Question

A bar magnet of magnetic moment $6J/T$ is aligned at ${{60}^{0}}$ with a uniform external magnetic field of $0.44T$. Calculate (a) the work done in turning the magnet to align its magnetic moment (i) normal to the magnetic field (ii) opposite to the magnetic field and, (b) the torque on the magnet in the final orientation in case (ii).

Answer 1

This problem can be solved by using the direct formula for the work done to rotate a bar magnet in an external magnetic field by a certain angle and the direct formula for the torque on a bar magnet in a magnetic field based upon its orientation.

Formula used:
$W=MB\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$
$\tau =MB\sin \theta$

Complete step-by-step answer:
Let us write the formula for the work done in rotating a bar magnet in a magnetic field through a certain angle.
The work done $W$ to rotate a bar magnet of magnetic moment $M$ in an external magnetic field of magnitude $B$ is given by
$W=MB\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$ ---(1)
Where ${{\theta }_{1}},{{\theta }_{2}}$ are the initial and final angles made by the axis of the magnetic moment with the direction of the magnetic field.
Also, the torque $\tau$ experienced by a bar magnet of magnetic moment $M$ when placed in a magnetic field of magnitude $B$ is given by
$\tau =MB\sin \theta$ --(2)
Where $\theta$ is the angle between the direction of the magnetic field and the axis of the magnetic moment.
Now, let us analyze the question.
The magnetic moment of the given bar magnet is $M=6J/T$.
The magnitude of the external magnetic field is $B=0.44T$.
The initial angle made by the axis of the bar magnet with the magnetic field is ${{\theta }_{1}} = {{60}^{0}}$.
a(i) When the bar magnet is turned normal to the magnetic field.
When the bar magnet is turned normal to the magnetic field, the final angle made by the axis of the bar magnet with the magnetic field is ${{\theta }_{2}}={{90}^{0}}$.
Let the work done in this case be ${{W}_{1}}$.
Now, using (1), we get
${{W}_{1}}=6\times 0.44\left( \cos {{60}^{0}}-\cos {{90}^{0}} \right)=2.64\left( \dfrac{1}{2}-0 \right)=2.64\times \dfrac{1}{2}=1.32J$ $\left( \because \cos {{60}^{0}}=\dfrac{1}{2},\cos {{90}^{0}}=0 \right)$
Therefore, the work done in this case is $1.32J$.

a(ii) When the bar magnet is turned opposite to the magnetic field.
When the bar magnet is turned opposite to the magnetic field, the final angle made by the axis of the bar magnet with the magnetic field is ${{\theta }_{2}}={{180}^{0}}$.
Let the work done in this case be ${{W}_{2}}$.
Now, using (1), we get
${{W}_{2}}=6\times 0.44\left( \cos {{60}^{0}}-\cos {{180}^{0}} \right)=2.64\left( \dfrac{1}{2}-\left( -1 \right) \right)=2.64\times \dfrac{3}{2}=3.96J$ $\left( \because \cos {{60}^{0}}=\dfrac{1}{2},\cos {{180}^{0}}=-1 \right)$
Therefore, the work done in this case is $3.96J$.

(b) Let the torque on the bar magnet when the magnet is placed opposite to the magnetic field be $\tau$.
In this case the angle made by the axis of the magnetic moment with the magnetic field is $\theta ={{180}^{0}}$.
Therefore, using (2), we get
$\tau =6\times 0.44\times \sin {{180}^{0}}=6\times 0.44\times 0=0$ $\left( \because \sin {{180}^{0}}=0 \right)$
Therefore, there is no torque on the bar magnet in this position.

Note: As we have seen in the answer, the bar magnet experiences no torque when it is placed opposite to the magnetic field. It also experiences no torque when it is placed along the magnetic field. Therefore, in these two positions it is in equilibrium. However, the position where it is opposite to the magnetic field is unstable equilibrium and if a slight external torque is given to it, it will move away from this position. However, the position where it is along the magnetic field, it is in stable equilibrium and if it is given a slight torque, it will try to return back to this position after the torque has been removed.