Question

A magnetic needle is placed parallel to a magnetic field. The amount of work done in rotating the coil by an angle of 60${}^0$ is W units. Then, the torque required to keep the needle in the displaced position is

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

Answer 1

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

Torque at an angle $\theta =ksin\theta$

work done to bring from zero to 60 degree $=W={\int }_0^{60}ksin\theta d\theta ={|-kcos\theta |}_0^{60}=\frac{k}{2}$

$\Rightarrow k=2W$

torque at 60 degree required to keep the needle stable $=ksin60=2W\frac{\sqrt{3}}{2}=\sqrt{3}W$