Inside a homogeneous long straight wire of circular cross-section, there is a circular cylindrical cavity whose axis is parallel to the conductor axis and displaced relative to it by a distance $l$. A direct current of density $\hat{j}$. flows along the wire. Find magnetic induction $\overrightarrow{B}$ inside the cavity.

By the principle of superposition, the required quantity can be given by

$\overrightarrow{B}=\overrightarrow{B_0}-\overrightarrow{B}\text{'},$

where $B_0$ is the magnetic field of the conductor without cavity, while $\overrightarrow{B}\text{'}$ is the magnetic field of the conductor which has removed.

Here, $\overrightarrow{B_0}=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{ir}{a^2}=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{j\left({\mathrm{\pi a}}^2\right)r}{a^2}=\frac{{\mu }_{0}r}{2}j$

This expression can be represented in vector form

$\overrightarrow{B_0}=\frac{{\mu }_0}{ }\left(\hat{j}\times \overrightarrow{r}\right)$

Similarly $\stackrel{\rightharpoonup }{B}\text{'}=\frac{{\mu }_0}{2}\left(\hat{j}\times \overrightarrow{r}\text{'}\right)$

Thus $\overrightarrow{B}=\overrightarrow{B_0}-\overrightarrow{B}\text{'}=\frac{{\mu }_0}{2}\left[\hat{j}\times \left(\overrightarrow{r}-\overrightarrow{r}\text{'}\right)\right]$

From the figure $\overrightarrow{r}=\overrightarrow{l}+\overrightarrow{r}\text{'}, \therefore \overrightarrow{r}-\overrightarrow{r}\text{'}=\overrightarrow{l.}$

$\therefore \stackrel{\rightharpoonup }{B}=\frac{{\mu }_0}{2}\left(\hat{j}\times \stackrel{\rightharpoonup }{l}\right)$

Therefore B = $\frac{1}{2}{\mu }_{0}jl$