Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity $\mathrm{\sigma }$ 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

$\mathrm{E}=\mathrm{\lambda }/2{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^{\mathrm{\prime }}$, where $\mathrm{\lambda }$ is the linear charge density of the inner cylinder.

And $\mathrm{V}={\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{Ed\ell }=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0}\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)$              …(i)

$\begin{array}{l}\mathrm{Now}, \mathrm{I}=\int \overrightarrow{\mathrm{J}}\cdot \mathrm{d}\overrightarrow{\mathrm{A}}=\mathrm{\sigma }\int \overrightarrow{\mathrm{E}}\cdot \mathrm{d}\overrightarrow{\mathrm{A}}\\ =\mathrm{\sigma }\int \frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}2\mathrm{\pi rdr}\end{array}$

Current per unit length will be

        $\mathrm{I}=\frac{\mathrm{\sigma \lambda }}{{\mathrm{\epsilon }}_0}$                        …(ii)

From Eq. (i), we get

       $\mathrm{I}=\frac{2{\mathrm{\sigma \pi \epsilon }}_0}{{\mathrm{\epsilon }}_0\ln (\mathrm{b}/\mathrm{a})}\mathrm{v}=\frac{2\mathrm{\pi \sigma }}{\ln {\displaystyle }(\mathrm{b}/\mathrm{a})}\mathrm{v}$

Alternatively,

$\begin{array}{l} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\text{ or }\mathrm{R}={\int }_{\mathrm{x}=\mathrm{a}}^{\mathrm{b}} \frac{1}{\mathrm{\sigma }}\frac{\mathrm{dx}}{2\mathrm{\pi x}1}=\frac{1}{2\pi \sigma }\ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right)\\ \therefore \mathrm{I}=\frac{2\mathrm{\pi \sigma V}}{\ln (\mathrm{b}/\mathrm{a})}\end{array}$

II method

$$\begin{gathered} dR=\frac{dx}{\sigma 2\pi xl} \\ \int dR={\int }_a^b\frac{dx}{\sigma 2\pi xl} \\ R=\frac{\log {\displaystyle \frac{b}{a}}}{\sigma 2\pi l} \\ i=\frac{V}{R}=\frac{V}{\frac{\log {\displaystyle \frac{b}{a}}}{\sigma 2\pi l}} \\ \frac{i}{l}=\frac{\sigma 2\pi V}{\log \frac{b}{a}} \end{gathered}$$