The inductance of the coil is given by the relation L = Nφ/I Where N = number of turns = 50 φ = flux = 200μWb I = current = 8 A L = 50 × 200 × 10−6 ⁄ 8 L = 1.25 mH
Given Inductance L = 2 H Rate of change of current di/dt = 5 A/sec Self induced EMF = − (Rate of change of current × Inductance) = −L(di/dt) = −(5 × 2) = −10V
EXAMIANS
