Question

From a cylinder of radius $R,$ a cylinder of radius $R/2$ is removed as shown in the figure. Current flowing in the remaining cylinder is I. Then, magnetic field strength is

• A. Zero at point $A$
• B. Zero at point $B$
• C. $$\frac{{\mu }_{0}I}{2\pi R}\text{at point}A$$
• D. $$\frac{{\mu }_{0}I}{3\pi R}\text{at point}B$$

Answer 1

The correct option is D $$\frac{{\mu }_{0}I}{3\pi R}\text{at point}B$$

First we can calculate current per unit area i.e., current density $\rho$.
$\rho =\frac{I}{\pi R^2-\pi {\left(R/2\right)}^2}$
$\rho =\frac{4I}{3\pi R^2}$

For magnetic feild strenght, we can solve this by superposition law

$B=B_{wholecylinder}-B_{cutpart}$

Now the magnetic strength at A is
$B_A=0-\frac{{\mu }_0{I}_{enclosed}}{2\pi r}$
$B_A=0-\frac{{\mu }_0{I}_{enclosed}}{2\pi \frac{R}{2}}$
From current density
$\begin{array}{cc}{I}_{enclosed}& =\rho \pi {\left(R/2\right)}^2\\ & =I/3\end{array}$

$\Rightarrow B_A=-\frac{{\mu }_{0}I/3}{2\pi \frac{R}{2}}$
$\Rightarrow B_A=-\frac{{\mu }_{0}I}{3\pi R}$


Now the magnetic strength at B is
$B=B_{wholecylinder}-B_{cutpart}$
$B_B=\frac{{\mu }_0{I}_{enclosed}}{2\pi r}-0$
$B_B=\frac{{\mu }_0{I}_{enclosed}}{2\pi \frac{R}{2}}$
From current density

$\begin{array}{cc}{I}_{enclosed}& =\rho \pi {\left(R/2\right)}^2\\ & =I/3\end{array}$

$\Rightarrow B_B=\frac{{\mu }_{0}I/3}{2\pi \frac{R}{2}}$
$\Rightarrow B_B=\frac{{\mu }_{0}I}{3\pi R}$