Question

If a bar magnet of magnetic moment M is freely suspended in a uniform magnetic field of strength B, the work done in rotating the magnet through an angle $\theta $ is
(A) $MB(1 - \sin \theta )$
(B) $MB\sin \theta $
(C) $MB\cos \theta $
(D) $MB(1 - \cos \theta )$

Answer 1

Hint:
In order to solve this question, we will use the general relation between magnetic moment, magnetic field, and the angle between the magnetic field and magnetic moment to find the work done in rotating it through some angle.

Formula used:
If M is the magnetic moment of a bar magnet, B is the strength of the magnetic field where the bar magnet is placed, and assume the bar magnet is rotated from initial angle ${\theta _1}$ to the final angle ${\theta _2}$ then work done in rotating the bar magnet in the magnetic field between these two angles is calculated as $W = MBcos({\theta _1}-{\theta _2})$



Complete step by step solution:
According to the question, we have given that a bar magnet of magnetic moment M is rotated in a magnetic field of strength B from initial angle ${\theta _1} = {0^o}$ as it was freely suspended and the initial angle is ${\theta _2} = \theta $ so the work done in rotating the bar magnet can be calculated using the formula $W = MBcos({\theta _1}-{\theta _2})$ on putting the values, we get
$
  W = MB(\cos 0 - \cos \theta ) \\
  W = MB(1 - \cos \theta ) \\
 $
Hence, the correct answer is option (D) $MB(1 - \cos \theta )$




Therefore, the correct option is D.




Note:
It should be remembered that when a bar magnet is freely suspended in a magnetic field it experiences zero torque which means the angle between the magnetic field and magnetic moment is zero also remember the basic trigonometric values such as $\cos {0^o} = 1$.