Question

A small bar magnet experiences a$0.020Nm$torque when the axis of the magnet is at $45°$to a $0.10T$magnetic field. What is the magnitude of its magnetic dipole moment?

Short answer

The magnitude of its magnetic dipole moment is$0.28A·m^2$.

Step-by-step solution

Step  1 : Given Information

We need to find the magnitude of its magnetic dipole moment.

Simplify

The torque $\tau$represents the magnitude of the force multiplied by the distance between the axis and the line of the action. A current loop inside a magnetic fields experiences a torque that is exerted by the magnetic force and the loop starts to rotate. For a square loop with side $l$, its torque is given as:

$\tau =\mu B\sin \theta \left(1\right)$

where $\mu$ represents the magnetic dipole moment of the loop, $l$is the current in the loop and $\theta$ is the angle between the magnetic field and the magnetic dipole moment of the loop.

When the dipole moment is parallel to the magnetic field, the torque is zero and it is maximum when the dipole moment is perpendicular to the magnetic field. Our target is to find $\mu$ Therefore we rearrange equation (1) for $\mu$to be in the form of :

role="math" localid="1650971813269" $\mu =\frac{\tau }{B\sin \theta }\left(2\right)$

Calculation

Now, We plug the values for $\tau ,Band\theta$into equation (2) to get $\mu$:

$$\begin{gathered} \mu =\frac{\tau }{B\sin \theta } \\ =\frac{0.020N·m}{(0.10T)\sin {45}^{°}} \\ =0.28A·m^2 \end{gathered}$$