Question

A long cylindrical conductor of radius $R$ carries current $i$ as shown below. The current density $J$ is a function of radius $r$ as $J=br$ where $b$ is a constant. The magnetic field at a distance $r_1(r_1<R)$ is:

• A. $\frac{{\mu }_{0}b{r}_1^2}{2}$
• B. Zero
• C. $\frac{{\mu }_{0}b{r}_1^3}{2}$
• D. $\frac{{\mu }_{0}b{r}_1^2}{3}$

Answer 1

The correct option is D $\frac{{\mu }_{0}b{r}_1^2}{3}$


Let us consider an elemental ring of thickness $dr$ at a distance $r$ from the centre.

So area of the element will be,
$dA=2\pi rdr$

Now the current flowing through this element,

$di=JdA=br\left(2\pi rdr\right)$

$di=2\pi b{r}^{2}dr$

Considering the direction of field lines as per current direction $⨀$, the enclosed current within the amperian loop $(r_1<R)$ will be

$I_{en}=\int di=\underset{r=0}{\overset{r_1}{\int }}2\pi b{r}^{2}dr$

$I_{en}=\frac{2\pi b{r}_1^3}{3}$

Applying ampere's circuital law for loop;

$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{en}$

$B\left(2\pi r_1\right)={\mu }_0\left(\frac{2\pi b{r}_1^3}{3}\right)$

$\therefore B=\frac{{\mu }_{0}b{r}_1^2}{3}$

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