A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60 is W. Now the torque required to keep the magnet in this new position is
- W 3
- 3W
- 3W2
- 2W 3
Answer
(2) 3W
Solution
Work done in rotating from 0 to 60 :
W = mB(1 - 60 ) = mB2 mB = 2W
Torque at 60 : = mB 60
= 2W 32 = 3W
Answer: (2)