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A toroid has major radius $R$ and small circular cross-section of radius $r$ ($r \ll R$), with $N$ turns. Its self-inductance is approximately:
A$\dfrac{\mu_0 N r}{R}$
B$\dfrac{\mu_0 N^2 R}{r^2}$
C$\dfrac{\mu_0 N^2 r^2}{2R}$
D$2\pi \mu_0 N R$
Answer & Solution
Correct answer: C. $\dfrac{\mu_0 N^2 r^2}{2R}$
Toroidal magnetic field $B = \mu_0 N i / (2\pi R)$; flux per turn $\phi = B \cdot \pi r^2 = \mu_0 N i r^2/(2R)$. $L = N\phi/i = \mu_0 N^2 r^2/(2R)$ (worked Example 12.5 in textbook).
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