Question

A short bar magnet placed with its axis at \( 30^\circ \) and a uniform external magnetic field of 0.5 T experiences a torque of magnitude equal to \( 4.5 \times 10^{-2} \) J. Then the magnitude of the magnetic moment of the magnet will be 

Hint:

The {torque} on a magnetic dipole in a uniform magnetic field is given by \( \tau = MB \sin \theta \). The magnetic moment \( M \) represents the strength of a magnet in an external field.

Answer 1

Step 1: Understanding Magnetic Torque 
- The torque (\(\tau\)) experienced by a magnetic dipole in a uniform magnetic field is given by: \[ \tau = MB \sin \theta \] where:
- \( M \) is the magnetic moment of the magnet, - \( B \) is the external magnetic field strength, - \( \theta \) is the angle between the magnetic moment and the field. 
Step 2: Given Values 
- \( \tau = 4.5 \times 10^{-2} \) J, - \( B = 0.5 \) T, - \( \theta = 30^\circ \), - \(\sin 30^\circ = \frac{1}{2} \). 
Step 3: Calculating \( M \) 
Rearranging the equation: \[ M = \frac{\tau}{B \sin \theta} \] \[ M = \frac{4.5 \times 10^{-2}}{(0.5) \times (\frac{1}{2})} \] \[ M = \frac{4.5 \times 10^{-2}}{0.25} \] \[ M = 18 \times 10^{-2} \] \[ M = 18 \times 10^{-2} { J T}^{-1} \] Thus, the magnetic moment of the magnet is \( 18 \times 10^{-2} \) J T\(^{-1}\).