Question

A magnet of magnetic moment $20$CGS units is freely suspended in a uniform magnetic field of intensity $0.3$CGS units. The moment of work done in deflecting it by an angle of $30^\circ $ in CGS units is
A. $6$
B. $3\sqrt 3 $
C. $3(2 - \sqrt 3 )$
D. $3$

Answer 1

Hint: The work done in deflecting the magnet through any angle in the magnetic field will be equal to the change in the potential energy associated with the magnet field between the initial and final position of the magnet.

Formula used:
 $U = - \overrightarrow M .\overrightarrow B = - MB\cos \theta $
Where,
$U$ is the potential energy of the magnet in the magnetic field,
$M$ is the magnetic moment of the magnet,
$B$ is the Intensity of the magnetic field,
$\theta $ is the angle between the magnetic moment and the magnetic field.

Complete step by step answer:
To find the work done in deflecting the magnet, we have to find the change in potential energy of the magnet.
A freely suspended magnet in the magnetic field always aligns itself parallel to the magnetic field and its north pole points towards the direction of the magnetic field.
In the question it is given that initially the magnet is suspended freely in the magnetic field. So, the magnet will be parallel to the magnetic field initially.
Thus, ${\theta _1} = 0^\circ $.
When the magnet is deflected by $30^\circ $,the angle between the field and the magnet is
${\theta _2} = 30^\circ $.
Thus, the work done is calculated as,
$W$= Final P.E – Initial P.E
$ \Rightarrow $$W = {U_2} - {U_1}$
$
   \Rightarrow W = - MB\cos {\theta _2} - ( - MB\cos {\theta _1}) \\
   \Rightarrow W = MB(\cos {\theta _1} - \cos {\theta _2}) \\
 $
Now substituting the values given in the question, we get
$
   \Rightarrow W = (20)(0.3)(\cos 0^\circ - \cos 30^\circ ) \\
   \Rightarrow W = 6(1 - \dfrac{{\sqrt 3 }}{2}) \\
   \Rightarrow W = 3(2 - \sqrt 3 ) \\
 $
Thus, option C is correct.

Note: The potential energy of the magnet is taken to be negative because in the free state of the magnet in the magnetic field, it should have the minimum energy possible. And also the work done will be the same irrespective of the direction of the deflection of the magnet because, $\cos ( - \theta ) = \cos (\theta )$.