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NEET UG PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1
An axle of truck is 2.5m\mathrm{2}\mathrm{.5 m} long. If the truck is moving due north at 30  ms(−1)30\,\,m{s^{( - 1)}} at a place where the vertical component of the earth’s magnetic field is 90  μT90\,\,\mu T, the potential difference between the two ends of the axle is
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Correct option: A
6.75  mV\mathrm{6}.\mathrm{75}\,\,\mathrm{mV} with west end positive
Option A (6.75 mV\mathrm{6}.\mathrm{75}\,\mathrm{mV} with west end positive) is correct because the motional EMF is calculated using ε=vBL\varepsilon = vBL, and the direction is determined by the right-hand rule.

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