Self Induction And Back Emf

As stated in NCERT Class 12, Chapter 6 (Electromagnetic Induction), a changing current in a coil induces a self-emf given by $\epsilon =-L\frac{dI}{dt}$, which by Lenz's law opposes the change in current that produces it. The magnitude of the rate of change of current is $\left|\frac{dI}{dt}\right|=\frac{10-4}{0.3}=\frac{6}{0.3}=20\text{A/s}$. Therefore $|\epsilon |=L\left|\frac{dI}{dt}\right|=0.6\times 20=12\text{V}$. The option 20 V mistakenly reports only the rate of current change without multiplying by inductance. The option 6 V uses the raw current change of 6 A without dividing by time or multiplying by $L$ properly. The option 36 V multiplies by an incorrect factor. A plausibility check confirms units: $\text{H}\times \text{A/s}=\text{(Vcdotps/A)}\times \text{(A/s)}=\text{V}$. Because the current is decreasing, the induced emf acts to maintain the current, illustrating the coil's opposition to any change in its own flux. This directional behaviour is a direct consequence of Lenz's law: had the current instead been increasing, the self-induced emf would have opposed the rise, always resisting whatever change is imposed. It is important to recognise that the induced emf depends on the rate of change of current and not on the instantaneous value of the current itself, so a large steady current produces no self-emf at all. This property is why inductors are used to smooth fluctuating currents and to protect circuits from sudden surges, since they inherently resist rapid changes and thereby stabilise the flow of charge in the circuit.