Question

A magnetic dipole aligned parallel to a uniform magnetic field requires a work of W units to rotate it through 60°. The torque exerted by the field on the dipole in this new position is:

• A. 2W
• B. W
• C. \( \frac{\sqrt{3}}{2} W \)
• D. 3​W

Answer 1

The Correct Option is C

The work done (W) in rotating a magnetic dipole from an angle \( \theta_1 \) to \( \theta_2 \) in a magnetic field B is given by:

\( W = mB (\cos \theta_1 - \cos \theta_2) \)

Here, the dipole is initially aligned parallel to the field, so \( \theta_1 = 0^\circ \) and it is rotated through 60°, so \( \theta_2 = 60^\circ \). Therefore:

\( W = mB (\cos 0^\circ - \cos 60^\circ) = mB (1 - \frac{1}{2}) = \frac{mB}{2} \)

Thus, \( mB = 2W \).

The torque (\( \tau \)) exerted on a magnetic dipole by a magnetic field is given by:

\( \tau = mB \sin \theta \)

In the new position, \( \theta = 60^\circ \). Therefore:

\( \tau = mB \sin 60^\circ = mB \cdot \frac{\sqrt{3}}{2} \)

Substituting \( mB = 2W \) into the torque equation, we get:

\( \tau = 2W \cdot \frac{\sqrt{3}}{2} = \sqrt{3} W \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} W \)

Therefore, the torque exerted by the field on the dipole in this new position is \( \frac{\sqrt{3}}{2} W \).

Answer 2

\(\text{ The torque } \tau \text{ on a magnetic dipole in a magnetic field is given by:}\\\)
\(\tau = M B \sin \theta\)
\(\text{where } M \text{ is the magnetic moment, } B \text{ is the magnetic field, and } \theta \text{ is the angle between the dipole and the field.}\\\)
\(\text{Given that the angle } \theta = 60^\circ, \text{ the torque becomes:}\\\)
\(\tau = M B \sin 60^\circ = M B \times \frac{\sqrt{3}}{2}\)
\(\text{Thus, the torque is } \frac{\sqrt{3}}{2} W, \text{ which corresponds to option 3.}\)