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A short bar magnet placed with its axis at $30^{o}$ with a uniform external magnetic field of 0.16 T experiences a torque of magnitude 0.032 Nm. The magnetic moment of the bar magnet will be

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0.23 J/T

0.40 J/T

0.80 J/T

Zero

answer is B.

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Detailed Solution

$τ = {MB}_{} sinθ \Rightarrow 0.032 = M x 0.16 x \sin 30^{o}$

$\Rightarrow M = 0.4 J / T$

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