Ques. 1JEE Main 2024, 4 April (Evening Shift)
The magnetic moment of a bar magnet is 0.5 Am2. It is suspended in a uniform magnetic field of 8 10-2 T. The work done in rotating it from its most stable to most unstable position is:
- 4 10-2 J
- 16 10-2 J
- 8 10-2 J
- Zero
Answer
(3) 8 10-2 J
Solution
U = -mB : most stable = 0 , most unstable = 180
W = U180 - U0 = mB - (-mB) = 2mB
= 2 0.5 8 10-2 = 8 10-2 J
Answer: (3)