Upon the insertion of an iron core with a cross-sectional area of 4 cm ^2 , the magnetic field inside a long solenoid having 50 turns cm ⁻¹ increases from 2.5 10⁻³ T to 2.5 T . Determine the current flowing in the solenoid.

Step-by-step solution

The magnetic field inside a long solenoid without a core is given by B₀ = ₀ n I . Given: Initial magnetic field, B₀ = 2.5 10⁻³ T Number of turns per unit length, n = 50 cm ⁻¹ = 5000 m ⁻¹ Permeability of free space, ₀ = 4 10⁻⁷ T m A ⁻¹ Substituting the values into the formula: 2.5 10⁻³ = 4 10⁻⁷ 5000 I 2.5 10⁻³ = 20000 10⁻⁷ I 2.5 10⁻³ = 2 10⁻³ I I = 2.5 2 2.5 6.28 0.4 A