A hollow cylinder has length L and inner and outer radii a and b. It is made of a material with resistivity $\rho$. A potential difference is set up between the inner and outer surfaces of the cylinder, each of which is an equipotential surface) so that current flows radially through the cylinder. What is the resistance to this radial current flow?
We recall that $J = \frac{I}{A}$, $E = \frac{dV}{dr}$. The area, A will be the surface area of the cylinder. Since
$$\begin{aligned} J &= \frac{1}{\rho} E \\ \frac{I}{A} &= \frac{1}{\rho} E \\ \frac{I}{2 \pi r L} &= \frac{1}{\rho} \frac{dV}{dr} \\ dV &= I \frac{\rho}{2 \pi r L} \, dr \end{aligned}$$
Since $dR = \frac{dV}{I}$,
$$ dR = \frac{\rho}{2 \pi r L} \, dr $$
Integrating from a to b,
$$\begin{aligned} R &= \int\limits_{a}^{b} dR \\ &= \int\limits_{a}^{b} \rho \frac{1}{2 \pi r L} \, dr \\ &= \frac{\rho}{2 \pi L} \int\limits_{a}^{b} \frac{1}{r} \, dr \\ &= \frac{\rho}{2 \pi L} \, \text{ln} \, \frac{b}{a} \end{aligned}$$
Next: Electromotive Force & Power In Circuits
Previous: Resistance And Resistivity
Back To Electromagnetism (UY1)
Thanks it helped way much