Question

A conductor having resistivity $\rho$ is bent in the shape of a hall cylinder as shown in figure. The inner and outer radii of the cylinder are $a$ and $b$ respectively and the height of the cylinder is $h$. A potential difference is applied across the two rectangular faces of the conductor. Calculate the resistance offered by the conductor.

• A. $\frac{\rho h}{\pi ba}$
• B. $\frac{\rho a}{\pi h^2}$
• C. $\frac{\rho \pi }{hln\left(b/a\right)}$
• D. $\frac{\rho \log b/a}{\pi h}$

Answer 1

The correct option is C $\frac{\rho \pi }{hln\left(b/a\right)}$

We know that

$R=\frac{\rho l}{A}$

To calculate total resistance first we have to calculate resistance from $a$ to $b$ at angle $\theta$

Small resistance at $r$ distance is

$dR=\frac{\rho dl}{dr\times h}=\frac{\rho rd\theta }{dr\times h}$

From $a$ to $b$ all resistance are connected in parallel therefore,

$\int \frac{1}{dR}={\int }_a^b\frac{hdr}{\rho rd\theta }$

$\int \frac{1}{dR}=\frac{h}{\rho d\theta }{\int }_a^b\frac{dr}{r}$

$R_{eq}=\frac{\rho d\theta }{hln\left(\frac{b}{a}\right)}$

Now total equivalent resistance from $0$ to $\pi$ is

$R_{eq}={\int }_0^{\pi }\frac{\rho d\theta }{hln\left(\frac{b}{a}\right)}=\frac{\rho \pi }{hln\left(\frac{b}{a}\right)}$

Correct Option is C