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Q.

The horizontal component of the earth's magnetic field at a place is $3 \times 10^{- 4} T$ and the dip is $\tan^{- 1} (\frac{4}{3}) .$ 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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a

$1 μ V$

b

$5 μ V$

c

$10 μ V$

d

Zero

answer is D.

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Detailed Solution

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Rod is moving towards east, so induced emf across its end will be

$e = B_{V} vl = (B_{H} \tan δ) vl$

$∴ e = 3 \times 10^{- 4} \times \frac{4}{3} \times (10 \times 10^{- 2}) \times 0 .25 = 10^{- 5} V = 10 μV$

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The horizontal component of the earth's magnetic field at a place is 3×10−4 T and the dip is tan−143. 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