A magnetic needle lying parallel to a magnetic field requires W unit of work to turn it through $60°$. The torque needed to maintain the needle in this position will be
$A. \sqrt {3}W$
$B. W$
$C. \left( { \sqrt { 3 } }/{ 2 } \right) W$
$D. 2W$

Hint: To solve this given problem, use the formula for work done on a magnetic needle in a magnetic field. Substitute the values in the above mentioned formula and find the value for work done. Now, use the formula for torque acting on the needle in terms of magnetic field and magnetic dipole moment. Substitute the values in this equation and find the value of torque in terms of work done.

Formula used:
$W=MB\left( \cos { { \theta }_{ 1 } } -\cos { { \theta }_{ 2 } } \right)$
$\tau= M \times B$

Complete answer:
Let M be the magnetic dipole moment of the needle
       B be the magnetic field
Work done on a magnetic needle in a magnetic field is given by,
$W=MB\left( \cos { { \theta }_{ 1 } } -\cos { { \theta }_{ 2 } } \right)$ …(1)
Where, ${ \theta }_{ 1 }$ is the initial position of the needle
            $ { \theta }_{ 2 } $ is the angle made after rotation of the needle
$\therefore { \theta }_{ 1 }={ 0 }^{ ° }$ and ${ \theta }_{ 2 }={ 60 }^{ ° }$
Substituting these values in the equation. (1) we get,
$W=MB\left( \cos { { 0 }^{ ° } } -\cos { { 60 }^{ ° } } \right)$
$\Rightarrow W=MB\left( 1-\dfrac { 1 }{ 2 } \right)$
$\Rightarrow W=\dfrac { MB }{ 2 }$
$\therefore MB=2W$ …(2)
Restoring torque acting on the needle is given by,
$\tau= M \times B$
$\Rightarrow \tau =MB\sin { { \theta }_{ 2 } }$
$\Rightarrow \tau= MB\sin { { 60 }^{ ° } }$
$\Rightarrow \tau =\dfrac { \sqrt { 3 } }{ 2 } MB$
Substituting equation. (2) in above equation we get,
$\tau =\dfrac { \sqrt { 3 } }{ 2 } \times 2W$
$\Rightarrow \tau= \sqrt {3}W$
Thus, the torque needed to maintain the needle in this position will be $\sqrt {3}W$.

So, the correct answer is option A i.e. $\sqrt {3}W$.

Note:
The torque acts to align the needle with magnetic dipole moment of the needle parallel to the magnetic field. Students must remember that the work done is calculated by subtracting initial potential energy from the final potential energy. They should also remember that the work done will be the same irrespective of the direction of the magnet.