Question

A potential difference is applied to the ends of a wire made up of an alloy and a current passes through the wire. The current density varies as $J=3+2r$, where $r$ is the distance of the point from the axis. If $R$ be the radius of wire, then the total current through any cross-section of the wire will be:

• A. $2\pi \left[(\frac{3R^2}{2}+\frac{2R^3}{3})\right]\text{units}$
• B. $\pi \left[(\frac{R^3}{3}+\frac{R^2}{4})\right]\text{units}$​​
• C. $\frac{\pi }{2}\left[(\frac{R^2}{2}+\frac{R^3}{3})\right]\text{units}$
• D. $2\pi \left[(R+\frac{R^2}{3})\right]\text{units}$

Answer 1

The correct option is A $2\pi \left[(\frac{3R^2}{2}+\frac{2R^3}{3})\right]\text{units}$

Let us consider a circular strip or radius $r$ and thickness $dr$.

The current through this small element will be,
$dI=\overrightarrow{J}\cdot \overrightarrow{ds}=Jds\cos 0^{\circ }$

$\Rightarrow dI=(3+2r)\left(2\pi rdr\right)$


(Here current is passing through cross-section hence angle between current density vector and area vector is $0^{\circ }$)


$\Rightarrow dI=2\pi (3r+2r^2)dr$

integrating both side

$I=\int dI={\int }_{r=0}^{R}2\pi (3r+2r^2)dr$

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

$\therefore I=2\pi [\frac{3R^2}{2}+\frac{2R^3}{3}]\text{units}$

Hence, option $\left(a\right)$ is correct.

$$\begin{array}{c}\text{why this question?}\\ \text{It intends to test your basic understanding}\\ \text{of current density and application of}\\ \text{calculus approach because current}\\ \text{density of wire is varying as we go}\\ \text{away from its centre.}\end{array}$$