Question

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

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

Answer 1

The correct option is A $\sqrt{3}W$

$W={\int }_{{\theta }_1=0^{\circ }}^{{\theta }_2={60}^{\circ }}MBsin\theta d\theta =MB[-cos\theta {]}_{0^{\circ }}^{{60}^{\circ }}$
$=-MB[cos{60}^{\circ }-cos{0}^{\circ }]=-MB[\frac{1}{2}-1]=\frac{MB}{2}$
$\therefore MB=2W$
Now, torque, $\tau =MBsin\theta =2Wsin{60}^{\circ }=2W\times \frac{\sqrt{3}}{2}=\sqrt{3}W$