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JEE Main 2018 (15 Apr, Shift 2) · Mutual Inductance

Mutual Inductance question from JEE Main 2018 (15 Apr, Shift 2)

Q57JEE Main 2018 · 15 APR · SHIFT 2Difficult

Mutual Inductance

At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity ω\omega about an axis along their common diameter. Calculate the emf induced in the smaller coil after a time t of its start of rotation.

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Answer: (a)

  1. Field at the centre: B=μ0I2RB = \frac{\mu_0 I}{2R}, uniform over the small coil.
  2. As the coil rotates: ϕ=Bπr2cos⁡ωt\phi = B\pi r^2\cos\omega t
  3. e=−dϕdt=μ0I2R ωπr2sin⁡ωte = -\frac{d\phi}{dt} = \frac{\mu_0 I}{2R}\,\omega\pi r^2 \sin\omega t
  4. So the answer is (a).
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