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MHT CET 2026 (13 Apr, Shift 2) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from MHT CET 2026 (13 Apr, Shift 2)

Q5MHT CET 2026 · 13 APR · SHIFT 2Easy

Magnetic Flux & Faraday's Law

A conducting loop of radius 10/[π]1/210/[\pi]^{1/2} cm is placed perpendicular to a uniform magnetic field 0.50.5 T. The magnetic field is decreased to zero in 0.50.5 s at a steady rate. The induced emf in the circular loop at 0.250.25 s is

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Answer: (b)

  1. A=πr2=π(10π)2=100A = \pi r^{2} = \pi\left(\frac{10}{\sqrt{\pi}}\right)^{2} = 100 cm2=10−2^{2} = 10^{-2} m2^{2}
  2. Steady decrease, so the emf is constant: e=AΔBΔt=10−2×0.50.5e = A\frac{\Delta B}{\Delta t} = 10^{-2}\times\frac{0.5}{0.5}
  3. e=10−2e = 10^{-2} V =10= 10 mV at any instant, including 0.250.25 s.
  4. So the answer is (b) 1010 mV.
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