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 $r=R$. Calculate the current in terms of $J_0$ and the conductors cross sectional area is $A=\pi R^2$.

• A. $J_{0}A$
• B. $J_{0}A/2$
• C. $J_{0}A/3$
• D. $2J_{0}A/3$

Answer 1

Answer: (C)

$J_{0}A/3$

As per the given problem, current density $J$ is:

$J=J_0(1-\frac{r}{R})$

And area is $\pi R^2$

The current density is given by:

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

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

$dI=J_0(1-\frac{r}{R})\left(2\pi rdr\right)$...........$(asdA=2\pi rdr)$

$dI=J_{0}2\pi (r-\frac{r^2}{R})dr$

Integrating-

$I=2\pi J_0{\int }_0^R(r-\frac{r^2}{R})dr$

$I=2\pi J_0{[\frac{r^2}{2}-\frac{r^3}{3R}]}_0^R$

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

$I=\frac{J_{0}A}{3}$ ....... (since A=$\pi R^2$)