Question

If $5\text{J}$ amount of work is required to rotate a current carrying magnetic coil in a uniform magnetic field by ${60}^{\circ }$, from a position parallel to the field. What is the torque required to maintain it in the new position.

• A. $10\sqrt{3}\text{J}$
• B. $5\text{J}$
• C. $10\text{J}$
• D. $5\sqrt{3}\text{J}$

Answer 1

The correct option is D $5\sqrt{3}\text{J}$

Work done in rotating a magnetic coil from $0^{\circ }$ to ${60}^{\circ }$ will be equal to the change in potential energy of the coil,

$\text{Work done}=U_2-U_1$

$\Rightarrow 5=-\mu B\cos {60}^{\circ }-(-\mu B\cos 0^{\circ })[\because U=-\mu B\cos \theta ]$

$\Rightarrow \mu B=10\text{J}$

Torque required to maintain in that position will be,

$\tau =\mu B\sin \theta$

$\Rightarrow \tau =10\times \sin {60}^{\circ }=5\sqrt{3}\text{J}$

Hence, option (d) is the correct answer.