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, $I$ in terms of $J_0$ if the conductor's cross sectional area is $A=\pi R^2$.

• A. $I=\frac{J_{0}A}{12}$
• B. $I=\frac{J_{0}A}{9}$
• C. $I=\frac{J_{0}A}{6}$
• D. $I=\frac{J_{0}A}{3}$

Answer 1

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


Lets, take a cylindrical element of radius $r$ and thickness $dr$.
Area of this element $dA=2\pi rdr$ .

Then current through this area $dA$ is,

$di=\overrightarrow{J}.\overrightarrow{dA}=JdA$

Substituting the value of $J$,

$\Rightarrow di=J_0(1-\frac{r}{R})2\pi rdr$

Integrating on both sides,

Total current through the cross-section, $i=\int di={\int }_0^R{J}_0(1-\frac{r}{R})2\pi rdr$

$\Rightarrow i=2\pi J_0[\frac{R^2}{2}-\frac{R^3}{3R}]$

$\Rightarrow i=\frac{J_0\pi R^2}{3}=\frac{J_{0}A}{3}$

Hence, option (a) is the correct answer.