A rod with a circular cross-section area 2 cm ^2 and length 40 cm is wound uniformly with 400 turns of an insulated wire. If a current 0.4 A flows in the wire winding, the total magnetic flux produced inside the winding is 4 10⁻⁶ Wb. The relative permeability of the rod is (Given permeability of vacuum ₀ = 4 10⁻⁷ N m A ⁻² )

Step-by-step solution

The magnetic flux inside the solenoid is given by = B A , where B is the magnetic field and A is the cross-sectional area. Substituting the given values, the magnetic field is: B = A = 4 10⁻⁶ 2 10⁻⁴ = 2 10⁻² T The magnetic field inside a solenoid with a core of relative permeability _r is also given by: B = ₀ _r n I = ₀ _r N l I Substituting the known values into the equation: 2 10⁻² = 4 10⁻⁷ _r 400 0.4 0.4 2 10⁻² = 4 10⁻⁷ _r 400 2 10⁻² = 16 10⁻⁵ _r Solving for _r : _r = 2 10⁻² 16 10⁻⁵ = 1000 8 = 125 Answer: 125