The current in an inductor is given by I = (3t + 8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self-inductance of the inducto r _______ mH.

Question

G1 Physics · Accepted Answer

Induced emf magnitude: ε = L(dI/dt) Given: I = 3t + 8 Therefore: dI/dt = 3 A/s Also: ε = 12 mV = 12 × 10⁻³ V Hence: L = ε/(dI/dt) = (12 × 10⁻³)/3 = 4 × 10⁻³ H = 4 mH