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Question

Physics Question on Magnetism and matter

A short bar magnet placed with its axis at 3030^{\circ} with a uniform external magnetic field of 0.16T0.16\,T experiences a torque of magnitude 0.032J0.032\,J . The magnetic moment of the bar magnet will be

A

0.23JT10.23\,JT^{-1}

B

0.40JT10.40\,JT^{-1}

C

0.80JT10.80\,JT^{-1}

D

zero

Answer

0.40JT10.40\,JT^{-1}

Explanation

Solution

Magnetic moment M=tBsinθ=0.0320.16×sin30oM=\frac{t}{B\sin \theta }=\frac{0.032}{0.16\times \sin {{30}^{o}}} =0.40JT1=0.40\,\,JT^{-1}