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A square loop of side 0.1 m moves with velocity 2 m/s perpendicular to a uniform field B = 0.5 T. The induced emf along the leading edge is
A1 V
B0.05 V
C0.01 V
D0.1 V
Answer & Solution
Correct answer: D. 0.1 V
1. Motional emf along a conductor of length L moving with velocity v in field B is ε = BLv.
2. Substitute B = 0.5, L = 0.1, v = 2.
3. ε = 0.5 × 0.1 × 2 = 0.1 V.
_Source: Maharashtra Balbharati Std XII Physics, Ch 12 "Electromagnetic Induction" §12.4_
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