Question

The work done in turning a magnet of magnetic moment M by an angle of $90^\circ $from the meridian is n times the corresponding work done to turn it through an angle of $60^\circ $
A. $n = \dfrac{1}{2}$
B. $n = 2$
C. $n = \dfrac{1}{4}$
D. $n = 1$

Answer 1

Hint: Concept of work done in moving a magnetic dipole from one angle to another angle in a uniform magnetic field.Also we have to use the formula $W = mB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$

Complete step by step answer:
Whenever a magnetic dipole or magnet is placed in a magnetic field, it experiences a torque.
Let is consider a magnet of length$2l$, placed in uniform magnetic field $\overrightarrow B $with magnetic dipole moment$\overrightarrow m $. Then, torque C acting on it is given by,
$\overrightarrow C = \overrightarrow m \times \overrightarrow B $Or $C = mB{\text{ sin}}\theta $…… (i)
Whose $\theta $ is the angle the magnet is placed with the magnetic field? This torque will tend to align the dipole in direction of $\overrightarrow B .$ If the dipole is rotated against torque action, work is to be done.
The work done in turning the dipole through a small angle do is given by
$dW = Cd\theta = mB\sin \theta d\theta $… (From (i))
If dipole is rotated from initial position $\theta = {\theta _1}$to $\theta = {\theta _2}$, then work done is
$W = \int {dW = \int\limits_{{\theta _1}}^{{\theta _2}} {mB\sin \theta d\theta = mB\left[ { - \cos \theta } \right]_{{\theta _1}}^{{\theta _2}}} } $
\[
 \implies W = - mB\left( {\cos {\theta _2} - \cos \theta } \right) \\
 \implies W = mB\left( {\cos {\theta _1} - \cos {\theta _2}} \right) \\
 \]
Now, at magnetic meridian, angle $ = 0^\circ $
Work done in rotating magnet of magnetic moment, M from magnetic moment
i.e. ${\theta _1} = 0^\circ $ to $\theta = 90^\circ $ is
${W_1} = MB\left( {\cos \theta - \cos \theta 90} \right)$
${W_1} = MB\left( {1 - 0} \right)$
${W_1} = MB$….. (ii)
Work done in rotating from magnetic meridian, ${\theta _1} = 0^\circ $ to $\theta = 60^\circ $ is
$
  {W_2} = MB\left( {\cos 0^\circ - \cos 60^\circ } \right) \\
  {W_2} = MB\left( {1 - 0.5} \right) \\
 $
${W_2} = 0.5MB$ ….. (iii)
According to question
${W_1} = n{W_2}$
$MB = n \times 0.5MB$….. (From (ii) and (iii))
$
  1 = n \times \dfrac{1}{2} \\
  n = 2 \\
 $

So, the correct answer is “Option B”.

Note:
 Magnetic meridian is the vertical plane passing through the magnetic axis of a freely suspended small magnet and the earth’s magnetic field acts in the direction of magnetic meridian. That’s why, ${\theta _1} = 0^\circ $not$90^\circ $.