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
- 2 3 J
- 3 J
- 3 J
- 32 J
Answer
(2) 3 J
Solution
Work done in turning from 0 to : W = mB(1 - )
3 = mB(1 - 60 ) = mB2
mB = 2 3
Torque at 60 : = mB 60 = 2 3 32 = 3
= 3 J (i.e. 3 N m)
Answer: (2)