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, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is -
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(c)
$V =\int_{ a }^{ b } E \cdot d \ell=\frac{\lambda}{2 \pi \varepsilon_0 r } \ln \left(\frac{ b }{ a }\right)$
Now, I $=\sigma \int \overrightarrow{ E } \cdot \overrightarrow{ dA }=\sigma \int \frac{\lambda}{2 \pi \varepsilon_0 r } \cdot 2 \pi dr =\frac{\sigma \lambda}{\varepsilon_0}$
From $(1)$: $I =\frac{2 \sigma \pi \varepsilon_0}{\varepsilon_0 \ln ( b / a )}=\frac{2 \pi \sigma}{\ln ( b / a )} v$
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