Show that the inductance of a toroid of rectangular cross-section is given by $$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$ where \(N\) is the total number of turns, \(a\) is the inner radius, \(b\) is the outside radius and \(H\) is the height of the toroid.

Solution:

Using the definition of the self inductance of a solenoid, we express \(L\) in terms of flux \(\phi\), \(N\) and \(I\):
$$ L=\frac{N\phi}{I}$$
We apply Ampere’s law to a closed path of radius \(a<r<b\):
$$ \oint \vec{B}.\vec{dl}=B(2\pi r)$$ $$=\mu_0NI$$ $$ \Rightarrow B=\frac{\mu_0NI}{2\pi r}$$ We express the flux in a strip of height \(H\) and width \(dr\):
$$ d\phi=BHdr=\frac{\mu_0NIH}{2\pi}\int_{a}^{b}\frac{dr}{r}$$ $$ =\frac{\mu_0NIH}{2\pi}ln(\frac{b}{a})$$
Substitute for flux \(\phi\) in the equation \(1\) we obtain the expression for \(L\)
$$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$