- #1
AmanWithoutAscarf
- 22
- 1
- Homework Statement
- Please check the problem below
- Relevant Equations
- .
Problem:
An annular disk consists of a sample of material with thickness d, inner radius ##\displaystyle R_{1}##, outer radius ##\displaystyle \ R_{2}##, and electrical conductivity ##\displaystyle \sigma##. Let the radial current ##I_0## flow from the inner periphery to the outer periphery of the disk.
Questions:
1. If ##n## is the conduction electron density, use Faraday's law to prove that the circular current is:
$$I_{c} =\frac{\sigma B}{2\pi ne} I_{0}\ln\frac{R_{2}}{R_{1}}$$
2. Prove that every circle with the center of origin in the disk is an equipotential line.
Here is my attemption of solving the first question:
I believe that the current through a cylindrical shell can be expressed as:
$$\displaystyle I_{c} =\int _{R_{1}}^{R_{2}} J_{\theta } .d.dr$$
To solve this, I considered the tangential and radial components of current density, ##\displaystyle J_{\theta }## and ##\displaystyle J_{r}##.
We have the following relationships:
$$\displaystyle ne.v_{r} =J_{r}$$
$$\displaystyle \sigma v_{r} B=J_{\theta }$$
$$\displaystyle J_{r} =\frac{ne}{\sigma B} J_{\theta }$$
Suppose ##\displaystyle J_{r}## is constant along the circular line with radius r
We have:
$$\displaystyle i_{r} =J_{r} .2\pi rd=I_{0} =const$$
Then we can write:
$$\displaystyle \frac{ne.2\pi rd}{\sigma B} J_{\theta } =I_{0}$$
$$\displaystyle I_{\theta } =I_{0} .\frac{\sigma B}{ne2\pi rd}$$
Integrate to find ##\displaystyle I_{c} :##
$$\displaystyle I_{c} =\int _{R_{1}}^{R_{2}} J_{\theta } .d.dr=\int _{R_{1}}^{R_{2}} I_{0} .\frac{\sigma B}{ne.2\pi r} dr$$
$$\displaystyle I_{c} =\frac{\sigma B}{2\pi ne} I_{0}\ln\frac{R_{2}}{R_{1}}$$
However, my solution is based on the assumption that is:
My question:
How is my answer correct? And can anyone help me with the last question, please?
An annular disk consists of a sample of material with thickness d, inner radius ##\displaystyle R_{1}##, outer radius ##\displaystyle \ R_{2}##, and electrical conductivity ##\displaystyle \sigma##. Let the radial current ##I_0## flow from the inner periphery to the outer periphery of the disk.
Questions:
1. If ##n## is the conduction electron density, use Faraday's law to prove that the circular current is:
$$I_{c} =\frac{\sigma B}{2\pi ne} I_{0}\ln\frac{R_{2}}{R_{1}}$$
2. Prove that every circle with the center of origin in the disk is an equipotential line.
Here is my attemption of solving the first question:
I believe that the current through a cylindrical shell can be expressed as:
$$\displaystyle I_{c} =\int _{R_{1}}^{R_{2}} J_{\theta } .d.dr$$
To solve this, I considered the tangential and radial components of current density, ##\displaystyle J_{\theta }## and ##\displaystyle J_{r}##.
We have the following relationships:
$$\displaystyle ne.v_{r} =J_{r}$$
$$\displaystyle \sigma v_{r} B=J_{\theta }$$
$$\displaystyle J_{r} =\frac{ne}{\sigma B} J_{\theta }$$
Suppose ##\displaystyle J_{r}## is constant along the circular line with radius r
We have:
$$\displaystyle i_{r} =J_{r} .2\pi rd=I_{0} =const$$
Then we can write:
$$\displaystyle \frac{ne.2\pi rd}{\sigma B} J_{\theta } =I_{0}$$
$$\displaystyle I_{\theta } =I_{0} .\frac{\sigma B}{ne2\pi rd}$$
Integrate to find ##\displaystyle I_{c} :##
$$\displaystyle I_{c} =\int _{R_{1}}^{R_{2}} J_{\theta } .d.dr=\int _{R_{1}}^{R_{2}} I_{0} .\frac{\sigma B}{ne.2\pi r} dr$$
$$\displaystyle I_{c} =\frac{\sigma B}{2\pi ne} I_{0}\ln\frac{R_{2}}{R_{1}}$$
However, my solution is based on the assumption that is:
without any valid support, and neither do I know how to solve the second question.
My question:
How is my answer correct? And can anyone help me with the last question, please?