A bar magnet having a magnetic moment of 2.0 × 10⁵ JT⁻¹, is placed along the direction of uniform magnetic field of magnitude B = 14 × 10⁻⁵ T. The work done in rotating the magnet slowly through 60° from the direction of field is :

Solution

Physics sir · 4 steps

Work done equals the gain in potential energy, so use W = MB(1 - cos θ), starting from θ = 0.

1. Energy of a dipole. A magnet at angle θ to the field has potential energy U = -MB cos θ. Rotating it slowly means no kinetic energy is gained, so the work you do equals the rise in U.

   W = U_(60°) - U_(0°) = MB(cos 0° - cos 60°) = MB(1 - cos 60°)

2. Find MB. MB = (2.0 × 10⁵ J T⁻¹) × (14 × 10⁻⁵ T) = 28 J.
3. Put in numbers. W = 28 × (1 - 1/2) = 28 × 1/2 = 14 J.
4. So. The work done is 14 J, which is option (a).

The idea to keep. Work to turn a magnet from θ₁ to θ₂ is MB(cos θ₁ - cos θ₂). From the field direction, θ₁ = 0, so it becomes MB(1 - cos θ).

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