Question

A magnetic needle lying parallel to a magnetic field requires $W$ unit of work to turn it through $60^{\circ }$ . What is the torque needed to maintain the needle in this position ?

• A. $\sqrt{3}W$
• B. $W$
• C. $\left(\frac{\sqrt{3}}{2}\right)W$
• D. $2W$

Answer 1

Answer: (A)

The work done in rotating the coil will be;
$W=MB\left(Cos{\left(\theta \right)}_1-Cos{\left(\theta \right)}_2\right)$
$=MB\left(Cos{0}^{\circ }-Cos60^{\circ }\right)$
$=MB\left(1-\frac{1}{2}\right)=\frac{MB}{2}$
The torque needed to maintain the coil in this position will be;
$\tau =MBSin\theta =MBSin60^{\circ }=\frac{\sqrt{3}}{2}MB$
$=\frac{\sqrt{3}}{2}(2W)\left[\because \frac{MB}{2}=W\right]$
$\Rightarrow \tau =\sqrt{3}W$