Question

An infinitely long hollow conducting cylinder with inner radius $\frac{R}{2}$ and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field $|B|$ as a function of the radial distance $r$ from the axis is best represented by

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

The cross-section of infinite wire is shown below, with direction of current perpendicularly inwards.



$r=$ distance of point from centre

For $r\le \frac{R}{2}$; using Ampere's circuital law,

Let us consider an Amperian loop of radius $r$

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

$\Rightarrow B\left(2\pi r\right)={\mu }_0{I}_{en}.....\left(i\right)$

$(\because I_{en}=0)$

$\therefore B=0$

Let $J$ be the current per unit area is wire.

For $\frac{R}{2}\le r\le R$,

$I_{en}=J\left[\right(\pi r^2)-\pi {\left(\frac{R}{2}\right)}^2]$

or, $I_{en}=J\pi [r^2-\frac{R^2}{4}]$

Again using eq. $\left(i\right)$

$B=\frac{{\mu }_0{I}_{en}}{2\pi r}$

$\Rightarrow B=\frac{{\mu }_0\pi [r^2-\frac{R^2}{4}]J}{2\pi r}$

or, $B=\frac{{\mu }_{0}J}{2r}(r^2-\frac{R^2}{4})......\left(ii\right)$

At $r=\frac{R}{2},B=0$

at $r=R,B=\frac{3{\mu }_{0}JR}{8}$

Here, for $\frac{R}{2}\le r\le R$, the value of $B$ will increase from $0$ to $\frac{3{\mu }_{0}JR}{8}$, but non - linearly as eq. (ii) also suggests.

For $r\ge R$;

$I_{en}=I_{total}=I$

Hence from eq $\left(i\right)$,

$B=\frac{{\mu }_{0}I}{2\pi r}$

$\Rightarrow B\propto \frac{1}{r}$

Hence, for $r\ge R$ the variation of B is hyperbolic.



$\therefore$ option $\left(d\right)$ is correct.