A metal conductor of length 1 m rotates vertically about one of its ends at an angular velocity of 5 rad/s. If horizontal component of earth's magnetic field is 0.2 × 10⁻⁴ T, then the e.m.f. developed between the two ends of the conductor is (a) 5 (b) 50 mV (c) 5 mV (d) 50

Answer: (d). Rod rotating about one end in a perpendicular field: e = 1/2Bω l² (vertical plane of rotation cuts the horizontal component). e = 1/2 × 0.2 × 10⁻⁴ × 5 × 1² = 5 × 10⁻⁵ V So the answer is (d) 50 μ V.

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Other: Induction & Displacement Current

A metal conductor of length $1\text{ m}$ rotates vertically about one of its ends at an angular velocity of $5\text{ rad/s}$ . If horizontal component of earth's magnetic field is $0.2 \times 10^{-4}\text{ T}$ , then the e.m.f. developed between the two ends of the conductor is

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Answer: (d)

Rod rotating about one end in a perpendicular field: $e = \frac{1}{2}B\omega l^2$ (vertical plane of rotation cuts the horizontal component).
$e = \frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1^2 = 5 \times 10^{-5}$ V
So the answer is (d) $50\ \mu$ V.

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