A bar magnet is held perpendicular to a uniform magnetic field. If the couple acting on the magnet is to be halved by rotating it, then the angle by which it is to be rotated is:
A. ${30^ \circ }$
B. ${45^ \circ }$
C. ${60^ \circ }$
D. ${90^ \circ }$

Hint:
When a bar magnet is placed in a uniform magnetic field, the two poles of the magnet experience a force. These experienced forces are equal in magnitude and opposite in direction. Thereby constituting a couple of forces that produce a torque. By applying the formula of torque we can calculate the desired angle.

Formula used:
The formula used here to calculate the angle by which the magnet is rotated is given as: -
$\tau = MB\sin \theta $
Here $\tau =$ Net force on magnet or Torque
$M=$ Uniform Magnetic field
$B=$ Magnetic flux density
$\theta =$ Angle between the magnetic field and magnetic axes of the bar magnet




Complete step by step solution:
We know that the formula for the torque acting on a magnet in a uniform magnetic field is given as:
$\tau = \overrightarrow M \times \overrightarrow B $
or, $\tau = MB\sin \theta $ … (1)
Let us consider both the 2 cases one by one:
Case 1. When ${\theta _1} = {90^ \circ }$, the couple acting on a magnet is ${\tau _1} = {\tau _1}$
Therefore, from eq. (1)
${\tau _1} = MB\sin {\theta _1}$ … (2)
Case 2. When the couple acting on a magnet is halved i.e., ${\tau _2} = \dfrac{{{\tau _1}}}{2}$ then the angle will be ${\theta _2} = ?$
Therefore, from eq. (1)
${\tau _2} = MB\sin {\theta _2}$
$ \Rightarrow \dfrac{{{\tau _1}}}{2} = MB\sin {\theta _2}$ … (3)
Divide eq. (2) by eq. (3), we get
$\dfrac{{{\tau _1}}}{{\dfrac{{{\tau _1}}}{2}}} = \dfrac{{MB\sin {\theta _1}}}{{MB\sin {\theta _2}}}$
$ \Rightarrow {\tau _1} \times \sin {\theta _2} = \sin {\theta _1} \times \dfrac{{{\tau _1}}}{2}$
On Further calculation, we get
$ \Rightarrow \sin {\theta _2} = \sin {90^ \circ } \times \dfrac{1}{2} = \dfrac{1}{2}$ $\left( {\therefore \sin {{90}^ \circ } = 1} \right)$
$ \Rightarrow \sin {\theta _2} = \sin {30^ \circ }$ i.e., ${\theta _2} = {30^ \circ }$
Thus, the angle by which a magnet is to be rotated so that the couple acting on it will become halved is
$ = {90^ \circ } - {\theta _2} = {90^ \circ } - {30^ \circ } = {60^ \circ }$
Hence, the correct option is (C) ${60^ \circ }$ .



Therefore, the correct option is C.




Note:
Generally a magnetic field consists of parallel magnetic field lines. Parallel lines are equally spaced apart. These fields have the same strength and direction at all points. The magnetic force experienced by a substance is same at all points.