Electromagnetic Induction

JEE Main level, numerical answer, medium

84 of 153

Question 84 of 153

A metallic cube of side 15cm15cm moving along yy-axis at a uniform velocity of 2ms−12m{s}^{-1}. In a region of uniform magnetic field of magnitude 0.5T0.5T directed along zz- axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be mVmV.

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Answer: 150

Using the equation
qE=qvB⇒E=vBqE=qvB \Rightarrow E=vB
Using the relation ∆V=Ed∆V=Ed
⇒∆V=vBd\Rightarrow ∆V=vBd
=2×12×0.15V=2\times \dfrac{1}{2}\times 0.15V
=0.15V=150mV=0.15V=150mV

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