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Q.

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.16 Tesla experiences a torque of magnitude 0.032 Joule. The magnetic moment of the bar magnet will be

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a

0.23 Joule/Tesla

b

0.40 Joule/Tesla

c

0.80Joule/Tesla

d

Zero

answer is B.

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Detailed Solution

τ=MBHsin⁡θ⇒0.032=M×0.16×sin⁡30∘⇒M=0.4J/ tesla
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