Question

The current density across a cylindrical conductor of radius $R$ varies according to the equation $J=J_0(1-\frac{r}{R})$, where $r$ is the distance from the axis. Thus the current density is a maximum $J_0$ at the axis $r=0$ and decreases linearly to zero at the surface $r=R$. Suppose that instead the current density is a maximum $J_0$ at the surface and decreases linearly to zero at the axis so that $J=J_0\frac{r}{R}$. If the current is given as $x{J}_{0}A/3$. Find $x$

Answer 1

As per the given problem, current density J is

$J=J_0\left(\frac{r}{R}\right)$

and area is $\pi R^2$

The current density is given by

$J=\frac{dI}{dA}$

$dI=J\left(dA\right)$

$dA=2\pi \times r\left(dr\right)$

$dI=\frac{J_0}{R}\left(2\pi r^2\right))dr$

Integrating with respect to r,

$I=\frac{J_0}{3R}\left(2\pi R^3\right))dr$

$I=\frac{J_0}{R}\left(\frac{2\pi R^3}{3}\right)$

$I=2J_0\frac{A}{3}$