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MHT CET 2026 (14 May, Shift 1) · Rotating Conductors & AC Generator

Rotating Conductors & AC Generator question from MHT CET 2026 (14 May, Shift 1)

Q1MHT CET 2026 · 14 MAY · SHIFT 1Easy

Rotating Conductors & AC Generator

A conductor of length 11 m rotates vertically about one of its ends at angular velocity 55 rad/s. If the horizontal component of earth's magnetic field is 0.2×10−40.2 \times 10^{-4} T, then the e.m.f. developed between the ends of the conductor is

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Answer: (c)

  1. Rod rotating about one end: e=12Bωl2e = \tfrac{1}{2}B\omega l^2, with BB the field normal to the plane of rotation (BHB_H here).
  2. e=12×0.2×10−4×5×12=5×10−5 Ve = \tfrac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1^2 = 5 \times 10^{-5}\text{ V}
  3. So the answer is (c) 50 μ50\ \muV.
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