An iron rod of volume 10 –3 m³ and relative permeability 1000 is placed as core in a solenoid with 10 turns/cm. If a current of 0.5 A is passed through the solenoid, then the magnetic moment of the rod will be :

Solution

Physics sir · 4 steps

Find H from the solenoid, multiply by susceptibility to get M, then multiply by volume to get the moment.

1. Field from solenoid. First convert the turns to per metre: 10 turns per cm is n = 1000 m⁻¹. The magnetising field is H = nI = 1000 × 0.5 = 500 A m⁻¹.
2. Magnetisation. Susceptibility is χ = μᵣ - 1 = 999. The magnetisation of the rod is M = χ H = 999 × 500 = 4.995 × 10⁵ A m⁻¹.

   m = M V = (μᵣ - 1) n I V

3. Magnetic moment. Magnetisation is magnetic moment per unit volume, so m = 4.995 × 10⁵ × 10⁻³ = 499.5 A m².
4. So. m ≈ 500 A m² = 5 × 10² A m², which is option (a).

The idea to keep. For a core in a solenoid, follow the chain H = nI, then M = χ H, then m = MV. Always put n in turns per metre before you start.

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