A circular coil has moment of inertia 0.8 kg m² around any diameter and is carrying current to produce a magnetic moment of 20 Am² . The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60ᵒ will be:

Solution

Physics sir · 4 steps

Use energy conservation: the drop in -mB cos θ is the rotational KE the coil gains.

1. Find the angles. The magnetic moment points along the normal to the coil's plane. The plane is vertical, so the moment starts out horizontal, at 90° to the vertical field. After the coil turns by 60°, the angle between m⃗ and B⃗ is 30°.
2. Energy conservation. The potential energy of a dipole is U = -mB cos θ. The coil starts from rest, so the potential energy it loses becomes rotational kinetic energy 1/2Iω².

   1/2Iω² = mB(cos 30° - cos 90°) = 20 × 4 × √3/2 = 40√3 J

3. Solve for omega. ω² = (2 × 40√3)/0.8 = 100√3, so ω = 10 × 3^(1/4) rad s⁻¹.
4. So. The coil reaches an angular speed of 10(3)^(1/4) rad s⁻¹, which is option (c).

The idea to keep. The rotation angle of the coil is not the angle in -mB cos θ. First find where m⃗ points, which is normal to the coil, then measure its angle from B⃗.

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