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)

$3\, J$

Answer 2

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 \bigg(1-\frac{1}{2}\bigg)=\frac{MB}{2}\hspace30mm ...(i)
The torque on the needle is
$\, \, \, \, \tau =\overline{M} \times \overline{B}$
In magnitude,
\tau = M B sin\, \theta =M B sin\, \, 60^\circ = \frac{\sqrt3}{2}MB \hspace15mm ...(ii)
Dividing (ii) by (i), we get
$\frac{\tau}{W} =\sqrt3$
$\tau=\sqrt3W =\sqrt3 \times \sqrt3\, J =3\, J$