||
| + | |||
| + | Now to calculate the potential difference: | ||
| + | |||
| + | :<math>V = - \int_b^a \vec{E} \cdot d \vec{l} = \int_a^b \vec{E} \cdot d \vec{l} = {\lambda \over \epsilon_0 2 \pi} \int_a^b {1 \over r} dr = {\lambda \over \epsilon_0 2 \pi} \ln(b/a)\;</math> | ||
| + | |||
| + | To bring V and I together, let's solve for linear charge density and plug it in to the equation for I above: | ||
| + | |||
| + | :<math>I = {\sigma \over \epsilon_0} {\epsilon_0 2 \pi \over \ln(b/a)} V L = {2 \pi L \over \ln(b/a)} \sigma V\;</math> | ||
| + | |||
| + | <br clear="all"/> | ||
---- | ---- | ||
| − | + | {{YouTubeRight|lyQymNCUMcY}} | |
| + | <br clear="all"/> | ||
| + | ---- | ||
| + | {{YouTubeRight|GSEtbW3DRjQ}} | ||
| + | '''Example 3''' | ||
| + | <br clear="all"/> | ||
| + | ---- | ||
| + | {{YouTubeRight|e_CDhwP41YE}} | ||
| + | <br clear="all"/> | ||
=== Problems === | === Problems === | ||
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=== Problems === | === Problems === | ||
| + | ---- | ||
| + | |||
| + | '''Example 4''' | ||
| + | ---- | ||
* [[../Problem/7|7.7]] | * [[../Problem/7|7.7]] | ||
* [[../Problem/8|7.8]] | * [[../Problem/8|7.8]] | ||
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* [[../Problem/11|7.11]] | * [[../Problem/11|7.11]] | ||
* [[../Problem/12|7.12]] | * [[../Problem/12|7.12]] | ||
| + | |||
| + | |||
| + | [[../2|7.2 Faraday's Law]] | ||
Latest revision as of 13:10, 19 October 2012
Contents
7.1 Electromotive Force
7.1.1 Ohm's Law
<html>Electrostatics and magnetostatics apply whenever <math>\rho\;</math> and <math>\vec{J}\;</math> are independent of time.
With steady currents, the charge density <math>\rho\;</math> remains constant. So you can have both steady currents and static charges at the same time.
One exception: <math>\vec{E} = 0\;</math> in a conductor. If this were so, you could not have a steady current inside a conductor.
For most substances,
- <math>\vec{J} = \sigma \vec{f}\;</math>
- <math>\sigma\;</math> is the conductivity of the material.
- <math>\rho = 1/\sigma\;</math> is the resistivity of the material.
- Insulators have a very small conductivity / very large resistivity, typically factor of 1,000,000,000,000,000,000!
- perfect conductors have infinite conductivity / zero resistivity.
- <math>\vec{f}\;</math> is the force per unit charge.
- Could be ANY force, even gravity, etc... "trained ants with tiny harnesses" (haha)
- We care about electromagnetic forces: <math>\vec{f} = \vec{E} + \vec{v} \times \vec{B}\;</math>
- Normally, the magnetic force is too small: <math>\vec{f} = \vec{E}\;</math>
Ohm's Law is <math>\vec{J} = \sigma \vec{E}\;</math>, the current is proportional to the electric field.
<html>
Example 1
A cylinder of constant cross-section has a potential put across the ends. How do the currents and potential relate?
First, we're going to assume that the electric field inside the cylinder is going to be constant. Example 3 explains why this is true.
Next, we relate current to the electric field:
- <math>\vec{J} = \sigma \vec{E}\;</math>
This is multiplied by the cross-sectional area to get the total current:
- <math>I = A J = \sigma A E\;</math>
The potential difference is equal to the electric field times the distance:
- <math>I = \sigma A V / L\;</math>
<html>
Example 2
We have two concentric cylinders of potential difference V. How much current flows between them over a distance L in terms of the conductivity and potential difference?
The field on the inside, thanks to Gauss's Law, is equal to the contained charge (linear charge density times the length) divided by the surface area.
- <math>\vec{E} = {1 \over \epsilon_0} {\lambda L \over 2 \pi r L} \hat r\;</math>
L's cancel, of course.
- <math>\vec{E} = {1 \over \epsilon_0} {\lambda \over 2 \pi r} \hat r\;</math>
The total current is:
- <math>I = J A = \sigma E A\;</math>
E A turns out to be the flux through a gaussian surface, so it equals the charge enclosed.
- <math>I = {\sigma \over \epsilon_0} \lambda L</math>
Now to calculate the potential difference:
- <math>V = - \int_b^a \vec{E} \cdot d \vec{l} = \int_a^b \vec{E} \cdot d \vec{l} = {\lambda \over \epsilon_0 2 \pi} \int_a^b {1 \over r} dr = {\lambda \over \epsilon_0 2 \pi} \ln(b/a)\;</math>
To bring V and I together, let's solve for linear charge density and plug it in to the equation for I above:
- <math>I = {\sigma \over \epsilon_0} {\epsilon_0 2 \pi \over \ln(b/a)} V L = {2 \pi L \over \ln(b/a)} \sigma V\;</math>
<html>
<html>
Example 3
<html>
Problems
7.1.2 Electromotive Forces
Problems
7.1.3 Motional emf
Problems
Example 4
