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JEE Main 2023 (24 Jan, Shift 1) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from JEE Main 2023 (24 Jan, Shift 1)

Q45JEE Main 2023 · 24 JAN · SHIFT 1Easy

Magnetic Flux & Faraday's Law

A conducting circular loop of radius 10π\frac{10}{\sqrt{\pi}} cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:

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Answer: (a)

  1. A=π(10π)2=100A = \pi\left(\frac{10}{\sqrt{\pi}}\right)^{2} = 100 cm2=10−2^{2} = 10^{-2} m2^{2}
  2. Steady rate, so emf is the same at every instant: 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
  4. So the answer is (a) emf = 10 mV.
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