A device called a toroid is often used to create an almost uniform magnetic field in some enclosed area. The device consists of a conducting wire wrapped around a ring (a torus) made of a non-conducting material. For a toroid having $N$ closely spaced turns of wire, calculate the magnetic field in the region occupied by the torus,a distance $r$ from the centre. 

To calculate this field, we must evaluate $\oint B.dI$ over the circle of radius $r$. By symmetry e see that the magnitude of the field is constant on this circle and tangent to it. So, $\oint B.dI=BI=B\left(2\mathrm{\pi r}\right)$

Furthermore, the circular closed path surrounds $N$ loops of wire, each of which carries a current $i$. Therefore, right side of equation is ${\mu }_{0}Ni$ in this case.

 $$\begin{gathered} \therefore \oint B.dI={\mu }_0{i}_{net} \\ B\left(2\mathrm{\pi r}\right)={\mu }_{0}Ni \\ B=\frac{{\mu }_{0}Ni}{2\mathrm{\pi r}} \end{gathered}$$

This results show that $B\propto \frac{1}{r}$ and hence is non-uniform in the region occupied by torus. However, if $r$ is very large compared with the cross-sectional radius of the torus, then the field is approximately uniform inside the torus. In that case, 

$\frac{N}{2\mathrm{\pi r}}=n=$number of turns per unit length of torus

$\therefore B={\mu }_{0}ni$