A long solenoid with 1000 turns/m has a core material with relative permeability 500 and volume 10³ cm³. If the core material is replaced by another material having relative permeability of 750 with same volume maintaining same current of 0.75 A in the solenoid, the fractional change in the magnetic moment of the core would be approximately ( x/499 ). Find the value of x.

Solution

Physics sir · 4 steps

The core's moment comes from its magnetisation, so track μᵣ - 1, not μᵣ itself.

1. Same H field. The field intensity inside a solenoid is H = nI = 1000 × 0.75 = 750 A m⁻¹. It depends only on the turns per metre and the current, so it is the same for both cores.
2. Moment of the core. The core's magnetisation is Iₘ = χ H with χ = μᵣ - 1. Its magnetic moment is magnetisation times volume, so M = (μᵣ - 1)HV, with V = 10³ cm³ = 10⁻³ m³.

   (Δ M)/M₁ = ((μᵣ₂-1)-(μᵣ₁-1))/(μᵣ₁-1) = (749-499)/499 = 250/499

3. Check with numbers. M₁ = 499 × 750 × 10⁻³ = 374.25 A m² and M₂ = 749 × 750 × 10⁻³ = 561.75 A m². The change is 187.5 A m², and 187.5/374.25 = 250/499, because H and V cancel in the ratio.
4. So. The fractional change is 250/499, so x = 250.

The idea to keep. Only the material's own response, χ = μᵣ - 1, magnetises the core. When H and volume stay fixed, the core's magnetic moment scales with χ.

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