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MHT CET 2024 (11 May, Shift 1) · Other: Induction & Displacement Current

Other: Induction & Displacement Current question from MHT CET 2024 (11 May, Shift 1)

Q3MHT CET 2024 · 11 MAY · SHIFT 1Easy

Other: Induction & Displacement Current

A boat is moving due east in a region where the earth's magnetic field is 3.6×10−5 N/Am3.6 \times 10^{-5}\text{ N/Am} due north and horizontal. The boat carries a vertical conducting rod 2 m long. If the speed of the boat is 2.00 m/s2.00\text{ m/s}, the magnitude of the induced e.m.f. in the rod is

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Answer: (d)

  1. Rod (vertical), velocity (east) and BB (north) are mutually perpendicular: e=Bvle = Bvl.
  2. e=3.6×10−5×2×2=1.44×10−4e = 3.6 \times 10^{-5} \times 2 \times 2 = 1.44 \times 10^{-4} V
  3. So the answer is (d) 0.1440.144 mV.
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