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MHT CET 2026 (11 Apr, Shift 1) · Motional EMF

Motional EMF question from MHT CET 2026 (11 Apr, Shift 1)

Q7MHT CET 2026 · 11 APR · SHIFT 1Medium

Motional EMF

A square loop of area 2525 cm2^{2} has a resistance of 10 Ω10\ \Omega. The loop is placed in uniform magnetic field of magnitude 4040 T. The plane of the loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in one second will be

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Answer: (b)

  1. Side =25= \sqrt{25} cm =0.05= 0.05 m; pulled out in 1 s, so v=0.05v = 0.05 m/s.
  2. ε=Blv=40×0.05×0.05=0.1\varepsilon = Blv = 40 \times 0.05 \times 0.05 = 0.1 V, I=0.110=0.01I = \frac{0.1}{10} = 0.01 A.
  3. Work done = heat produced =εIt=0.1×0.01×1=1.0×10−3= \varepsilon I t = 0.1 \times 0.01 \times 1 = 1.0\times10^{-3} J
  4. So the answer is (b) 1.0×10−31.0\times10^{-3} J.
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