The current in an inductor is given by $\mathrm{I}=(3 \mathrm{t}+8)$ where $\mathrm{t}$ is in second. The magnitude of induced emf produced in the inductor is $12 \mathrm{~mV}$. The self-inductance of the inductor _________ $\mathrm{mH}$.

<p>The induced emf ($\varepsilon$) in an inductor is given by Faraday's law of electromagnetic induction, which in its differential form for an inductor can be expressed as:</p> <p>$\varepsilon = L \frac{dI}{dt}$</p> <p>where:</p> <ul> <li>$\varepsilon$ is the induced emf in the inductor,</li> <li>$L$ is the inductance of the inductor,</li> <li>$\frac{dI}{dt}$ is the rate of change of current through the inductor.</li> </ul> <p>Given that the current $I = (3t + 8)$, where $t$ is in seconds, we can find the rate of change of current by differentiating $I$ with respect to $t$.</p> <p>$\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3$</p> <p>The given magnitude of induced emf is $12 \, \text{mV} = 12 \times 10^{-3} \, \text{V}$ (since $1\,\text{mV} = 10^{-3} \, \text{V}$).</p> <p>Now, plug these values into the formula to find $L$:</p> <p>$12 \times 10^{-3} = L \cdot 3$</p> <p>Solving for $L$ gives:</p> <p>$L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H} = 4 \, \text{mH}$</p> <p>Therefore, the self-inductance of the inductor is <strong>4 mH</strong>.</p>