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A bar magnet of magnetic moment $\mathrm{M}$, is placed in a magnetic field of induction B. The torque exerted on it is
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Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$
$\tau=\mathrm{MB} \sin \theta=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}$
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