Question

In a long cylindrical wire of radius $R$, magnetic induction varies with the distance $r$ from axis as $B=C{r}^{\alpha }$, where $C\&\alpha$ are constant. Find the function of current density in wire with the distance from the axis of wire.

• A. $\left[\frac{(\alpha +1)C}{{\mu }_0}\right]r^{\alpha -1}$
• B. $\left[\frac{(\alpha -1)}{{\mu }_0}\right]r^{1-\alpha }$
• C. $\left[\frac{(\alpha +1)C}{{\mu }_0}\right]r^{\alpha +1}$
• D. $\left[\frac{(\alpha -1)}{{\mu }_0}\right]r^{\alpha +1}$

Answer 1

The correct option is A $\left[\frac{(\alpha +1)C}{{\mu }_0}\right]r^{\alpha -1}$


At a distance $r$ from the axis of wire, we consider a closed path $\text{M}$ as shown in the figure.

Applying Ampere's circuital law on this path

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0{I}_{\text{enclosed}}..........\left(1\right)$

If $J\left(x\right)$ is the current density inside the wire as a function of distance $x$ from the axis of wire, the enclosed current within the closed path $\text{M}$ is given by integrating the current in the elemental ring of radius $x$ and width $dx$ considered in the cross-section of wire as shown in the figure.

$I_{\text{enclosed}}={\int }_0^{r}J\left(x\right)\cdot 2\pi xdx........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we have

$\oint Bdl={\mu }_0{\int }_0^{r}J\left(x\right)\cdot 2\pi xdx$

$B\times 2\pi r={\mu }_0{\int }_0^{r}J\left(x\right)\cdot 2\pi xdx$

$C{r}^{\alpha +1}={\mu }_0{\int }_0^{r}J\left(x\right)xdx(\because B=C{r}^{\alpha })$

Differentiating the above expression with respect to $x$ gives

$(\alpha +1)C{r}^{\alpha }={\mu }_{0}J\left(r\right)\cdot r$

$\therefore J\left(r\right)=\left[\frac{(\alpha +1)C)}{{\mu }_0}\right]r^{\alpha -1}$

Hence, option $\left(\text{C}\right)$ is correct.