A magnet of length 10cm and pole strength
0.5Am is placed normal to a uniform magnetic field
of strength 5 x 10 weber m. The moment of the
couple acting on it is
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Answer:
Answer:2.5×10^-6
Explanation:d=10cm=10×10^-2m
m^2=F.d
=B.m^2.d
=5×10^-5×0.5×10×10^-2
=2.5×10^-6 N-m
Explanation:
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