Q. A magnetic needle suspended parallel to a magnetic field requires √3 J of work to turn it through 60°. The torque needed to maintain the needle in this position will be
A) 2√3 J
B) 3 J
C) √3 J
D) (3/2) J
Explanation
[Solution from source] Work to turn from 0° to 60° is W = MB(1 − cos 60°) = MB/2 = √3 J, so MB = 2√3 J. The torque to hold it there is τ = MB sin 60° = 2√3 × √3/2 = 3 (in N m). Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.