Electrostatic potential difference across a self-inductor

Question

This question is related to the much more vague closed question asked by someone else, but I am asking something more specific.

Suppose we have a circuit lying inside a three-dimensional region $D$ that is isolated from all electromagnetic fields coming from outside the circuit itself. Suppose that the circuit includes one inductor (with inductance $L$), and there is a three-dimensional region $R$ around the inductor such that the time-derivative of the inductor's magnetic field strength is negligible at each point in space outside of $R$. Assume that the region $R$ is small enough to not intersect with any other circuit components (besides ideal wire coming from the two terminals of the inductor) nor with any junctions in the circuit. Suppose that the rest of the circuit has negligible inductance.

We know that the voltage "across" the inductor is $-L\frac{\mathrm{d}I}{\mathrm{d}t}$, in the sense that [cf. Chapter 22 of the Feynman Lectures] if you connect an ideal voltmeter across the inductor using ideal connecting wires, and the voltmeter and its connecting wires stay outside of $R$, then the value shown by the voltmeter will agree with $-L\frac{\mathrm{d}I}{\mathrm{d}t}$ (or $+L\frac{\mathrm{d}I}{\mathrm{d}t}$ if you connect it in the opposite direction to that in which the current $I$ is being measured).

But what I'm wondering is whether this matches the "electrostatic potential difference" across the inductor; people seem to implicitly assume that it does [see my "Motivation" below], but I haven't seen this justified. More specifically, let us define the electrostatic potential difference between two points $\mathbf{a}$ and $\mathbf{b}$ within $D$ by $$V_{\mathbf{a},\mathbf{b}}^\text{static} = \frac{1}{4\pi} \iiint_D (\mathrm{div}\,\mathbf{E})(\mathbf{r})\left( \frac{1}{|\mathbf{r}-\mathbf{b}|} - \frac{1}{|\mathbf{r}-\mathbf{a}|} \right) \, \mathrm{d}^3\mathbf{r}$$ where $\mathbf{E}$ is the electric field strength; if we take points $\mathbf{a}$ and $\mathbf{b}$ on either side of the inductor, such that $\mathbf{a}$ and $\mathbf{b}$ are joined by just ideal wire and the inductor, and they both lie outside of $R$, do we necessarily have that $$V_{\mathbf{a},\mathbf{b}}^\text{static} \,\approx\, -L\frac{\mathrm{d}I}{\mathrm{d}t}$$ (where the current $I$ through the inductor is measured in the direction from $\mathbf{a}$ to $\mathbf{b}$)?

If so, how do we know this?

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Motivation.

The above $L\frac{\mathrm{d}I}{\mathrm{d}t}$ formula is typically expressed as being something like "the induced emf in the coil" or "the induced emf across the inductor". I am wanting to know if the implied picture of the physics conveyed by this kind of phrasing is really correct.

To be more specific:

Imagine that in place of the inductor were an ideal, negligible-inductance chemical battery with an "emf" of 9V. Then in particular, the electrostatic potential difference $V_{\mathbf{a},\mathbf{b}}^\text{static}$ would be 9V, and this would coincide with $\,-\!\int \mathbf{E}(\mathbf{r}) \boldsymbol{\cdot} \mathrm{d}\mathbf{r}$ along any path in $D$ from $\mathbf{a}$ to $\mathbf{b}$. In very simplified terms: the battery uses its storage of chemical energy to do work on the body of mobile charge, and the electrostatic fields in the circuit do (ideally) exactly opposing work. The idea of the phrase "electromotive force" is that it's describing the "driving effect", so to speak, of the force provided by the battery on the body of mobile charge.

Now in the case of the inductor, as the current changes, the electric field $\mathbf{E}_{\mathrm{ind}}$ of the mobile charge in the inductor is itself doing work on the body of mobile charge; and ideally, inside the inductor, $\mathbf{E}_{\mathrm{ind}}$ is exactly balanced out by the electrostatic field $\mathbf{E}_{\mathrm{static}}$ of the circuit.

Now let us define "the inductor's emf" along a path as $\int \mathbf{E}_{\mathrm{ind}}(\mathbf{r}) \boldsymbol{\cdot} \mathrm{d}\mathbf{r}$ along that path; this seems to me the most accurate formalisation of the intuitive picture of "induced electromotive force". For the path from $\mathbf{a}$ to $\mathbf{b}$ that goes through the inductor itself, since $\mathbf{E}_{\mathrm{ind}}$ and $\mathbf{E}_{\mathrm{static}}$ balance each other out inside the conductive material, the inductor's emf along that path will be the same as $\,-\!\int \mathbf{E}_{\mathrm{static}}(\mathbf{r}) \boldsymbol{\cdot} \mathrm{d}\mathbf{r}$, which is the electrostatic potential difference $V_{\mathbf{a},\mathbf{b}}^\text{static}$.

Now if we use the phrasing that "the induced emf in the coil is $L\frac{\mathrm{d}I}{\mathrm{d}t}$", what I think we are really implying is that $\int \mathbf{E}_{\mathrm{ind}}(\mathbf{r}) \boldsymbol{\cdot} \mathrm{d}\mathbf{r}$ along the path from $\mathbf{a}$ to $\mathbf{b}$ through the inductor is $-L\frac{\mathrm{d}I}{\mathrm{d}t}$, and hence by implication $V_{\mathbf{a},\mathbf{b}}^\text{static}$ is equal to $-L\frac{\mathrm{d}I}{\mathrm{d}t}$. In other words, in short, the implication behind the typical phrasing is that

as far as the electrostatics and the emfs outside the region $R$ are concerned, everything is the same as if we had instead a chemical battery of emf equal to $-L\frac{\mathrm{d}I}{\mathrm{d}t}$.

But I don't see an obvious reason why this should be so; to be more precise:

Couldn't we have that the inductor's emf $\int \mathbf{E}_{\mathrm{ind}}(\mathbf{r}) \boldsymbol{\cdot} \mathrm{d}\mathbf{r}$ along the path from $\mathbf{a}$ to $\mathbf{b}$ through the inductor itself is equal to some arbitrary value $X$, and then the inductor's emf $\int \mathbf{E}_{\mathrm{ind}}(\mathbf{r}) \boldsymbol{\cdot} \mathrm{d}\mathbf{r}$ along every single other path round the circuit from $\mathbf{b}$ back to $\mathbf{a}$ is equal to $-L\frac{\mathrm{d}I}{\mathrm{d}t}-X$?

Answer 1

Things will get simpler if you choose to define voltage as the line integral of the electric field (the 'total' electric field, not just its conservative component). Voltage will then have an 'electrostatic' component and an 'induced' component.
Along the paths that jump 'across' the inductor's terminal you will have a nonzero voltage due almost entirely to the electrostatic component, while along path going inside the coil, the electrostatic and induced components will almost exactly compensate each other leaving a near-zero contribution corresponding to the ohmic loss in the non-perfect conductor.

You need to renounce to KVL, though. Because if you go around a loop formed by the coil and the jump in the space between its terminals, you will get a non-zero result (it's the EMF).

That is, the voltage across the coil (which in the quasi-static approximation is -L dI/dt) is different from the voltage along the coil (which is r I, and is usually approximated as zero if the coil is a very good conductor).

It is easier to see in the case of mutual inductance, because there the variable magnetic field is generated by a separated circuit and you won't fall into the trap of circular reasoning.

To address your added motivation, consider that the electric field in the coil and the space around it has not only to obey Maxwell's equations, but it must also satisfy the constitutive relation in the material and the space around it, and also boundary conditions at the interfaces between materials. Don't forget the continuity equation.
When you consider all these equations and constraints, it emerges the need for a surface and interface charge. This is the charge that 'steers' the electric field inside the conductors, and makes sure the Ohm's law is obeyed in the higher resistance portions of the circuit for example. In your specific example, the surface charge creates the Coulombian electric field that superposes to the induced electric field that is there regardless of the presence of the coil.
This superposition will lead to a nearly complete cancellation of the electric field inside the coil (where $\vec{E}_{coul}$ and $\vec{E}_{ind}$ are almost equal and opposite). In the (supposedly small) gap between terminals almost all the field is due to the coulombian field.

The partition of the field is kind of trivially explained in the case of the mutual inductor: here the induced electric field - if we neglect loading of the 'primary' - is independent from the presence of the secondary coil. So, we will have a field whose circulation around the core will give us the EMF. Add to this the knowledge that by Ohm's law the resultant electric field $\vec{E}_{coul} + \vec{E}_{ind}$ is ideally zero (in a perfect conductor) and it follows that the line integral of $\vec{E}_{coul}$ inside the coil - gap excluded - has to be equal to almost the entire EMF. But the Coulombian field is conservative, so its circulation needs to be zero, hence the line integral of the gap must be equal in magnitude and opposite in sign to the its line integral inside the coil.

Now, why we know there must be charges along the coil surface and in particular on the ends facing in the gap?
I will give here a derivation for the case of a conductor carrying a current in steady-state (DC) or quasi-static conditions (AC at low enough frequency that won't show retardation effect across the entire circuit - and don't even think about radiation!)

I like circuits where KCL works, so I'd start with the continuity equation where the divergence of the current density is zero (closed current lines)

$$ \nabla \cdot \vec j = 0 \\ \nabla \cdot (\sigma \vec E) = 0 $$

The latter equation is clearly the result of the incorporation the local form of Ohm's law. Both the electric field and the conductivity can vary with the spatial coordinates, and we can expand the divergence using the product rule

$$ \sigma \nabla \cdot \vec E + \vec E \cdot \nabla \sigma = 0 $$

from which follows

$$ \nabla \cdot E = -\frac{\vec E}{\sigma}\cdot \nabla \sigma $$

We have another equation that tells us something about the divergence of the electric field. For reasons that will be immediately clear, it is best to use the form of Gauss' theorem that links the electric displacement to the charge density,

$$ \nabla \cdot (\epsilon_0 \epsilon_r \vec E) = \rho $$

Mind you, the relative permittivity will in general be dependent from the spatial coordinates, therefore by expanding the divergence as we did before, we get this expression for charge density :

$$ \rho = \epsilon_0 \epsilon_r \nabla \cdot \vec E + \epsilon_0 \vec E \cdot \nabla \epsilon_r $$

Now, let's plug in the expression of the divergence of E in terms of the conductivity we obtained above, and we get:

$$ \rho = - \epsilon_0 \epsilon_r \frac{\vec E}{\sigma} \cdot \nabla \sigma + \epsilon_0 \vec E \cdot \nabla \epsilon_r $$

Last but not least, if we substitute

$$ \vec E = \frac{\vec j}{\sigma} $$

we get the following expression for the charge density in terms of the current flowing in the circuit and the gradients in permittivity and conductivity of the material.

$$ \rho = - \frac{\epsilon_0 \epsilon_r}{\sigma ^ 2} \vec j \cdot \nabla \sigma + \frac{\epsilon_0}{\sigma} \vec j \cdot \nabla \epsilon_r $$

We can compact the formula by applying the product rule in reverse and by making use of $\epsilon = \epsilon_0 \epsilon_r$ to get:

$$ \rho = \vec j \cdot \nabla \frac{\epsilon}{\sigma} $$

This nice compact equation tells us that we must have a nonzero charge density $\rho$ wherever there are gradients in the permittivity $\epsilon$ and the conductivity $\sigma$ of the material carrying a current.

In steady-state and quasi-static conditions (where instant after instant we have a succession of steady state solutions), the current density is constant throughout the conductors, but the electrical properties present an abrupt change when we transition from the conductor to the air or vacuum around it, or to a different type of conductor (for example from copper to carbon powder inside a resistor). That's where the surface and interface charge will build up. It's usually a ridiculously small amount of charge, but classical electrodynamics requires it to be there.

For a different approach, you might want to have a look at how Sommerfeld solved the case of an infinitely long cylindrical conductor carrying a constant current I. It finds - making large use of the symmetry of the system - that the electric field has a discontinuity at the lateral surface of the conductor, thereby requiring the build up of a non-zero surface charge. It is example 17 ("Detailed treatment of the field of a straight wire and a coil") on page 125 in volume III ("Electrodynamics") of his theoretical physics course.

If you feel that a self inductor can be quantitatively different, in terms of the role of the surface charge, from a mutual inductor, I suggest you to solve your problem numerically and see what values you get for the charge distribution and contributes to voltage. Note, though, that a single coil will not allow you to consider its magnetic field fully confined within its boundaries; go for a long inductor, instead.

Answer 2

EDIT: My answer below is wrong!

In the comments, we have uncovered your true confusion:

The $\int\vec E\cdot\vec{\mathrm d\ell}$ following the inside of the inductor is NOT zero.

Go and watch Lewin's lecture here: https://youtu.be/nGQbA2jwkWI?t=325

If you are in a time-crunch, skip to 33:40 for the main event.