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{J\,dA} $ 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\int\limits_{R/4}^{R/2}{{{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} $