Question

A magnetic dipole aligned parallel to a uniform magnetic field requires a work of \(W\) to rotate it through \(60^\circ\). The torque exerted by the field on the dipole in this new position is

• A. \(2W\)
• B. \(W\)
• C. \(\sqrt{3}\,W\)
• D. \(\dfrac{\sqrt{3}}{2}W\)

Hint:

For a magnetic dipole: \[ U=-MB\cos\theta \] \[ \tau=MB\sin\theta \] A common shortcut: If the dipole is rotated from \(0^\circ\) to \(\theta\), \[ W=MB(1-\cos\theta) \] which can be used directly to find \(MB\).

Answer 1

The Correct Option is C

Step 1:Calculate the work done. Initially, \[ \theta_1=0^\circ \] Finally, \[ \theta_2=60^\circ \] Therefore, \[ W=U_2-U_1 \] \[ W= \left(-MB\cos60^\circ\right) - \left(-MB\cos0^\circ\right) \] \[ W= -\frac{MB}{2}+MB \] \[ W=\frac{MB}{2} \] Hence, \[ MB=2W \]

Step 2: Find the torque at \(60^\circ\). \[ \tau=MB\sin60^\circ \] Substituting \[ MB=2W \] \[ \tau=(2W)\left(\frac{\sqrt3}{2}\right) \] \[ \tau=\sqrt3\,W \]

Step 3: State the answer. \[ { \tau=\sqrt3\,W } \] Hence, the correct option is \[ {(C)} \]