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JEE Main 2023 (29 Jan, Shift 1) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2023 (29 Jan, Shift 1)

Q50JEE Main 2023 · 29 JAN · SHIFT 1Medium

Magnetic Flux & Faraday's Law

A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B = 0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cms −1^{-1}. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

  1. e=B⋅2πrdrdte = B \cdot 2\pi r\frac{dr}{dt}
  2. e=0.8×2π×0.1×0.02=3.2π×10−3e = 0.8 \times 2\pi \times 0.1 \times 0.02 = 3.2\pi \times 10^{-3} V
  3. e≈10.05×10−3e \approx 10.05 \times 10^{-3} V
  4. So the induced emf is about 1010 mV.
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