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JEE Main 2019 (10 Jan, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2019 (10 Jan, Shift 1)

Q46JEE Main 2019 · 10 JAN · SHIFT 1Medium

Motional EMF

A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is -

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Answer: (b)

Force on charges $q\vec v\times\vec B$ : $\hat j \times \hat k = \hat i$ , so charges separate along $x$ .
In equilibrium the faces normal to $x$ have $V = Bvd$ , with $d$ = edge.
$V = 0.1 \times 6 \times 0.02 = 0.012$ V
$= 12$ mV. So the answer is (b).

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