Question

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

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

Answer 1

Answer: (A)

Work done to rotate a magnetic needle from ${\theta }_1$ to ${\theta }_2$ is
$W=MB\left(cos{\theta }_1-cos{\theta }_2\right)$
where, $M$ is magnetic dipole moment and $B$ is magnetic field.
So, $W=MB\left(cos{0}^{\circ }-cos60^{\circ }\right)$
$W=MB\left(1-\frac{1}{2}\right)=\frac{MB}{2}$
$\Rightarrow MB=2W$
Restoring torque on the needle $\tau =MBsin\theta$
$\Rightarrow \tau =MBsin60^{\circ }$
$\Rightarrow \tau =2W\times \frac{\sqrt{3}}{2}$
$\Rightarrow \tau =\sqrt{3}W$