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It's commonly taught that the electric field in a conductive material is zero. Hence the voltage through a perfectly conductive material is zero. I however when learning about inductors in physics assumed the presence of changing magnetic fields to be an exception to this rule. An ideal inductor is assumed to have a resistance of zero but has a voltage of $-L \frac{\mathrm{dI} }{\mathrm{d} t}$. This means the magnetic fields must produce electric fields inside the coil. We get this from: $\nabla \times E = -\frac{\mathrm{dB} }{\mathrm{d} t}$. I, however, read the Feynman Lectures on the topic and he says:

As we have seen before, there can be no electric fields inside a perfect conductor. (The smallest fields would produce infinite currents.) Therefore the integral from a to b via the coil is zero. The whole contribution to the line integral of E comes from the path outside the inductance from terminal b to terminal a. Since we have assumed that there are no magnetic fields in the space outside of the “box,” this part of the integral is independent of the path chosen and we can define the potentials of the two terminals.

My question is how does zero voltage through the inductor coil amount to a voltage between the ends? At what point in the wire is this voltage introduced?

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    $\begingroup$ There is that pesky magnetic field to consider... $\endgroup$ – Jon Custer Apr 18 at 21:11
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This means the magnetic fields must produce electric fields inside the coil.

Yes, but this electric field, called induced electric field $\mathbf E_i$ , is only one component of total field in the conductor of the coil. It is not the total field.

Total electric field in an ideal wire of the coil is zero, but this does not imply that voltage on the terminals has to be zero. Voltage in AC circuits is not defined by total electric field, but only by its Coulomb ("electrostatic") component. This is because voltage comes from electrostatics and only makes sense for electrostatic field.

In an ideal conductor, total electric field has to be zero, so any induced electric field has to be counteracted and cancelled by some other contribution to the field, of same magnitude but opposite direction in space. Electric field that is always present in electric circuits is the Coulomb field $\mathbf E_{Coulomb}$ due to charges on conductors' surfaces. This field is zero inside conductors in the special case where everything is in static equilibrium, but as soon as charges accelerate (when electric current changes), the Coulomb field will be non-zero inside conductors, so that it can cancel the induced field. So, in the conductor of coil with changing current, the induced electric field due to accelerated charges in the coil is cancelled by the Coulomb field of all the charges in the circuit (mostly on the surfaces of the conductors in the circuit).

Voltage is integral of this Coulomb, or "electrostatic" component of the field. It is electrostatic in the sense it is the potential Coulomb field. Its integral is therefore independent of the path, it only depends on the endpoints. To calculate voltage between coil terminals A and B, one can use the path that the charge carriers actually follow, but if one chose a path that went out of that path and then back, as long as the endpoints are the same, the result would be the same.

Induced electromotive force (emf), on the other hand, is integral of the induced electric field, and does depend not only on the endpoints, but also on the path. However, usually we are interested only in the value of emf for the path the charge carriers in the coil go through.

Let motion "from A to B" be the positive sense of circulation in the circuit. Then the value of the emf for this oriented path is, in the common convention,

$$ emf(A~to~ B) = \int_A^B \mathbf E_i\cdot d\mathbf s = -L\frac{dI}{dt}.~~~(1) $$

Since induced electric field is everywhere in the ideal coil cancelled by the Coulomb electric field, the integral of this Coulomb field has to have the opposite value: $$ \int_A^B \mathbf E_{Coulomb}\cdot d\mathbf s = + L\frac{dI}{dt}.~~~(2) $$ This integral is also the difference of potentials $\varphi_B - \varphi_A$, i.e. voltage of B with respect to A.

In real coils, this cancellation of electric field components is not complete and total electric field is not necessarily zero. Thus voltage does not cancel emf exactly. Emf is still given by the general formula (1), but real case voltage has no such general formula as (2), that formula is valid only in the ideal case.

However, if the conductor is Ohmic, we can find a different relation between the emf and voltage. We can write the generalized Ohm law for the coil:

$$ \int_{A,~through~the~path~of~the~current}^B (\mathbf E_i + \mathbf E_{Coulomb})\cdot d\mathbf s = RI $$ where $R$ is Ohmic resistance of the coil conductor from $A$ to $B$. Using emf and voltage, this is

$$ emf(A~to~B) + (\varphi_B-\varphi_A) = RI $$

So emf and voltage no longer cancel each other completely, but their sum is "net active force" that pushes the current against the resistance; the greater the resistance, the greater the difference of emf and voltage magnitudes has to be to maintain the same current.

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The voltage across any conductor is zero in electrostatics. If you have a dynamically changing magnetic field, then the concept of "voltage" doesn't really even make sense any more. The electric field inside of a perfect conductor will indeed be zero, but the presence of a changing magnetic field means that we can no longer guarantee that $\oint_C \vec E \cdot d\vec l = 0$ around a closed curve $C$. To make it so that the circuit rule $\sum_i \Delta V_i = 0$ for voltage differences around a closed loop still holds for circuits with inductors in them, we define the "voltage" across the inductor so that it cancels out the effect of the changing magnetic field. The electric field inside the conductor is still 0 though. (assuming it's a perfect conductor)

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In my opinion:

The charge is accelerating through the coil because there is clearly a di/dt term. This causes charge to not be evenly distributed and uneven charge distributions create voltages I think.

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