Question

The work done in rotating a bar magnet of magnetic moment $M$ from its unstable equilibrium position to its stable equilibrium position in a uniform magnetic field is $B$ is
$A)\text{ }2MB$
$B)\text{ }MB$
$C)\text{ -}MB$
$D)\text{ -}2MB$

Answer 1

This problem can be solved by using the direct formula for the work done in rotating a bar magnet in a magnetic field through a certain angle. In the unstable equilibrium position, the magnetic moment of the bar magnet is anti parallel to the magnetic field and in the stable equilibrium position, it is parallel to the magnetic field.

Formula used:
$W=MB\left( \cos {{\theta }_{i}}-\cos {{\theta }_{f}} \right)$

Complete step by step answer:
We will use the direct formula for the external work required to rotate a bar magnet in a magnetic field in terms of its magnetic moment, magnitude of the magnetic field and angle made by the magnetic moment with the magnetic field in the initial and final positions.
The work done $W$ in rotating a bar magnet of magnetic moment $M$ in a magnetic field of magnitude $B$ is given by
$W=MB\left( \cos {{\theta }_{i}}-\cos {{\theta }_{f}} \right)$ --(1)
Where ${{\theta }_{i}}$ is the angle made by the magnetic moment with the magnetic field in the initial orientation and ${{\theta }_{f}}$ is the angle made by the magnetic moment with the magnetic field in the final orientation.
Now, let us analyze the question.
The magnetic moment of the given bar magnet is $M$.
The magnitude of the magnetic field is $B$.

In the initial orientation, the magnet is in the unstable equilibrium position. In the unstable equilibrium position, the magnetic moment is anti parallel with the magnetic field.
Therefore, the angle made by the direction of the magnetic moment with the direction of the magnetic field is ${{\theta }_{i}}={{180}^{0}}$.
In the final orientation, the magnet is in the stable equilibrium position. In the stable equilibrium position, the magnetic moment is parallel with the magnetic field.
Therefore, the angle made by the direction of the magnetic moment with the direction of the magnetic field is ${{\theta }_{f}}={{0}^{0}}$.
Let the work done in rotating the bar magnet through the initial orientation to the final orientation in the magnetic field be $W$.
Therefore, using (1), we get
$W=MB\left( \cos \left( {{180}^{0}} \right)-\cos \left( {{0}^{0}} \right) \right)$
$\therefore W=MB\left( -1-\left( 1 \right) \right)=MB\left( -1-1 \right)=MB\left( -2 \right)=-2MB$ $\left( \because \cos {{180}^{0}}=-1,\cos {{0}^{0}}=1 \right)$
Therefore, we have got the required work done as $-2MB$.
Therefore, the correct option is $D)\text{ -}2MB$.

Note:
Students should take note that since electricity and magnetism are closely related, a similar analogous formula is present for the work done to rotate an electric dipole in an electric field through an angle. This formula can be obtained by replacing the magnetic field in formula (1) with the magnitude of the electric field and replacing the magnetic moment in formula (1), with the magnitude of the electric dipole moment. Hence, by remembering any one of the formulae, students can solve similar problems of electric dipoles also.