Inside a long solenoid that carries a current of 5.00 A , the magnetic field B is found to be 3.14 10⁻² T . Determine the number of turns per unit length for this solenoid.

Step-by-step solution

The magnetic field inside a long solenoid is given by the formula: B = ₀ n I Rearranging for the number of turns per unit length n : n = B ₀ I Substituting the given values B = 3.14 10⁻² T , I = 5.00 A , and ₀ = 4 10⁻⁷ T m/A : n = 3.14 10⁻² 4 10⁻⁷ 5.00 Using 3.14 : n = 3.14 10⁻² 4 3.14 10⁻⁷ 5.00 n = 10⁻² 20 10⁻⁷ n = 10⁻² 2 10⁻⁶ n = 0.5 10⁴ = 5000 turns/m