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Q.

A magnet of length 0.1 m and pole strength 50 Am is kept in a magnetic field 4 ×10-5  T at an angle 300 to it. The couple acting on it is

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a

10-4 Nm

b

10-1 Nm

c

10-2 Nm

d

10-3 Nm

answer is A.

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Detailed Solution

M=50×0.1 = 5B = 4×10−5Tτ = MBSinθ = 4×10−5×5×12=10×10−5 = 10−4N−m
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