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A short bar magnet placed with its axis at $30 °$ with a uniform external magnetic field of 0.25T experiences a torque of magnitude equal to $4.5 \times 10^{- 2} Nm,$ what is magnitude of magnetic moment?

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$0.36 A m^{2}$

$3.6 A m^{2}$

$360 A m^{2}$

$3600 A m^{2}$

answer is A.

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Detailed Solution

$T = M B \sin θ$

$4.5 \times 10^{- 2} = M \times \frac{1}{4} \times \frac{1}{2}$

$M = 36 \times 10^{- 2} A m^{2}$

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