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MHT CET 2024 (9 May, Shift 2) · Motional EMF

Motional EMF question from MHT CET 2024 (9 May, Shift 2)

Q20MHT CET 2024 · 9 MAY · SHIFT 2Medium

Motional EMF

A square loop ABCD is moving with constant velocity ' $\vec{v}$ ' in a uniform magnetic field ' $\vec{B}$ ' which is perpendicular to the plane of paper and directed outward. The resistance of coil is ' $R$ ', then the rate of production of heat energy in the loop is [ L - length of side of loop]

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Answer: (b)

Only the side cutting the field edge gives motional emf $e = BLv$ .
Current $I = \dfrac{BLv}{R}$ , heat rate $P = I^2R = \dfrac{e^2}{R}$ .
$P = \dfrac{B^2L^2v^2}{R}$
So the answer is (b) $\dfrac{B^2L^2V^2}{R}$ .

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