Question

A long conductor of circular cross-section has radius $r$ and length $l$ as shown in the figure. The conductivity of the material near the axis is $\sigma_1$ and increases linearly with the distance from axis and becomes $\sigma_2$ near the surface. Find the resistance of the conductor if the current enters from the one end and leaves from the other end.

Answer 1

The resistance of the conductor is $\frac{3l}{\pi r^2 (\sigma_1 + 2\sigma_2)}$.

Answer 2

The resistance of a conductor with non-uniform conductivity can be found by considering the total current flowing through it under a uniform electric field along the direction of current flow. The conductivity $\sigma$ varies linearly with the radial distance $x$ from the axis, from $\sigma_1$ at the axis ($x=0$) to $\sigma_2$ at the surface ($x=r$). This can be expressed as: $\sigma(x) = \sigma_1 + \frac{\sigma_2 - \sigma_1}{r} x$

The current enters from one end and leaves from the other, meaning the current flows along the length $l$ of the conductor. The electric field $E$ is uniform along the length, so $E = V/l$, where $V$ is the potential difference across the conductor.

The current density $J(x)$ at a radial distance $x$ is given by $J(x) = \sigma(x) E$. The infinitesimal cross-sectional area of a cylindrical shell at radius $x$ with thickness $dx$ is $dA = 2\pi x \, dx$.

The total current $I$ flowing through the conductor is the integral of the current density over the entire cross-sectional area: $I = \int_{0}^{r} J(x) \, dA = \int_{0}^{r} \sigma(x) E \, (2\pi x \, dx)$ Substituting the expression for $\sigma(x)$ and $E=V/l$: $I = \int_{0}^{r} \left(\sigma_1 + \frac{\sigma_2 - \sigma_1}{r} x\right) \frac{V}{l} (2\pi x \, dx)$ $I = \frac{2\pi V}{l} \int_{0}^{r} \left(\sigma_1 x + \frac{\sigma_2 - \sigma_1}{r} x^2\right) dx$ Evaluating the integral: $I = \frac{2\pi V}{l} \left[ \sigma_1 \frac{x^2}{2} + \frac{\sigma_2 - \sigma_1}{r} \frac{x^3}{3} \right]_{0}^{r}$ $I = \frac{2\pi V}{l} \left( \frac{\sigma_1 r^2}{2} + \frac{\sigma_2 - \sigma_1}{r} \frac{r^3}{3} \right)$ $I = \frac{2\pi V r^2}{l} \left( \frac{\sigma_1}{2} + \frac{\sigma_2 - \sigma_1}{3} \right)$ $I = \frac{2\pi V r^2}{l} \left( \frac{3\sigma_1 + 2\sigma_2 - 2\sigma_1}{6} \right)$ $I = \frac{2\pi V r^2}{l} \left( \frac{\sigma_1 + 2\sigma_2}{6} \right)$ $I = \frac{\pi V r^2 (\sigma_1 + 2\sigma_2)}{3l}$

The resistance $R$ of the conductor is given by $R = V/I$. $R = \frac{V}{\frac{\pi V r^2 (\sigma_1 + 2\sigma_2)}{3l}}$ $R = \frac{3l}{\pi r^2 (\sigma_1 + 2\sigma_2)}$