Work done rotating a bar magnet – Rankers Physics

Bar Magnet: Practice Problem & Solution

A bar magnet of magnetic moment \(\vec{M}\) is placed in a uniform magnetic field \(\vec{B}\) such that \(\vec{M}\) and \(\vec{B}\) are antiparallel. If the magnet is rotated through \(90^\circ\), then the work done on the magnet will be
\(-MB\)
\(\frac{-MB}{2}\)
\(2MB\)
\(\frac{MB}{2}\)

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:

Work done in rotating a magnetic dipole is \(W = MB(\cos \theta_1 - \cos \theta_2)\). Since initially \(\theta_1 = 180^\circ\) and finally \(\theta_2 = 90^\circ\), \(W = MB(\cos 180^\circ - \cos 90^\circ) = MB(-1 - 0) = -MB\).

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