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JEE Main 2017 (8 Apr, Shift 1) · Mutual Inductance

Mutual Inductance question from JEE Main 2017 (8 Apr, Shift 1)

Q58JEE Main 2017 · 8 APR · SHIFT 1Medium

Mutual Inductance

A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos (ω\omegat). The emf induced in the smaller inner loop is nearly:

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Answer: (a)

  1. Field at the centre of the big loop: B=μ0I2bB = \frac{\mu_0 I}{2b}, nearly uniform over the small loop.
  2. Flux: ϕ=μ0I0cos⁡ωt2b πa2\phi = \frac{\mu_0 I_0 \cos\omega t}{2b}\,\pi a^2
  3. e=−dϕdt=πμ0I02⋅a2b ωsin⁡ωte = -\frac{d\phi}{dt} = \frac{\pi \mu_0 I_0}{2}\cdot\frac{a^2}{b}\,\omega \sin\omega t
  4. So the answer is (a).
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