Torque on a Bar Magnet – Rankers Physics

Bar Magnet: Practice Problem & Solution

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
\( \frac{mBl}{2} \)
\( \frac{\sqrt{3}mBl}{2} \)
\( mBl \)
\( 2mBl \)

Solution Explained:

To solve this problem, we apply the core principles of Bar Magnet. Understanding the underlying formula is key to arriving at the correct answer below:

The magnetic dipole moment of the bar magnet is \( M = m \cdot l \). The torque experienced in a magnetic field is \( \tau = M B \sin \theta = m l B \sin 60^\circ = \frac{\sqrt{3} mBl}{2} \).

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