Q103 / 120 · VITEEE 2016 · Physics

The current density varies with radial distance $r$ as $J=a r^{2}$, in a cylindrical wire of radius $R$. The current passing through the wire between radial distance $R / 3$ and $R / 2$ is

  • Correct answer
    A
    $\frac{65 \pi a R^{4}}{2592}$
  • B
    $\frac{25 \pi a R^{4}}{72}$
  • C
    $\frac{65 \pi a^{2} R^{3}}{2938}$
  • D
    $\frac{81 \pi a^{2} R^{4}}{144}$

Solution

Given; $J={ar}^{2}$. $\begin{array}{l}i=\int_{1}^{2} J \times 2 \pi r d r=\int_{R / 3}^{R / 2} a r^{2} \times 2 \pi r d r \\ =2 \pi a \int_{R / 3}^{R / 2} r^{3} d r=2 \pi a\left|\frac{r^{4}}{4}\right|_{R / 3}^{R / 2} \\ \quad=\frac{\pi a}{2}\left[\left(\frac{R}{2}\right)^{4}-\left(\frac{R}{3}\right)^{4}\right] \\ =\frac{\pi a R^{4}}{2} \times \frac{65}{81 \times 16}=\frac{65 \pi a R^{4}}{2592}\end{array}$

Hence, the correct answer is option: 'A'.

Relevant Questions