JEE Main 2024 Physics question
Not in JEE Advanced syllabus
The magnetic moment of a bar magnet is 0.5 Am. It is suspended in a uniform magnetic field of 8 × 10 T. The work done in rotating it from its most stable to most unstable position is:
Answer: (B)
Solution
W = 2MB = 2 × 0.5 × 8 × 10⁻² = 8 × 10⁻² J.