A rod with circular cross-section area 2 cm² and length 40 cm is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside windings is 4 π × 10⁻⁶ Wb. The relative permeability of the rod is
(Given : Permeability of vacuum μ₀=4 π × 10⁻⁷ NA⁻²)

Solution

Physics sir · 4 steps

Flux gives you B inside the rod. Divide that by the vacuum field μ₀ nI and you have μᵣ.

1. Field from flux. The flux through the rod's cross-section is Φ = BA. With A = 2 cm² = 2 × 10⁻⁴ m², this gives B = (4π × 10⁻⁶)/(2 × 10⁻⁴) = 2π × 10⁻² T.
2. Field formula. Inside a long winding filled with a material, B = μ₀ μᵣ n I. Here n is the number of turns per metre: n = 400/(0.4 m) = 1000 m⁻¹.

   μᵣ = B/(μ₀ n I) = (2π × 10⁻²)/(4π × 10⁻⁷ × 1000 × 0.4)

3. Put in numbers. The denominator is 1.6π × 10⁻⁴ T, so μᵣ = (2 × 10⁻²)/(1.6 × 10⁻⁴) = 125.
4. So. The relative permeability of the rod is 125, option (b).

The idea to keep. For a filled solenoid, B = μ₀ μᵣ nI with n in turns per metre. The flux through the cross-section is BA; multiply by N only when the question asks for flux linkage.

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