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MHT CET 2025 (20 Apr, Shift 2) · Motional EMF

Motional EMF question from MHT CET 2025 (20 Apr, Shift 2)

Q15MHT CET 2025 · 20 APR · SHIFT 2Medium

Motional EMF

A simple pendulum with bob of mass m and conducting wire of length $L$ swings under gravity through an angle $\theta$ . The component of earth's magnetic field in the direction perpendicular to swing is B. Maximum e.m.f. induced across the pendulum is ( $g =$ acceleration due to gravity)

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Answer: (a)

Maximum speed at the lowest point: $v = \sqrt{2gL(1-\cos\theta)} = 2\sqrt{gL}\sin\frac{\theta}{2}$ .
Taking $\varepsilon_{max} = BLv_{max}$ (the usual exam approach):
$\varepsilon_{max} = 2BL\sqrt{gL}\sin\frac{\theta}{2}$
So the answer is (a).

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