Electromagnetic Induction

JEE Main level, single correct, medium

97 of 153

Question 97 of 153

A solid metal cube of edge length 2cm2cm is moving in the positive yy-direction, at a constant speed of 6ms−16m{s}^{-1}. There is a uniform magnetic field of 0.1T\text{0.1}T in the positive zz-direction. The potential difference between the two faces of the cube, perpendicular to the xx-axis, is

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Answer: (A) 12mV12mV

emf developed is given by, e=Bvle=Bvl
∴e=0.1(2×10−2)×6⇒e=12×10−3=12mV∴e=\text{0.1}\left(2\times {10}^{-2}\right)\times 6 \Rightarrow e=12\times {10}^{-3}=12mV

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