If I have a magnetic field and I place a coil that is a perfect conductor in this field that I'm rotating.

This would mean that there is a change in the flux through the area of the coil which would induce an emf in the coil.

But shouldn't the electric field inside a perfect conductor be zero?

How is this interpreted?


6 Answers 6


There are no real perfect conductors, but if one were to exist, the electric field would indeed be zero. Currents would flow in the wire to cancel out the change in the magnetic field (the currents produce their own magnetic field which counteracts the change in the one you are producing).

  • $\begingroup$ I essentially came to this analysis in my answer, but it is important to note that the electric field only becomes zero because of the magnetic field the moving charges produce in response to the applied magnetic field. $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 22:44
  • 3
    $\begingroup$ A superconductor is a real perfect conductor (up to some current limit), and current flows cancelling out the magnetic field is exactly what happens - it is called the "Meissner effect". $\endgroup$
    – Richold
    Commented Feb 7, 2021 at 1:38
  • $\begingroup$ But induced electric field forms closed loops .But how can the electrostatic field set up the the rearranging electrons form closed loops to cancel out the induced field? $\endgroup$ Commented Jul 14, 2023 at 2:14

The electric field inside the perfect conductor is zero. This is why, regarding the situation described in OP the physics books usually speak not of inducing electric field, but of inducing an electromotive force. EMF is more abstract concept that does not really imply existence of an electric field, although the OP seems to confuse it with the voltage bias created by the battery.

A less formal approach to understanding such idealized situations is thinking about a real conductor, which has finite resistance and non-zero electric field inside. It may be more difficult to describe mathematically, but it facilitates things in terms of mental reasoning.

Much of the confusion related to this question is about interpreting the meaning of the terms like perfect conductor and electromotive firce (emf). Like many terms in physics these are clearly defined concepts, rather than the things that their names suggest ("work" is the classical example of what I mean). Although clearly defined in physics books, these definitions are often overlooked due to the suggestive nature of the names. Specifically:

  • A perfect conductor is assumed to have infinite amount of highly mobile charges, so that it immediately screens any electric field in it. In other words, the electric field inside such a conductor is zero by definition. In this sense superconductors are not perfect conductors, but rather the materials with zero resistance.
  • Electromotive force is an action, frequently understood as an amount of work. It does not imply existence of an electric field.
  • $\begingroup$ This does not seem right. Faraday's law still holds. No e.m.f. is induced. $\endgroup$
    – Tony
    Commented Feb 5, 2021 at 10:08
  • 1
    $\begingroup$ @Tony This is how it is presented in elementary physics textbooks, e.f., Holliday&Resnik. And this is probably the only way to reconcile the concept of a perfect conductor with Faraday's law. $\endgroup$
    – Roger V.
    Commented Feb 5, 2021 at 10:12
  • $\begingroup$ @Vadim I tried to look into your second suggestion in my answer. Let me know if you think I got the analysis right. $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 22:46
  • $\begingroup$ @Tony isn't e.m.f induced because of Faraday's law? In any case, electrons have to change their average speed in order to cancel the flux going inside (to maintain zero electric field), and to change average speed implies an acceleration, which implies some initial electric field penetration. You need the electric field to speed up the electrons. (In Lorentz force terminology, a magnetic field only curves electron paths, but never increases their kinetic energy.) $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 22:49
  • 1
    $\begingroup$ @JonathanJeffrey I think it is unfair that your answer received so many downvotes. $\endgroup$
    – Roger V.
    Commented Feb 7, 2021 at 6:44

Note: I initially considered the case of a changing magnetic field instead of a rotating conductor. (I guess the phrase "field that I'm rotating" in the initial question was unclear.)

If you're spinning the conductor instead, I still believe much of the intuition can be gleaned from this, as long as you limit to low angular accelerations, at least.

To be slightly more accurate for the case of the spinning conductor, you should consider the Lorentz force and back out the response of the electrons from that. I believe you will find the motion of the electrons relative to the loop to be the same as my discussion below.

Let's look at the Maxwell–Faraday equation $$\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}.$$

A changing magnetic field produces a swirling electric field that can accelerate charges around a loop. Let look at this in the cases of a normal conductor, a "perfect conductor" (I mean this loosely, meaning in the limit of a normal conductor approaching infinite conductance), and a superconductor.

Normal conductor

In a conductor with finite conductance (a "normal conductor"), we already know the story that the a little bit of the electric field can penetrate inside and also induce a drift current in the bulk electrons. (I've made multiple answers on this sort of topic, so you can see some of my recent answers for intuition in this area.)

"Perfect" conductor

In general, we like to say a perfect conductor doesn't allow electric field lines in. This is a little inaccurate, first of all, because there are no perfect "normal" conductors, and second of all, because this catch-all explanation misses the important physics of why conductors typically cancel much of the incoming field, and better conductors cancel more of the field.

Let's first take a step back.

What cancels (either all or most of) the electric field in conductors?

Main Process: Charge Build-Up

For most circuits, it all has to do with charge build-up somewhere in the material. In a conductor, charges move in response to whatever electric field penetrates, which usually often leads to an electrostatic case of essentially all the electric field being canceled in the conductor, even for the case of a normal conductor.

But a characteristic of this situation is that there is "an electromotive force," which is fancy word for saying "we are consistently changing something about our situation that is not letting these charges settle in the way they normally do."

For example, in a circuit, we have a battery. Charges move to attempt to cancel the electric field the battery produces, but each the charges accumulate on the battery to try to alter the voltage on the terminals, the battery simply plucks that charge and puts it on the other terminal, sustaining the motion. Thus in this case, there ends up being a steady-state of nonzero current in a conductor, and nonzero electric field in the conductor due to needing to overcoming the resistance of the conductor to sustain the current.

For the case of a battery, in the limit that that conductor is perfect, in order to completely cancel out the field, the charges still need to accumulate somewhere in the circuit, e.g. on a resistor. Now the electric field in the conductor will be zero, but charges will still flow, keeping the momentum they initially had from being accelerated the instant the field wasn't zero in the near-perfect conductor.

What cancels (part of) a changing magnetic field in conductors?

Main Process: Eddy currents:

For this particular case, for a loop exposed to a changing magnetic field, there will actually turn out to be another important effect: eddy currents. These occur because of Faraday's law: charges in a changing magnetic field want to swirl. They swirl locally in eddy currents then tend to counteract the magnetic field, and thus prevent some field lines from entering.

However, the case is precisely the same as before, in the sense that any resistance present prevents electrons from swirling fast enough to stop all entering field lines. So the external, changing magnetic field penetrates some distance inside, in any case, although this penetration depth may become very small as the conductance approaches infinity.

Applying to the original scenario

@Orpheus was asking about a coil formed by a perfect conductor in a changing magnetic field. Let us simplify to discuss a simple loop.

The answer is simply: eddy currents in response to the changing magnetic field keep the changing external magnetic field relatively close to the surface, but it still penetrates some distance that decreases as conductance increases. The impinging, changing magnetic field still has a total EMF around the loop, and there is a swirling electric field (confined more and more tightly to the surface in the limit of high conductance) that accelerates charges around the loop. This acceleration around the loop ends when the external EMF balances with the "reverse EMF" produced by accelerating electrons producing their own flux, but for finite conductance, the acceleration will end sooner because the resistance will help slow the electrons down.

You can also think of the net flow around the loop as one big eddy current, if you want. Now, let me explain some finer points about the previous paragraph in a few more words.

Since there is no resistor in the loop, there is no place for charges to build up to counteract the loop EMF from Faraday's law in the charge-build-up way. So the charges accelerate. This produces an magnetic field that curls around the loop, a back-action on the applied flux. But this flux also obeys Faraday's law, and produces a reverse EMF. Thus, electrons will continue to accelerate until the reverse EMF from this produced electric field equals the applied EMF.

The question is, does that equality ever occur? I might try to go back and do some calculations, but, in any case, this is already a bit of a trick question, because perfect conductors don't exist, so in fact charges don't continue accelerating forever, and instead the drag force from the non-zero resistance adds with the self EMF to balances out the applied EMF. However, in this case, we are using a drag force to balance out the EMF. This drag force is a phenomenological model we use because we don't want to model to microscropic electric fields in a material. If we still want to use our space-averaged electric field $\mathbf{E}$, we better keep this drag force separate from Maxwell's laws, and just include it as another force.

Thus, the electric field penetrates even a very good conductor with finite conductance, because before the self EMF stops the charge accelerations, self EMF + resistance does. But, as stated before, this penetration is confined tightly to the surface due to eddy currents, so the current is almost all on the surface as well.

Now we get to the fun part: superconductors.


Now, what is interesting to me is understanding how this works in the closest thing to a real-world perfect conductor, which are superconductors. I don't know much about superconductors, which have other properties like the trapping of magnetic flux lines inside of itself. Thankfully, how a superconducting loop responds to a applied magnetic flux has already been answered here on Physics StackExchange, for the case of a superconducting loop.

To paraphrase @Alfred Centauri's answer, the magnetic flux through a superconducting loop never changes; but to sustain this constant magnetic flux requires the current inside the loop to perfectly counteract any flux you try to push though. Because superconductors can only sustain a certain amount of current before becoming a normal conductor, this implies a high enough magnetic field will rupture superconductivity in the loop.

Note I make no mention of where in the superconductor this current is flowing, because I don't know much about superconductors. However, you should note that the case of superconductors already matches the intuition we gained from thinking about a limit of a perfect conductor: in order to stop accelerating, charges have to move fast enough to cancel the magnetic field.


Thus, to summarize:

  • In a normal conductor with fixed loop geometry, current in a loop flows only while the magnetic flux through the loop (from an external magnetic field) is changing, and dies after this applied magnetic flux stops changing, because the electrons will dissipate their energy due to the resistance of a loop. Eddy currents prevent this some of the external magnetic field from entering the conductor.
  • In a normal conductor with very, very high conductance, a penetrating electric field accelerates the charges, but the self EMF of the electrons is almost enough to cancel all of the electric field to stop the acceleration. Surface eddy currents prevent almost all magnetic field from entering the material. Because of this, the the loop current remains tightly confined to the surface, and can be thought of as a surface current. But because we rely on resistance a little bit to slow the electrons, some electric field must penetrate a shallow penetration depth.
  • In a superconductor, such flowing charges perfectly counteract the applied magnetic flux, and thus the flux through the superconducting loop is constant.

In at least in the first two cases, the electric field penetrates the loop a little bit; and we can use the second case to reason about the third. There is also a similar notion of penetration depth for superconductors, which may or may not match this intuitive notion of penetration.Also, note for a lot of the time when I talking about "stopping the acceleration," I mean for the case of a constant derivative of applied flux.

Edit note: My original explanation did not include eddy currents, which I realized are important in this situation. In this scenario, these are what keep the impinging magnetic field (and thus the electric field) confined to near the surface.

  • $\begingroup$ Could those downvoting me actually explain what they believe is wrong with my analysis? $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 22:50
  • 1
    $\begingroup$ do realise tht the equatons in the accepted anser - for emf is derive from maxwells eqation? $\endgroup$
    – user276724
    Commented Feb 6, 2021 at 22:50
  • $\begingroup$ I understand that, but they emphasized Gauss's law in the second part of the answer, and used Gauss's law incorrectly. The coil (or even a loop) does not have a simple geometry that allows Gauss's law to be used effectively here, as far as I can see. $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 22:53
  • $\begingroup$ The equation you use is the differential form of the integral Maxwell equation used to get the equation I had for the EMF. You're answer is getting down voted because you fail to see this connection, and thus your entire answer is based on an incorrect assumption. Your contention that the Maxwell equation I used (Gauss's law) is not applicable is also incorrect. $\endgroup$
    – joseph h
    Commented Feb 6, 2021 at 23:29
  • 1
    $\begingroup$ Dude, I can probably literally point you to the pages of Griffiths that explains that getting zero for the Gaussian integral does not imply zero electric field in the region. To be clear, this is because charges outside the Gaussian surface cam source field lines that pass in and out of the surface, canceling in the integral. $\endgroup$
    – prolyx
    Commented Feb 7, 2021 at 0:45

Consider how you would do the experiment. To determine the EMF, you must connect a voltmeter to the coil. A voltmeter works by sensing an electric field in its sense element. The electric field in the conductor is zero, but the sense element must have an electric field to function. That's where the EMF induces an electric field.


This concept has been covered many times (see here, here, etc). But the answers provided here are a bit unsatisfactory, and I thought I'd try to explain some of the confusion and also provide an explanation of the phenomenon briefly.

Short answer: The electric field inside an ideal conductor is zero (sometimes by definition). The magnetic field is the only component of the electromagnetic force present, so it is responsible for any and all currents in the conductor. However, by the nature of a perfect conductor (and free charges in general) the effects will be limited to the surface. Note that, once the current is started, no force is required to maintain it, meaning the current will continue to flow indefinitely unless acted upon by an outside force (hey Newton).

Source of Confusion

  1. The definition of voltage and EMF
  2. The definition of conductivity
  3. Surface conditions of a general interface, and of a dielectric and an ideal conductor in particular

Let's take these in turn. Electromotive force (EMF) is a concept which helps extend the concept of electrostatic potential (voltage) to cases where there is a changing magnetic field. When a changing magnetic field is present, voltage is no longer uniquely defined (see my question here for details). But we can include the magnetic field's current-generating nature to define an equivalent force which obeys Ohm's law, hence why we can calculate the current flow in a circular conducting ring, where it does not make sense to talk about voltage between two points. This is the EMF, which to be clear, is often a magnetic effect, not an electric effect. But see my question that I linked, where I put this on solid footing.

One of the major issues here is the connection between the general form of Maxwell's equations, and the macroscopic world of materials. A conductor is not a simple system of charges in free space, its behavior is modeled on average, a macroscopic model of microscopic phenomena. There is no simple way to derive the concept of conductivity $\sigma$ from Maxwell's equations alone (it would require a model of electromagnetic conduction which itself requires quantum electrodynamics). That being said, we can describe the effects.

A conductor is a macroscopic material which has free electric charges constrained to the bounds of the material. These charges move under the influence of the electromagnetic force, which means they mutually repel, they can be acted upon by outside forces, etc. Because the charges are free, it is possible to have movement of the charges as a group, equivalent to conduction of a charge density in a vacuum, but constrained to the material. Experimental evidence has shown that, given a material, the ratio of the electric field to the current density is constant, expressed by the equation: $$ \mathbf{J} = \sigma \mathbf{E} $$ Where $$ \mathbf{J} = \rho \mathbf{v} $$ For charge volume density $\rho$ and average velocity of the charges $\mathbf{v}$. Note that $\mathbf{E}$ and $\mathbf{J}$ are in the same direction.

The value of this model is that we can substitute $\mathbf{J}$ terms in Maxwell's equations with $\sigma\mathbf{E}$, making analysis of conductive materials a bit easier. In differential (point) form, Maxwell's equations can now be written:

$$ \begin{matrix} \nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0}\rho & \nabla \cdot \mathbf{B} = 0 \\ \nabla \times \mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t} & \nabla \times \mathbf{B} = \mu_0 \sigma \mathbf{E} + \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} \end{matrix} $$ Using Maxwell's equations, it's possible to define interface conditions which must hold when two dissimilar materials share an interface. In the general case, these are: $$ \begin{matrix} \hat{\mathbf{n}} \times (\mathbf{E}_1 - \mathbf{E}_2) = 0 & \hat{\mathbf{n}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_s \\ \hat{\mathbf{n}} \cdot (\mathbf{D}_1 - \mathbf{D}_2) = \rho_s & \hat{\mathbf{n}} \cdot (\mathbf{B}_1 - \mathbf{B}_2) = 0 \end{matrix} $$ Where $\mathbf{E}_1$ is the electric field at the surface in material 1, and $\mathbf{E}_2$ is the electric field at the surface in material 2, and other terms are defined similarly. The term $\mathbf{J}_s$ is the surface current, and $\rho_s$ is the surface charge density. The unit vector $\hat{\mathbf{n}}$ is a normal unit vector at the surface.

The surface current density is equivalent to the tangential electric field scaled by the conductivity. But we find our big issue when we try to take the limit $\sigma \rightarrow \infty$, and we try to apply Ohm's law. For if we assume (as we have) $$ \mathbf{J} = \sigma \mathbf{E} $$ Then as $\sigma \rightarrow \infty$, and if $\mathbf{E} \neq 0$, then the current density goes to infinity. But this is in violent disagreement with physical reality, unless $\mathbf{E} \rightarrow 0$. But in such a case, what becomes of $\mathbf{J}$? It is not valid to assume a particular value of $\mathbf{J}$ from this relation alone, we must instead use Maxwell's equations above with $\mathbf{J}$ substituted back in for $\sigma \mathbf{E}$.

We require at a minimum that $\mathbf{E} \rightarrow 0$, and so we have: $$ \begin{matrix} 0 = \frac{1}{\epsilon_0}\rho & \nabla \cdot \mathbf{B} = 0 \\ 0 = -\frac{\partial\mathbf{B}}{\partial t} & \nabla \times \mathbf{B} = \mu_0 \mathbf{J} \end{matrix} $$ Thus the relation $$ \nabla \times \mathbf{B} = \mu_0 \mathbf{J} $$ Is the only one that defines the magnitude of the current $\mathbf{J}$. Moreover, it implies that a current density can exist without the presence of an electric field, if it is induced by the rotational components of the magnetic field.

Returning to the interface conditions, setting $\mathbf{E}_1$ and $\mathbf{D}_1$ to zero (but not $\mathbf{E}_2$ and $\mathbf{D}_2$) the interface relations become: $$ \begin{matrix} \hat{\mathbf{n}} \times (-\mathbf{E}_2) = 0 & \hat{\mathbf{n}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_s \\ \hat{\mathbf{n}} \cdot (-\mathbf{D}_2) = \rho_s & \hat{\mathbf{n}} \cdot (\mathbf{B}_1 - \mathbf{B}_2) = 0 \end{matrix} $$ In the case where there is no magnetic source within the perfect conductor, then, the only magnetic field which can be present is the normal component (which much be continuous across the interface), while the tangential component (which is discontinuous when surface currents exist) is still somewhat undetermined.

This is actually an interesting question, is the magnetic field inside an ideal conductor zero? It's not quite possible to answer on strictly classical grounds as we've described above, i.e. from Maxwell's equations and Ohm's law alone. The resolution to the problem classically often uses a concept of least energy, which would have both electric and magnetic fields zero to minimize energy (see: the second link I provided in the first sentence).

Anyway, the magnetic field just outside the ideal conductor can still induce surface currents, hence you can have currents "in" (or rather "on") a perfect conductor, even in the absence of an electric field. You see from the above considerations why the electric field being zero is not a problem, and hopefully you can appreciate the subtleties of this sort of problem and how to approach questions like this.


Yes. The electric field in a conductor is zero but because the coil is being rotated in a magnetic field, there is a change in magnetic flux and the emf that’s generated is given by

$$\epsilon = - \frac{d \phi}{dt}$$

where $\phi$ is the magnetic flux.

Since the magnetic field is assumed to be constant (and obviously the area of the loop is constant) the angle between the magnetic field and the loop is constantly changing.

Now you ask, well if there is a current (due to emf) in the coil, then there must be an electric field to move the charges, so how is the electric field zero in the wire conductor?

The answer is, since the net charge in the wire is always zero, then therefore by Gauss’s law, the electric field is zero. That is

$$\int_S \vec E \cdot \vec dS \propto Q$$

where $Q$ is the net charge enclosed by the surface $S$, such that if $Q$ is zero, then so to is $\vec E$.

As an example, consider a wire that is not connected to a voltage source. It is obviously electrically neutral. If we now connected it to a battery, no additional electrons will be added to the circuit. But the electrons are in motion, and the negative charge moving in one direction “corresponds” to opposite charge moving in the other direction, and the net charge is still zero. Thus the electric field is still zero.

Also note that this law refers the electric field due to charge enclosed in a specific region. While you have zero net charge inside the conductor, this does not mean that there cannot be a surface charge density - charged particles on the surface of the wire - which can moved along the wire's surface creating a current.

  • $\begingroup$ I don't think this is right. No e.m.f. is generated. If it were, then Faraday's law would imply an electric field, and there can't be one in a perfect conductor. $\endgroup$
    – Tony
    Commented Feb 5, 2021 at 10:09
  • $\begingroup$ There is a current in the coil because an emf is induced in the coil since there’s a changing magnetic flux. This is consistent with Maxwell’s equation/Faraday’s law. $\endgroup$
    – joseph h
    Commented Feb 5, 2021 at 10:18
  • $\begingroup$ The application of Gauss's law is completely wrong. You can only use Gauss's law to imply the electric field is zero in cases with certain symmetries, and there is no such symmetry guarantee for an arbitrary coil. You also do not answer the OP's question of what causes the electrons to move: "emf" does not accelerate electrons directly; "emf" is a way to think about electric and magnetic fields that do accelerate electrons. Also, surface charges on a conductor are perfectly capable of moving to carry the current with response to the electric field produced by the varying magnetic flux. $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 21:18
  • $\begingroup$ @josephh I'm not sure the fix you made fixes the fundamental problem, and I'm not sure how Gauss's law helps here. You can e.g. have a polarized dielectric object that is neutral but had non-zero electric field inside due to the surface charges, and the Gauss's law argument (at least without modification) does not explain how this differs from a conductor with surface charges. $\endgroup$
    – prolyx
    Commented Feb 6, 2021 at 22:57
  • 1
    $\begingroup$ You can only use Gauss's law to imply the electric field is zero in cases with certain symmetries Yes this is true and there is nothing wrong with it's application here. Why do you think we cannot apply it to a wire (with cylindrical symmetry)? And you are severely mistaken about EMF. EMF is short for "electro-motive force". That is, a force that moves electrons. $\endgroup$
    – user280085
    Commented Feb 6, 2021 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.