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Question: A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.16 Tesla ...

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

A

0.23 Joule/Tesla

B

0.40 Joule/Tesla

C

0.80Joule/Tesla

D

Zero

Answer

0.40 Joule/Tesla

Explanation

Solution

τ=MBHsin⁡θ⇒0.032=M×0.16×sin⁡30∘\tau = M B _ { H } \sin \theta \Rightarrow 0.032 = M \times 0.16 \times \sin 30 ^ { \circ }