Question

Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity s and a constant potential difference V is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is –

• A. $\frac{4\pi\sigma}{\mathcal{l}n(b/a)}$V
• B. $\frac{4\pi\sigma}{(b + a)}$V
• C. $\frac{2\pi\sigma}{\mathcal{l}n(b/a)}$V
• D. $\frac{2\pi\sigma}{(b/a)}$V

Answer 1

Answer: (C)

$\frac{2\pi\sigma}{\mathcal{l}n(b/a)}$V

Answer 2

E = $\frac{\lambda}{2\pi \in_{0}r}$ where l is the linear charge density on the inner cylinder.

and V = $\int_{a}^{b}{E.d\mathcal{l}}$ =$\frac{\lambda}{2\pi \in_{0}}$ $\mathcal{l}n$ $\left\lbrack \frac{b}{a} \right\rbrack$ …(1)

Now I = $\int_{}^{}{\overrightarrow{J}.\overset{\rightarrow}{dA}} = \sigma\int_{}^{}{\overrightarrow{E}.}\overset{\rightarrow}{dA}$= s$\int_{}^{}\frac{\lambda}{2\pi \in_{0}r}$.2prdr

Current per unit length will be :

I = $\frac{\sigma\lambda}{\in_{0}}$…(2)

From (1) : I = $\frac{2\sigma\pi \in_{0}}{\in_{0}\mathcal{l}n(b/a)}V$ = $\frac{2\pi\sigma}{\mathcal{l}n(b/a)}V$