Solution:
The magnetic dipole moment of the bar magnet is \( M = m \cdot l \). The torque experienced in a magnetic field is \( \tau = M B \sin \theta = m l B \sin 60^\circ = \frac{\sqrt{3} mBl}{2} \).
A bar magnet of length \( l \) and pole strength \( m \) is placed in uniform magnetic field \( B \) at an angle of \( 60^\circ \) with field. The torque on the bar magnet at this instant will be
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