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A horizontal conducting rod 0.5 m long moves at 10 m/s perpendicular to a 0.4 T magnetic field. The induced emf is
A0.5 V
B4.0 V
C1.0 V
D2.0 V
Answer & Solution
Correct answer: D. 2.0 V
ε = Blv = 0.4 × 0.5 × 10 = 2.0 V.
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