Question

A magnetic needle suspended parallel to a magnetic field requires $\sqrt 3 \,J$ of work to turn it through $60^{\circ}.$ The torque needed to maintain the needle in this position will be

• A. $2 \sqrt3\, J$
• B. $\sqrt3\, J$
• C. $3\, J$
• D. $\frac{3}{2}\, J$

Answer 1

Answer: (C)

Work done in changing the orientation of a magnetic needle of magnetic moment $M$ in a magnetic field $B$ from position $\theta_1$ to $\theta_2$ is given by
$W=MB(\cos\theta_1-\cos\theta_2)$
Here, $\theta_1=0^{\circ}, \theta_2 = 60^{\circ}$
$=MB (1-\frac{1}{2})=\frac{MB}{2}$ ...(i)
The torque on the needle is
$\vec {\tau} =\vec {M} \times \vec {B}$ In magnitude,
$ \tau = M B \sin\, \theta =M B \sin\, \, 60^{\circ} = \frac{\sqrt3}{2}MB$ ...(ii)
Dividing (ii) by (i), we get
$ \frac{\tau}{W} =\sqrt3$
$ \tau=\sqrt3W =\sqrt3 \times \sqrt3\, J =3\, J$