JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
A small bar magnet placed with its axis
at 30ᵒ with an external field of 0.06 T
experiences a torque of 0.018 Nm. The
minimum work required to rotate it from its
stable to unstable equilibrium position is :
- (a) 6.4 × 10⁻² J
- (b) 9.2 × 10⁻³ J
- (c) 7.2 × 10⁻² J
- (d) 11.7 × 10⁻³ J
Answer and solution
Answer (c) 7.2 × 10⁻² J
Solution
Physics sir · 4 steps
Use the torque to find the magnetic moment first. Then the flip from stable to unstable costs exactly 2MB.
- Find the moment. Torque on a dipole is τ = MB sin θ. So M = 0.018/(0.06 × sin 30°) = 0.018/0.03 = 0.6 A m².
- Two equilibrium positions. Potential energy is U = -MB cos θ. Stable equilibrium is θ = 0°, where U = -MB. Unstable equilibrium is θ = 180°, where U = +MB.
W = U_(180°) - U_(0°) = MB - (-MB) = 2MB
- Put in numbers. W = 2 × 0.6 × 0.06 = 0.072 J.
- So. The minimum work is 7.2 × 10⁻² J, which is option (c).
The idea to keep. The 30° only helps you find M. Stable to unstable means 0° to 180°, and that always costs 2MB, whatever angle the magnet started at.
Answer from the NTA final answer key. Solution by PrepWiser.