Question

Two long coaxial and conducting cylinders of radius $a$ and $b$ are separated by a material of conductivity $\sigma$ and a constant potential difference $V$ is maintained between them. Then the current, per unit length flowing in between cylinders is

• A. $\frac{4\pi \sigma }{ln\left(\frac{b}{a}\right)}V$
• B. $\frac{4\pi \sigma }{(b+a)}V$
• C. $\frac{2\pi \sigma }{ln\left(\frac{b}{a}\right)}V$
• D. $\frac{2\pi \sigma }{(b+a)}V$

Answer 1

The correct option is C $\frac{2\pi \sigma }{ln\left(\frac{b}{a}\right)}V$

Resistance of the small cylindrical element of thickness $dx$ is $dR=\frac{dx}{\sigma dA}=\frac{dx}{\sigma 2\pi xl}$
As each element is in series
$R_{eff}=\int dR={\int }_a^b\frac{dx}{\sigma 2\pi xl}=\frac{1}{2\pi \sigma l}ln\left(b/a\right)$
From Ohms law,
$I=\frac{V}{R}=\frac{V}{\frac{1}{2\pi \sigma l}ln\left(b/a\right)}$
$I=\frac{2\pi \sigma lV}{ln\left(\frac{b}{a}\right)}$
Current per unit length is
$\frac{I}{l}=\frac{2\pi \sigma V}{ln\left(\frac{b}{a}\right)}$

Why this Question? Tip: Use of integration for finding effective resistance is significant for JEE.