A metallic conductor of length 1m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad/s. If the horizontal component of earth's magnetic field is 0.2 × 10^–4 T, then emf induced between the two ends of the conductor is

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50mV

5µV

5mV

50µV

answer is B.

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Detailed Solution

emf induced between the two ends $= \frac{B_{H} {ωl}^{2}}{2}$
$\frac{0 .2 \times 10^{- 4} \times 5 \times 1}{2} = 0 .5 \times 10^{- 4} = 50 \times 10^{- 6} V = 50 μV$

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