Question

The current density across a cylindrical conductor of radius $R$ varies in magnitude according to the equation $J=J_0(1-\frac{r}{R})$ where $r$ is the distance from the central axis. Thus, the current density is maximum $J_0$ at the axis ($r=0$) and decreases linearly to zero at the surface $(r=R)$. The current in terms of $J_0$ and conductor's cross-sectional area $A$ is :-

• A. $\frac{J_{0}A}{3}$
• B. $\frac{J_{0}A}{6}$
• C. $\frac{J_{0}A}{2}$
• D. $\frac{J_{0}A}{5}$

Answer 1

The correct option is A $\frac{J_{0}A}{3}$

$J=J_0(1-\frac{r}{R})$
Taking a shell element at distance $r$ from the centre and having thickness $dr$
$di=JdA=J\left(2\pi rdr\right)$
$i=\underset{0}{\overset{R}{\int }}{J}_0(1-\frac{r}{R})2\pi rdr$
$i=2\pi J_0[\frac{R^2}{2}-\frac{R^3}{3R}]=\frac{J_{0}A}{3}$