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, $W=MB\left(cos 0 ^\circ - cos 60 ^\circ \right)$
$W=MB\left(1 - \frac{1}{2}\right)=\frac{M B}{2}$
$\Rightarrow MB=2W$
Restoring torque on the needle $\tau=MBsin\theta $
$\Rightarrow \tau=MBsin60^\circ $
$\Rightarrow \tau=2W\times \frac{\sqrt{3}}{2}$
$\Rightarrow \tau=\sqrt{3}W$