Question

The current density varies with radial distance r as $j=a{r}^2$ (where, a is a constant) in a cylindrical wire of radius R. The current passing through the wire between radial distance $\frac{R}{4}$ and $\frac{R}{4}$ is

• A. $\frac{20\pi a{R}^4}{331}$
• B. $\frac{15\pi a{R}^4}{512}$
• C. $\frac{\pi a{R}^4}{332}$
• D. $\frac{3\pi a{R}^4}{864}$

Answer 1

Answer: (B)

Given $J=a{r}^2$ As $I=\int JdA$ Here, $dA=2\pi rdr$ $[\because A=\pi r^2]$ So $I={\int }_{R/4}^{R/2}a{r}^2(2\pi r)dr$ $I=2\pi a\underset{R/4}{\overset{R/2}{\int }}{r}^{3}dr=2\pi a{\left[\frac{r^4}{4}\right]}_{R/4}^{R/2}$ $=\frac{\pi a}{2}\left[{\left(\frac{R}{2}\right)}^4-{\left(\frac{R}{4}\right)}^4\right]$ $=\frac{\pi a{R}^4}{2}\left[\frac{1}{16}-\frac{1}{256}\right]$ $=\frac{\pi a{R}^4}{2}\left[\frac{256-16}{4096}\right]$ $=\frac{15\pi a{R}^2}{512}$