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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, 2019 at 21:11

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This means the magnetic fields must produce electric fields inside the coil.

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

Total electric field in an ideal wire of the coil is zero (because it is perfect conductor). However, this does not imply that voltage on the terminals has to be zero. Voltage on a component in AC circuits is not defined by total electric field, but only by its conservative (Coulomb, electrostatic) component. In AC circuits "voltage" usually means difference of electric potential, and this difference is defined based on the conservative field component.

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

Voltage or potential difference is integral of this conservative component of the field $\mathbf E_C$. 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 $\mathbf E_i$, and depends not only on the endpoints, but also on the integration path. However, usually we are interested only in the value of emf for the special path where the charge carriers in the coil go.

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 conservative electric field, the integral of this conservative field has to have the opposite value: $$ \int_A^B \mathbf E_{C}\cdot d\mathbf s = + L\frac{dI}{dt}.~~~(2) $$ This integral gives the difference of potentials $\varphi_B - \varphi_A$.

In real coils, this cancellation of electric field components is not complete and total electric field is not necessarily zero. Thus potential difference does not counteract emf exactly. Emf is still given by the general formula (1), but potential difference is no longer equal to $LdI/dt$; this expression is valid only in the case of ideal inductor.

However, if the conductor is Ohmic, we can find a different relation between the emf and potential difference. The Kirchhoff 2nd circuital law (generalized Ohm's law) for the coil from terminal A to terminal B implies:

$$ \int_{A,~through~the~path~of~the~current}^B \mathbf E_{total}\cdot d\mathbf s = RI. $$

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

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

Since the right-hand side is not zero in general, emf and potential difference do not cancel each other completely in real inductors. Their algebraic sum can be understood as the "net active force" that pushes the current against the Ohmic resistance. The emf usually acts against the potential difference, so the greater the Ohmic resistance, the greater the difference of magnitudes of emf and potential difference has to be to maintain the same current.

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    $\begingroup$ In low frequency regime, the induced field is conservative ($\nabla\times \mathbf E_i = \mathbf 0$) inside the ideal conductor. It is not necessarily conservative outside. $\endgroup$ Jun 6, 2020 at 14:04
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    $\begingroup$ I am not sure why you think outside the component the field is conservative. In general, it is not. For example, if current changes in time in a solenoid made of ideal conductor, electric field outside the conductor (in the vacuum) is not conservative (except for special instants such as when rate of change of current is zero, so electric field vanishes in some spatial region). $\endgroup$ Jun 6, 2020 at 21:47
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    $\begingroup$ I think I understand what you mean now - we consider a closed path that goes through the ideal conductor from one terminal to another, and then back through the outside region (vacuum). You are right that when the second part of the path is far enough so that induced electric field there is negligible, electric field there is almost Coulombic. So in both regions the electric field is conservative: in the first, it vanishes, in the second it does not but is almost electrostatic. So we can say electric field in the region where the closed loop is is conservative? $\endgroup$ Jun 6, 2020 at 22:09
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    $\begingroup$ We actually can - in the sense $\nabla \times \mathbf E =0$ anywhere in that region. However, this is local property of electric field at any point of that region. It does not follow that line integral for such a field and closed loop in that region has to be zero. $\endgroup$ Jun 6, 2020 at 22:13
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    $\begingroup$ Your expectation is right - there are discontinuities and the region where the conservativeness holds isn't simply connected (it is a torus). $\endgroup$ Jun 6, 2020 at 22:21
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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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