Question

The figure shows the cross-section of a long conducting cylinder of inner radius a and outer radius b The cylinder carries a current whose current density $J=C{r}^2$ where C is a constant. What is the magnitude of the magnetic field $B$ at a point $r$, where $a<r<b$ ?

• A. $B=\frac{{\mu }_0\pi C}{4r^2}\left(r^2-a^2\right)$
• B. $B=\frac{{\mu }_{0}C}{4r^2}\left(r^4-a^4\right)$
• C. $B=\frac{{\mu }_0\pi C}{4r}\left(r^2-a^2\right)$
• D. $B=\frac{{\mu }_{0}C}{4r}\left(r^4-a^4\right)$

Answer 1

Answer: (D)

Given: Current density $J=C{r}^2$
For a region $a<r<b$. The Amperian loop is a circle labelled as 1 According to Ampere's circuital law,
$\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$
$B(2\pi r)={\mu }_0\int JdA={\mu }_0\underset{a}{\overset{r}{\int }}C{r}^{\prime }2\left(2\pi r^{\prime }d{r}^{\prime }\right)$
$={\mu }_{0}C2\pi \underset{a}{\overset{r}{\int }}{r}^{\prime }3d{r}^{\prime }={\mu }_{0}C2\pi {\left[\frac{r^{\prime }4}{4}\right]}_a^r$
$B2\pi r=\frac{{\mu }_{0}C2\pi }{4}\left[r^4-a^4\right],B=\frac{{\mu }_{0}C}{4r}\left[r^4-a^4\right]$