Question

A magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through ${60}^{\circ }$, the torque needed to maintain the needle in this position will be

• A. $W$
• B. $\sqrt{3}W$
• C. $\frac{\sqrt{3}W}{2}$
• D. $2W$

Answer 1

The correct option is B $\sqrt{3}W$

Torque acting on a magnetic dipole of dipole moment $\overrightarrow{M}$ in a magnetic field $\overrightarrow{B}$ is given by
$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

and potential of a magnetic dipole of dipole moment $\overrightarrow{M}$ in a magnetic field $\overrightarrow{B}$ is given by
$U=-\overrightarrow{M}\cdot \overrightarrow{B}$



Work done to turn the magnet through ${60}^{\circ }$
$W=U_f-U_i=-MB\cos {60}^{\circ }-(-MB\cos 0^{\circ })$
$=\frac{-MB}{2}+MB=\frac{MB}{2}$

Torque on magnet in final position
$\tau =MB\sin {60}^{\circ }=MB\frac{\sqrt{3}}{2}$
$=\sqrt{3}W$
which is the torque that must be applied to maintain the needle in this position.