JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
A bar magnet with a magnetic moment 5.0 Am² is placed in parallel position relative to a magnetic field of 0.4 T. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _____________.
- (a) zero
- (b) 1 J
- (c) 2 J
- (d) 4 J
Answer and solution
Answer (d) 4 J
Solution
Physics sir · 4 steps
Work done equals the change in potential energy, and parallel to antiparallel is a jump from -mB to +mB.
- Potential energy. A magnetic dipole in a uniform field has potential energy U = -mB cos θ, where θ is the angle between the moment and the field.
- Two positions. Parallel means θ = 0°, so U₁ = -mB. Antiparallel means θ = 180°, so U₂ = +mB.
W = U₂ - U₁ = mB - (-mB) = 2mB
- Put in numbers. W = 2 × 5.0 × 0.4 = 4 J.
- So. The work needed to turn the magnet from parallel to antiparallel is 4 J, option (d).
The idea to keep. To turn a dipole from θ₁ to θ₂, the work is W = mB(cos θ₁ - cos θ₂). A full flip from parallel to antiparallel always costs 2mB.
Answer from the NTA final answer key. Solution by PrepWiser.