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.



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 \(L\)
$$ L=\frac{\mu_0N^2Hln(b/a)}{2\pi}$$