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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:

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