I'm having a little bit of a hard time understanding some concepts to do with self-inductance in class.
I understand mutual inductance. That's when a loop has some time-dependent current going through it, and because the current is time-dependent, the magnetic field the loop produces is also time-dependent, and the magnetic flux in a nearby loop is time-dependent.
With self-inductance, though, is the idea that a single loop carrying a time-dependent current gives rise to a time-dependent magnetic field, and that very same time-dependent magnetic field gives rise to a change in flux in the single loop itself, which then generates an emf?
If I'm correct in this reasoning, then that should mean that the new emf gives way to a current, which opposes the previously existing current. Why doesn't this just result in an infinite loop creating an opposing current, and then one that opposes that, and then one that opposes THAT…etc.?
Best Answer
We can get a general idea for this by considering the case of a battery connected to a series combination of a resistor and a solenoid, the latter of which acts as an inductor.
Let our solenoid have radius $r$, length $\ell$, and consist of $N$ turns of wire. If a current $I$ is flowing through it, then there will be a magnetic field in the interior of the solenoid given by $$B = \frac{\mu_0 N I}{\ell}$$
The total magnetic flux through the solenoid is $\Phi_B = 2\pi r^2 NB$. If we apply Faraday's law to the circuit, we obtain the following equation:
$$V_0-IR=-\frac{d}{dt}\Phi_B = -2\pi r^2 N \frac{dB}{dt} = -\left[\frac{2\pi \mu_0 r^2 N^2}{\ell}\right]\frac{dI}{dt} \equiv -L\frac{dI}{dt}$$ where $V_0$ is the voltage across the battery and $L$ is called the (self) inductance of the solenoid. Unlike the mutual inductance between two coils - where the current through one coil induces an EMF around the other - here we see that the current through the solenoid induces an EMF across the same solenoid. This is why the effect is called self-inductance.
The differential equation takes the form $$L\frac{dI}{dt} + IR=-V_0$$
Assuming that the circuit is closed at $t=0$ (so no current is flowing yet, $I(0)=0$), the solution to this equation is $$I(t) = \frac{V_0}{R}\left(1-e^{-tR/L}\right)$$ which you can straightforwardly check.