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 inductor ______ mH.

We are given that the current in an inductor is $$I = (3t + 8)$$ A (where $$t$$ is in seconds) and the magnitude of the induced EMF is $$12$$ mV, and we need to find the self-inductance.

Since the EMF induced in an inductor is given by $$|emf| = L \frac{dI}{dt}$$ where $$L$$ is the self-inductance and $$\frac{dI}{dt}$$ is the rate of change of current, we first compute this rate.

For $$I = 3t + 8$$, differentiating with respect to time gives $$\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3 \text{ A/s}$$, so the rate of change of current is constant at 3 A/s.

Substituting into the EMF formula yields $$12 \times 10^{-3} = L \times 3$$, which gives $$L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \text{ H} = 4 \text{ mH}$$.

The answer is 4 mH.