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The horizontal component of earth's magnetic field at a place is 3 X 10-4 T and the dip is tan^-1 $(\frac{4}{3})$ . A metal rod of length 0.25 m is placed in N-S direction and is moved at a constant speed of 10 ems^-1 towards the east. The emf induced in the rod will be [BCECE Mains 2012]

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1 $μ V$

7 $μ V$

10 $μ V$

5 $μ V$

answer is D.

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Detailed Solution

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

$B = B_{ν}, v l = (B_{H} \tan θ) v l$ $[∴ \tan θ = B_{V} / B_{H}]$

$= 3 \times 10^{- 4} \times \frac{4}{3} \times 10 \times 10^{- 2} \times 0.25$

$= 10^{- 5} V = 10 μ V$

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