Question

A long cylinder of uniform cross section and radius $R$ is carrying current $i$ along its length and current density is uniform. There is a cylindrical cavity or uniform cross section and radius $r$ in the cylinder parallel to its length. The axis of the cylindrical cavity is separated by a distance $d$ from the axis of the cylinder. Calculate the magnetic field at the axis of the cylinder.

Answer 1

Current density, $j=\frac{i}{\pi (R^2-r^2)}$
Current through the smaller cross section: $i^{\prime }=j\pi r^2$ (in absence of cavity)
Magnetic field due to this current at a point on the axis of bigger cylinder:
$B_1=\frac{{\mu }_0{i}_1^{\prime }}{2\pi d}=\frac{{\mu }_{0}i}{2\pi d}\times j\pi r^2=\frac{{\mu }_0{r}^2}{2d}\times \frac{i}{\pi (R^2-r^2)}$

Let $B_2$ be the magnetic field due to remaining portion. Then
$B_1+B_2=0$ or $B_2=-B_1$ or $\left|B_2\right|=\left|B_1\right|$
$B_2=\frac{{\mu }_{0}i}{2\pi d}\times \frac{r^2}{(R^2-r^2)}$