Question 157: Electromagnetic Induction

Question

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 _________ $$\mathrm{mH}$$.

Accepted Answer

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:
$$\varepsilon = L \frac{dI}{dt}$$
where:
$\varepsilon$ is the induced emf in the inductor,
$L$ is the inductance of the inductor,
$\frac{dI}{dt}$ is the rate of change of current through the inductor.
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 $t$.
$$\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3$$
The given magnitude of induced emf is $12 \, 	ext{mV} = 12 	imes 10^{-3} \, 	ext{V}$ (since $1\,	ext{mV} = 10^{-3} \, 	ext{V}$).
Now, plug these values into the formula to find $L$:
$$12 	imes 10^{-3} = L \cdot 3$$
Solving for $L$ gives:
$$L = \frac{12 	imes 10^{-3}}{3} = 4 	imes 10^{-3} \, 	ext{H} = 4 \, 	ext{mH}$$
Therefore, the self-inductance of the inductor is
4 mH
.

Electromagnetic Induction

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

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