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MHT CET 2023 (10 May, Shift 1) · Motional EMF

Motional EMF question from MHT CET 2023 (10 May, Shift 1)

Q30MHT CET 2023 · 10 MAY · SHIFT 1Medium

Motional EMF

A square loop of area 25 cm225\text{ cm}^{2} has a resistance of 10 Ω10\ \Omega. This loop is placed in a uniform magnetic field of magnitude 40 T40\text{ T}. The plane of 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 L=0.05 mL = 0.05\text{ m}; pulled out in 1 s, so v=0.05 m/sv = 0.05\text{ m/s}.
  2. e=BLv=40×0.05×0.05=0.1 Ve = BLv = 40 \times 0.05 \times 0.05 = 0.1\text{ V}
  3. P=e2R=0.0110=10−3 WP = \dfrac{e^2}{R} = \dfrac{0.01}{10} = 10^{-3}\text{ W}
  4. W=Pt=1.0×10−3 JW = Pt = 1.0 \times 10^{-3}\text{ J}
  5. So the answer is (b) 1.0×10−3 J1.0 \times 10^{-3}\text{ J}.
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