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MHT CET 2026 (17 Apr, Shift 2) · Mutual Inductance

Mutual Inductance question from MHT CET 2026 (17 Apr, Shift 2)

Q12MHT CET 2026 · 17 APR · SHIFT 2Easy

Mutual Inductance

Two planar concentric rings of metal wire having radii 'r1r_1' and 'r2r_2' (with r1>r2r_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 ( μ0\mu_0 = permeability of free space)

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Answer: (b)

  1. Field of larger ring at centre: B=μ0I2r1B = \dfrac{\mu_0 I}{2r_1}, taken uniform over the smaller ring
  2. Flux through smaller ring: Φ=μ0I2r1πr22\Phi = \dfrac{\mu_0 I}{2r_1}\pi r_2^2
  3. M=ΦI=μ0πr222r1M = \dfrac{\Phi}{I} = \dfrac{\mu_0\pi r_2^2}{2r_1}
  4. So the answer is (b) μ0πr222r1\dfrac{\mu_0\pi r_2^2}{2r_1}.
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