A coil having 100 turns, area of 5 × 10⁻³ m², carrying current of 1 mA is placed in uniform magnetic field of 0.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90° is _________ μ J.

Solution

Physics sir · 4 steps

Plane perpendicular to B means the area vector lies along B, so the coil starts in its lowest-energy position.

1. Magnetic moment. For a coil, M = NIA = 100 × (1 × 10⁻³) × (5 × 10⁻³) = 5 × 10⁻⁴ A m².
2. Starting angle. The area vector, and so M⃗, is normal to the plane of the coil. The plane is perpendicular to B⃗, so M⃗ is parallel to B⃗ and θ₁ = 0°. Turning through 90° gives θ₂ = 90°.

   W = MB(cos θ₁ - cos θ₂) = MB(1 - 0) = MB

3. Put in numbers. W = 5 × 10⁻⁴ × 0.20 = 1 × 10⁻⁴ J.
4. So. 1 × 10⁻⁴ J = 100 μJ, so the work done is 100.

The common slip: 1. Leaving out the 100 turns and using M = IA gives 1 μJ, not 100.

The idea to keep. Work to rotate a dipole in a field is MB(cos θ₁ - cos θ₂), with θ measured from the area vector, not the plane. For a coil, M = NIA, so never drop the turns.

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