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MHT CET 2024 (4 May, Shift 2) · Mutual Inductance

Mutual Inductance question from MHT CET 2024 (4 May, Shift 2)

Q32MHT CET 2024 · 4 MAY · SHIFT 2Easy

Mutual Inductance

The planar concentric rings of metal wire having radii $r_1$ and $r_2$ (with $r_1 > r_2$ ) are placed in air. The current $I$ is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( $\mu_0 =$ permeability of free space)

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Answer: (d)

Field of larger ring at centre: $B = \dfrac{\mu_0 I}{2r_1}$ , taken uniform over the smaller ring
Flux through smaller ring: $\Phi = \dfrac{\mu_0 I}{2r_1}\pi r_2^2$
$M = \dfrac{\Phi}{I} = \dfrac{\mu_0\pi r_2^2}{2r_1}$
So the answer is (d) $\dfrac{\mu_0\pi r_2^2}{2r_1}$ .

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