Question

A short bar magnet placed with its axis at $30^\circ$ with a uniform external magnetic field of $0.25\,{\text{T}}$ experiences a torque of magnitude equal $4.5 \times {10^{ - 2}}\,{\text{J}}$ . What is the magnitude of magnetic moment of the magnet

Answer 1

First of all, we will find a formula which relates torque, magnetic moment and magnetic field. We will use the trigonometric ratio sine in the formula. We will substitute the required values and manipulate accordingly to obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
The short bar magnet is inclined at an angle of $30^\circ$ with a uniform magnetic field.
The strength of the magnetic field is $0.25\,{\text{T}}$ .
The magnitude of torque which is experienced by the bar magnet is $4.5 \times {10^{ - 2}}\,{\text{J}}$ .
We are asked to find the magnitude of the magnetic moment of the magnet.

To begin with, we know that a magnetic field exerts a force on a current-bearing straight wire; it exerts a torque on a current-bearing wire loop. Torque allows a fixed axis to rotate around an object.

To find the magnetic field, we will use an equation which relates torque, magnetic moment and magnetic field, which is given by:
$\tau = MB\sin \theta$ …… (1)
Where,
$\tau$ indicates the torque.
$M$ indicates the magnetic moment.
$B$ indicates the strength of the magnetic field.
$\theta$ indicates the angle between magnetic moment and the magnetic field.

Now, we substitute the required values in the equation (1), and we get
$\tau = MB\sin \theta \\\ \Rightarrow 4.5 \times {10^{ - 2}} = M \times 0.25 \times \sin 30^\circ \\\ \Rightarrow 4.5 \times {10^{ - 2}} = M \times 0.25 \times 0.5 \\\ \Rightarrow M = \dfrac{{4.5 \times {{10}^{ - 2}}}}{{0.25 \times 0.5}} \\\$
Again, we simplify further,
$\Rightarrow M = 0.36\,{\text{J}}\,{{\text{T}}^{ - 1}}$

Hence, the magnitude of magnetic moment of the magnet is $0.36\,{\text{J}}\,{{\text{T}}^{ - 1}}$.

Note: In the given problem, many students seem to have confusion regarding the use of trigonometric ratio in the formula, which is sine but not cosine. As we must know that torque is the cross product of magnetic moment and the magnetic field. We can just remember that the dipole always wants to line up according to the magnetic field.