A current is passing through a cylindrical conductor with a hole (or cavity) inside it. Find its magnitude of the magnetic field. Consider that the uniform current density is J and the distance between the centre of the original cylinder and the centre of the small the cylinder is b

Let us find the magnetic field at point $P$ inside the cavity at a distance $r_1$ from $O$ and $r_2$ from $C$.

$J$=current per unit area

$R$=radius of cylinder

$\alpha$=radius of cavity

$i_1=$whole current from cylinder $=J\left({\mathrm{\pi R}}^2\right)$

$i_{2=}$current from hole$=J\left({\mathrm{\pi a}}^2\right)$

At point $P$ magnetic field due to $i_1$ is $B_1$(perpendicular to $OP$) and is $B_2$ due to $i_2$ (perpendicular to $CP$) indirection shown. Although, $B_1$ and $B_2$ are actually at $P$, but for better understanding they are drawn at $O$ and $C$ respectively, Let $B_x$ be the x- component of resultant of  $B_1$ and $B_2$ and $B_y$ its y-component. Then,

$$\begin{gathered} B_x=B_1\sin \alpha -B_2\sin \beta \\ =\left(\frac{{\mu }_0}{2\mathrm{\pi }}\frac{i_1}{R^2}{r}_1\right)\sin \alpha -\left(\frac{{\mu }_0}{2\mathrm{\pi }}\frac{i_2}{R^2}{r}_2\right)\sin \beta \\ =\left(\frac{{\mu }_0}{2\mathrm{\pi }}\frac{J{\mathrm{\pi R}}^2}{R^2}{r}_1.\sin \alpha \right)-\left(\frac{{\mu }_0}{2\mathrm{\pi }}\frac{J\pi a^2}{a^2}{r}_2.\sin \beta \right) \\ =\frac{{\mu }_{0}J}{2}\left(r_1.\sin \alpha -r_2.\sin \beta \right)=0 \end{gathered}$$

Because in $∆OPC$, $\frac{r_1}{\sin \beta }=\frac{r_2}{\sin \alpha }=h$ or $r_1\sin \alpha -r_2\sin \beta =0$

Now,

 $$\begin{gathered} B_y=-\left(B_1 \cos \alpha +B_2 \cos \beta \right) \\ =-\frac{{\mu }_{0}J}{2}\left(r_1 \cos \alpha + r_2\cos \beta \right) \end{gathered}$$

From $∆OPC$,, we can say that

$r_1 \cos \alpha + r_2\cos \beta =b$ or $B_y=-\frac{{\mu }_{0}Jb}{2}=$constant

Thus we can say that net magnetic field at point $P$ is along negative y-direction and constant in magnitude.