A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in

Question

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of 1.0 T. The field lines make an angle of 60 degrees with the normal to the coil. What is the magnitude of the counter torque that must be applied to prevent the coil from turning?

Professor PC Thomas Classes · Answer

Torque $	au = NIAB \sin	heta = 30 	imes 6.0 	imes (\pi 	imes 0.08^2) 	imes 1.0 	imes \sin(60^\circ) = 30 	imes 6.0 	imes 0.0201 	imes 0.866 \approx 3.13	ext{ N m}$.