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Q.

A bar magnet of length 10 cm and having the pole strength equal to 10–3 weber is kept in a magnetic field having magnetic induction (B) equal to 4π×10−3 Tesla. It makes an angle of 30o with the direction of magnetic induction. The value of the torque acting on the magnet isμ0=4π×10−7weber/amp×m

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a

2π×10−7N×m

b

2π×10−5N×m

c

0.5N×m

d

0.5×102N×m

answer is A.

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Detailed Solution

Torque τ=MBHsinθ=0.1×10−3×4π×10−3×sin⁡30∘=10−7×4π×12=2π×10−7N×m
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