Question

From a cylinder of radius $R$, a cylinder of radius $\frac{R}{2}$ is removed as shown in the figure. The current flowing in the remaining portion of cylinder is $I$. Then magnetic field strength at point $\text{B}$ is

• A. $\frac{{\mu }_{0}I}{2\pi R}$
• B. $\frac{{\mu }_{0}I}{4\pi R}$
• C. $\frac{{\mu }_{0}I}{3\pi R}$
• D. $\frac{{\mu }_{0}I}{\pi R}$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{3\pi R}$

Given, $I$ current flows through the remaining part.

Let us consider the total cylinder and removed part of cylinder as two independent solid cylinders having current densities $J$ and $-J$ respectively and radii be $R$ and $\frac{R}{2}$ respectively as shown in the figure.



Now, $J=\frac{I}{\pi R^2-\pi {\left(\frac{R}{2}\right)}^2}=\frac{4I}{3\pi R^2}$

Let the current through total cylinder and removed part be $I_1$ and $I_2$ respectively

$I_1=J{A}_1=\frac{4I}{3\pi R^2}\pi R^2=\frac{4I}{3}$

Similarly ,

$I_2=-J{A}_2=-\frac{4I}{3\pi R^2}\pi {\left(\frac{R}{2}\right)}^2=\frac{-I}{3}$

Now, field at point $B$ is,

As the point $\text{B}$ is lying on the axis of cylinder with radius $\frac{R}{2}$, so its field $B_1$ is given by,

$B_1=0(\because r=0)$

Now, at point $\text{B}$ the field due to complete cylinder of radius $R$ is $B_2$(say) and is given by,

$B_2=\frac{{\mu }_0{I}_1}{2\pi R^2}r[\because r=\frac{R}{2}]$

$B_2=\frac{{\mu }_0\left(\frac{4I}{3}\right)}{2\pi R^2}\left(\frac{R}{2}\right)$

$\therefore B_2=\frac{{\mu }_{0}I}{3\pi R}$

Now, $B_{net}=B_1+B_2=$ $\frac{{\mu }_{0}I}{3\pi R}$

Why this Question ? Note: Field along the axis of cylinder is zero, and the field at a distance $r$ from the axis of a current carrying clinder is given by, $B_{in}=\frac{{\mu }_{0}I}{2\pi R^2}\left(r\right)$