JEE Main 2024 · 4 April · evening · Single correct · +4 / −1
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:
- (a) 4 × 10⁻² J
- (b) 16 × 10⁻² J
- (c) 8 × 10⁻² J
- (d) Zero
Answer and solution
Answer (c) 8 × 10⁻² J
Solution
Physics sir · 4 steps
Work done equals the change in potential energy, and going from -MB to +MB is a jump of 2MB.
- Potential energy. A magnetic dipole at angle θ to a uniform field has U = -MB cos θ. The work you do in turning it slowly equals the gain in U.
- The two positions. Most stable is θ = 0°, where U = -MB. Most unstable is θ = 180°, where U = +MB.
W = U_(180°) - U_(0°) = MB - (-MB) = 2MB
- Put in numbers. W = 2 × 0.5 × 8 × 10⁻² = 8 × 10⁻² J.
- So. The work done is 8 × 10⁻² J, which is option (c).
The idea to keep. For a dipole in a uniform field, W = MB(cos θ₁ - cos θ₂). Stable to unstable means 0° to 180°, so the work is always 2MB.
Answer from the NTA final answer key. Solution by PrepWiser.