Question

Current density in a wire varies with $r$ $($distance from the centre$)$ according to the relation, $J=k{r}^2$. Here, $k$ is a constant. What will be the magnitude of magnetic field at a distance $R(R<a)$ from the centre?

$[a$ is the radius of the wire$]$

• A. $\frac{{\mu }_{o}k{R}^2}{12}$
• B. $\frac{{\mu }_{o}k{R}^2}{4}$
• C. $\frac{{\mu }_{o}k{R}^3}{12}$
• D. $\frac{{\mu }_{o}k{R}^3}{4}$

Answer 1

The correct option is D $\frac{{\mu }_{o}k{R}^3}{4}$

Consider an elemental ring of radius $r$ having a radial thickness $dr$.

The current flowing through this thin ring will be,

$dI=JdA=k{r}^2\left(2\pi rdr\right)$

The total current flowing through a circular cross-section of radius $R$ will be,

$\int dI=I={\int }_0^{R}k{r}^2\left(2\pi rdr\right)$

$=2\pi k{\int }_0^R{r}^{3}dr$

$=\frac{\pi k{R}^4}{2}$

According to Ampere's law,

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{o}I$

$\Rightarrow B\left(2\pi R\right)={\mu }_o\frac{\pi k{R}^4}{2}$

$\therefore B=\frac{{\mu }_{o}k{R}^3}{4}$

Hence, option $\left(D\right)$ is the correct answer.