Microscopic picture of an inductor

Question

I have a good understanding of how inductors behave in electrical circuits, and a somewhat rough-and-ready understanding of how this behaviour arises from Maxwell's equations. However, what I don't have a good mental picture of is how electromagnetic induction works on the microscopic level, i.e. in terms of forces experienced by individual electrons in the wire. Ideally I would like to be able to understand the operation of an inductor, at least qualitatively, in terms of the statistical mechanics of the electron gas in the coil and the electron spins in the core.

To clarify: in Maxwell's equations there is a term for $I$, the electric current. But current is a macroscopic quantity - it's the expected number of charges passing through a surface, with the expectation taken over an ensemble. So Faraday's law is a macroscopic relation, just like the gas equation. For the gas equation, we can understand how it arises from the microscopic motion of molecules. I want to understand how Faraday's law arises from the microscopic motion of electrons.

Would anyone be able to provide an explanation of how induction works in microscopic terms, or point me towards somewhere I can read up on it?

Answer 1

(Upgrading my comment to an answer as per @Nathanial's request but with more detail...and an animation)

Mutual induction (e.g. in a transformer) is easier to understand than self-induction (the inductance of a coil). Mutual induction separates the magnetic field source from the motion of the electrons in the pickup coil. A time varying magnetic field generates an e-field. Each conduction electron in the pickup coil then sees this e-field, the field from any electron density gradients along the wire, and the confining force at the boundary of the wire. Imbalances in these forces cause the electron to move.

We can simulate the above by (1) treating the pickup coil as being formed from a chain of electrons, (2) limit the interaction between electrons to nearest neighbors (and linearizing the force vs displacement), and (3) assuming a cylindrical time varying B-field co-linear with the coil. The last assumption allows us to quickly calculate the generated electric field on the electrons using the integral form of the Faraday equation and symmetry. For an open single wire loop, a constant $dB/dt$ will result in an equilibrium condition (assuming a finite resistance in the wire) where the charge pushed by the e-field accumulates in the end of the wire (pretty much like a capacitor plate). Throwing this into Mathematica to calculate the equilibrium locations and then varying $dB/dt$ sinusoidally:

There are no dynamics going on here except the changing $dB/dt$. Effectively, by using the equilibrium locations for each time step, I have assumed that the coil resistance is large enough to damp out any ringing in the coil. Should be possible to include such dynamics and see LC ringing.

Answer 2

Take a loop of wire (the simplest inductor), carrying a steady current $I_0$ (to start with). Let's say it's a superconductor, so the current is really staying constant. You understand that there is also a steady magnetic field created by this current, $B_0$.

So far, there is no force on the electrons--or at any rate, no force that we care about (only a force in the direction along the wire can produce an EMF, and this magnetic field cannot create such a force.)

Now we warm up the wire above the superconducting transition. Now there's a resistance, and the current starts to go down, from $I_0$ towards, at first, $0.99I_0$. As the current goes down, the magnetic field gets smaller: Now it's $0.99B_0$. If you believe Maxwell's equations, a changing magnetic field creates an electric field looping around the wire. That electric field pushes on the electrons (an EMF). The sign of the EMF, you can check, is in the same direction as $I_0$ (Lenz's law). So this is an inductor: The faster the current changes, the faster the magnetic field changes, the stronger the electric field is created, the more EMF opposes the change in current.

Answer 3

You do not need neither a coil, nor core. The vacuum is a pretty good conductor of the magnetic field. Cores and coils only complicate the things. Consider a single loop of current. Surprise: It also has a magnetic field!

Individual electrons are susceptible to the electric field, that is induced by the changing magnetic field in the usual way: $F = q * E$. $\vec B$ circles around the wire, $E$ is in the direction of wire.

Might be you need to look at quantum electrodynamics?