Question

A short magnet placed in a uniform magnetic field making an angle with the field experiences a torque. If the angle made by the magnet with the field is changed from \(30^\circ\) to \(45^\circ\), the torque on the magnet?

• A. increases by \(50\%\)
• B. decreases by \(50\%\)
• C. decreases by \(41.4\%\)
• D. increases by \(41.4\%\)

Hint:

The torque on a magnetic dipole in a uniform magnetic field depends on \( \sin \theta \). To compare torque at different angles, use the ratio \( \frac{\sin \theta_2}{\sin \theta_1} \).

Answer 1

The Correct Option is D

Step 1: Understanding the torque on a magnet in a magnetic field 
The torque (\( \tau \)) experienced by a magnet in a uniform magnetic field is given by: \[ \tau = MB \sin \theta \] where: - \( M \) is the magnetic moment of the magnet, - \( B \) is the magnetic field strength, - \( \theta \) is the angle between the magnetic moment and the field. 

Step 2: Calculate the torque ratio 
Initially, the angle is \( \theta_1 = 30^\circ \), so the initial torque is: \[ \tau_1 = MB \sin 30^\circ \] Since \( \sin 30^\circ = \frac{1}{2} \), we get: \[ \tau_1 = MB \times \frac{1}{2} \] When the angle is changed to \( \theta_2 = 45^\circ \), the new torque is: \[ \tau_2 = MB \sin 45^\circ \] Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} = 0.707 \), we get: \[ \tau_2 = MB \times 0.707 \] 

Step 3: Find the percentage increase 
The percentage increase in torque is given by: \[ \frac{\tau_2 - \tau_1}{\tau_1} \times 100 \] Substituting the values: \[ \frac{0.707 MB - 0.5 MB}{0.5 MB} \times 100 \] \[ = \frac{0.207 MB}{0.5 MB} \times 100 \] \[ = 41.4\% \] Thus, the torque increases by 41.4\% when the angle changes from \( 30^\circ \) to \( 45^\circ \).