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

The correct option is D $MB(1-\cos \theta )$

We know that,

$W_{field}=-\Delta U$

$W_{field}=U_i-U_f$

$U=$ Magnetic potential energy of dipole in external field, which is given by,

$U=-\overrightarrow{M}.\overrightarrow{B}$

Initially the magnet is alligned to field,
so that angle $\theta$ between
$\overrightarrow{M}and\overrightarrow{B}=0^{\circ }$

$U_i=-MB\cos 0^{\circ }=-MB$

Finally, ${\theta }_f=\theta \Rightarrow U_f=-MB\cos \theta$

So $W_{field}=-MB-(-MB\cos \theta )$

$=MB(\cos \theta -1)$

Since, there is no change in kinetic energy,

$W_{ext}=-W_{field}=MB(1-\cos \theta )$

Hence, option $\left(D\right)$ is the correct answer.

Why this Question? Tip: The work done by conservative forces and the change in potential energy of the system is related by, $W_{cons}=-\Delta U$ Also, If there is no change in the kinetic energy of the system, $W_{ext}=-W_{cons}$