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