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MHT CET 2026 (13 May, Shift 1) · Magnetic Flux & Faraday's Law

Magnetic Flux & Faraday's Law question from MHT CET 2026 (13 May, Shift 1)

Q16MHT CET 2026 · 13 MAY · SHIFT 1Medium

Magnetic Flux & Faraday's Law

A circular ring of diameter 2020 cm has a resistance of 0.01 Ω0.01\ \Omega. The charge that will flow through the ring, if it is turned from a position perpendicular to a uniform magnetic field of 22T to a position parallel to the field is (in coulomb) (∵cos⁡0∘=1, cos⁡90∘=0\because \cos 0^\circ = 1,\ \cos 90^\circ = 0)

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Answer: (b)

  1. Radius =0.1= 0.1 m, area A=π(0.1)2=0.01πA = \pi (0.1)^2 = 0.01\pi m2^2.
  2. Perpendicular to parallel: flux goes from BABA to 00, so Δϕ=2×0.01π=0.02π\Delta\phi = 2 \times 0.01\pi = 0.02\pi Wb.
  3. q=ΔϕR=0.02π0.01=2πq = \frac{\Delta\phi}{R} = \frac{0.02\pi}{0.01} = 2\pi C
  4. So the answer is (b) 2π2\pi.
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