Question

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

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

Answer 1

Answer: (B)

$\sqrt{3}W$

$W=MB(1-\cos 0^o)=0$ ....(1)

and $W=MB(1-\cos {60}^o)$ ... [as $\theta ={60}^o$]

or, $W=\frac{MB}{2}$

$MB=2W$....(2)

Also, torque , $T=\frac{\sqrt{3}}{2}MB$

$T=\frac{\sqrt{3}}{2}\times 2W$ ..... [from equation (2)]

$T=\sqrt{3}W$