A circular coil having 200 turns, 2.5 × 10⁻⁴ m² area and carrying 100 μ A current is placed in a uniform magnetic field of 1 T. Initially the magnetic dipole moment (M⃗) was directed along B⃗. Amount of work, required to rotate the coil through 90° from its initial orientation such that M⃗ becomes perpendicular to B⃗, is ________ μJ.

Solution

Physics sir · 3 steps

Work done is the rise in potential energy, -MB cos θ, going from 0° to 90°.

1. Dipole moment. For a coil, M = NIA = 200 × (100 × 10⁻⁶ A) × (2.5 × 10⁻⁴ m²) = 5 × 10⁻⁶ A m².
2. Energy view. A dipole in a field has potential energy U = -MB cos θ. The work you must do to turn it slowly equals the change in U.

   W = U_(90°) - U_(0°) = MB(cos 0° - cos 90°) = MB

3. So. W = (5 × 10⁻⁶) × 1 = 5 × 10⁻⁶ J, which is 5 μJ.

The common slip: 0.025. Leaving out the 200 turns and using M = IA gives 0.025 μJ. Each turn adds its own moment, so multiply by N.

The idea to keep. To rotate a dipole from θ₁ to θ₂, the work is MB(cos θ₁ - cos θ₂). For a coil, always build M as NIA and count every turn.

Answer from the NTA final answer key. Solution by PrepWiser.