The horizontal component of the earth’s magnetic field at a place is 3 × 10^–4T and the dip is ${\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$ . A metal rod of length 0.25 m placed in the north-south position and is moved at a constant speed of 10 cm/s towards the east. The emf induced in the rod will be

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10 μV

1 μV

5 μV

Zero

answer is D.

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Detailed Solution

Rod is moving towards east, so induced emf across its end will be e = BVvl $= ({B_H}\tan \varphi )\,vl$
$\therefore e = 3 \times {10^{ - 4}} \times \frac{4}{3} \times (10 \times {10^{ - 2}}) \times 0.25 = {10^{ - 5}}V = 10\mu V$

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