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Figure shows a coil of $N_{2}$ turns wound as shown around part of a toroid of $N_{1}$ turns. The toroid's inner radius is $a$ , its outer radius is $b$ , and its height is $h$ . Then the mutual inductance M for the toroid-coil combination is .

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$M = \frac{μ_{0} N_{1} N_{2} h}{2 π} \ln \frac{b}{a}$

$M = \frac{μ_{0} N_{1} N_{2} h}{2 π}$

$M = \frac{μ_{0} h}{2 π} \ln \frac{b}{a}$

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answer is A.

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Detailed Solution

The magnetic field of the toroid a distance r from the centre,

$B = \frac{μ_{0} N_{1} i}{2 π r}$

The total flux through the area $(h d r)$

$ϕ_{t o t a l} = \int_{a}^{b} N_{2} B (h dr) = N_{2} \int_{a}^{b} \frac{μ_{0} N_{1} i}{2 π r} (h dr) = \frac{μ_{0} N_{1} N_{2} h}{2 π} \ln (\frac{b}{a}) i$

$∴$ $M = \frac{μ_{0} N_{1} N_{2} h}{2 π} \ln \frac{b}{a}$

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