Question

A small bar magnet of moment M is placed in a uniform field of H. If magnet makes an angle of ${30^o}$ with field, the torque acting on the magnet is
(A) MH
(B) $\dfrac{{MH}}{2}$
(C) $\dfrac{{MH}}{3}$
(D) $\dfrac{{MH}}{4}$

Answer 1

Hint:
In order to solve this question, we will use the general formula of torque acting on the bar magnet in terms of magnetic moment M and magnetic field H and the angle between them and then determine the correct option.


Formula used:
If a bar magnet having magnetic moment M and placed in a uniform magnetic field H and the angle between the magnetic moment and magnetic field is $\theta $ then torque acting on the bar magnet can be calculated using $\tau = MB\sin \theta $



Complete step by step solution:
According to the question, we have given that a bar magnet has a magnetic moment of M and it’s placed in a uniform magnetic field H and the angle between magnetic moment and the magnetic field is given as $\theta = {30^o}$, So the torque acting on the bar magnet can be calculated using $\tau = MB\sin \theta $ on putting the values of parameters we get,
$
  \tau = MB\sin {30^o} \\
  \tau = \dfrac{{MB}}{2} \\
 $
So, the value of torque acting on the bar magnet is $\dfrac{{MB}}{2}$
Hence, the correct answer is option (B) $\dfrac{{MH}}{2}$


Therefore, the correct option is B.




Note:
It should be remembered that torque is a vector quantity and in actuality, the torque is the vector product between the magnetic moment of the bar magnet and the magnetic field in which it’s placed remember the basic trigonometric values such as $\sin {30^o} = \dfrac{1}{2}$ and this formula of torque is only applicable in a uniform magnetic field, in non-uniform magnetic field bar magnet experiences a torque and a net force as well.