My questions are related to understanding how emf can exist in circuit with zero resistivity.

I will always assume that all metallic wire are resistance-less.

First scenario: metallic loop

I use a perfect metallic wire and I make a loop with that. I can compute the electromotive force:

$$e=\oint \mathbf{E}.\mathbf{dl}=\iint \mathbf{\nabla} \wedge \mathbf{E}.\mathbf{dS}=-\frac{d}{dt}\iint \mathbf{B}.\mathbf{dS}=-\frac{d \Phi}{dt}$$ Where $\Phi$ is thus the magnetic flux through this loop.

Now, as I assumed my conductor being perfect, I have no electric field inside the metallic wire: $\mathbf{E}=\mathbf{0} \Rightarrow e=0$.

There is thus no emf in this particular case. As explained in some of the answers, if someone puts an magnetic field inducing a flux inside my loop, a current will be induced to create another flux exactly compensate this field such that in the end we will still have $\frac{d \Phi}{d t}=0$

Second scenario: inductor

Now, let me consider a circuit composed of an inductor (some coils) in serie with (for example) a current generator. I assume no external magnetic field has been put.

If my generator generates a fluctuating current, the flux will also vary as here :$\Phi=L I$. Thus, an emf will be created in my coils.

On the other hand, my wire are perfect, thus there is no electric field inside and no emf can be present in the coils because $e=\oint_{\text{coils}}\mathbf{E}.\mathbf{dl}=\oint_{\text{coils}}\mathbf{0}.\mathbf{dl}=0$.

How to precisely solve this confusion ?

  • $\begingroup$ Isn't it the total electric field (the non-conservative induced field plus the conservative field due to charge distribution) that must be zero? $\endgroup$ Apr 17, 2021 at 0:00
  • $\begingroup$ @AlfredCentauri I am not sure to see what you mean. Here E is indeed the total electric field inside the material. And I use maxwell equations on it which gives the relationship with magnetic field. $\endgroup$
    – StarBucK
    Apr 17, 2021 at 0:07
  • $\begingroup$ Related but not quite a duplicate: Can current be induced in a superconductor? $\endgroup$ Apr 17, 2021 at 12:35

5 Answers 5


In a perfect conductor, by definition, mobile charge within the conductor instantly redistributes in such a way as to make the electric field inside the conductor zero.

Generally speaking, for a perfect conductor, it's not that the induced electric field is zero (it isn't), it's that the mobile charge within the perfect conductor is distributed, at all times, such that the induced electric field is precisely cancelled by the electric field of the charge distribution. For example, an inductor made of perfect conductor has zero electric field within the conductor yet, can have a voltage across due to the induced emf separating charge to its ends.

In the case of a closed loop of perfect conductor, as another answer points out, the flux threading the loop must be constant, i.e., the right hand side of the equation must be zero. When there is an external, changing flux threading the loop, the current of mobile charges changes in just the way required to produce a changing flux that precisely cancels the external change.

Update to address the edited in second scenario:

Simplify the scenario as follows: you first have a closed loop of perfect conductor. You cut the conductor and insert an ideal current generator.

Before you cut the closed loop, you could find a closed path along the loop entirely within the conductor for the closed line integral of the electric field. Since the perfect conductor, by definition, has zero electric field inside, the closed line integral along this path must give zero which implies that the (total) magnetic flux threading the loop must be constant.

But after you cut the loop and connect the current generator, you can no longer find a closed path along the loop that is entirely within the perfect conductor. Thus, the fact that there is zero electric field within the conductor doesn't constrain the closed lined integral of the electric field to be zero, i.e., the emf associated with the closed path is not constrained to be zero.

Put another way, when the current produced by the current generator is changing, there will be a voltage across the generator that is due to the charge distribution.

What charge distribution you ask? As I mentioned earlier in my answer, the mobile charge within the perfect conductor will always be distributed such that the electric field due to the charge distribution precisely cancels (within the conductor) the induced electric field due to the changing flux threading the loop.

But outside the conductor, the conservative electric field (due to the charge distribution) doesn't generally cancel the non-conservative induced electric field and so, over that part of the path, there is non-zero electric field such that the closed line integral of the electric field is non-zero.

  • $\begingroup$ Thank you for your answer. But I still dont understand. If you agree that the total electric field vanishes inside the conductor, then the emf is actually vanishing as well as it is based on this total emf. If I would have to only take the induced electric field I could not relate it to the magnetic as I did because Maxwell equations adress the total fields. I hope it clarify my confusion $\endgroup$
    – StarBucK
    Apr 17, 2021 at 9:44
  • $\begingroup$ @StarBucK, I don't follow your reasoning, it appears wrong to me. $\endgroup$ Apr 17, 2021 at 11:56
  • $\begingroup$ My point is that if the total electric field vanishes inside the conductor, by definition of what emf is, the emf necesserally vanish as it is the integral of this electric field. For instance I then dont get your last paragraph. $\endgroup$
    – StarBucK
    Apr 17, 2021 at 11:58
  • $\begingroup$ @StarBucK, I'll try to work on that then. $\endgroup$ Apr 17, 2021 at 12:05
  • 1
    $\begingroup$ @StarBucK, I think you've got it. Just remember that emf, in this context, isn't located anywhere, it's a number associated to a closed path. But I think know what you're getting at in a circuits context where we think of, e.g., a battery as a source of emf (generated by a chemical process rather than induction). $\endgroup$ Apr 17, 2021 at 19:26

There are two possible answers to your question.

First, if you assume a hypothetical ideal conductor with no limits or non-ideal details. Then there is no limit to how much current can go through the wire and no need for an EMF to drive a current. So when the flux tries to change then there will be a current induced in the wire which, by Lenz’ law will oppose the change in flux. Because there can be no EMF, the current will be exactly the size to completely cancel the change in flux.

Second, if you assume a real superconductor then there is actually some resistance to AC even if it is superconducting at DC. This is because there are some Cooper pairs that get broken by thermal fluctuations. So for a real superconductor you get a little EMF in the wire and a large induced current, but not quite as large as with the ideal conductor.

  • $\begingroup$ Thank you for your answer. To be fair im a little bit confused because it seems to imply that you indeed need some resistance. But if I generalize the discussion to general inductor that I plug to circuit we never need to add this little resistance in the models. However I dont know what question to ask precisely to clarify this yet. Also it is a different explanation than the other provided so I am a little bit lost now $\endgroup$
    – StarBucK
    Apr 17, 2021 at 10:02
  • $\begingroup$ I don’t know why you would think some resistance is needed. The first section indeed deals with the theoretical no-resistance scenario. The second part describes a real superconductor where there is no DC resistance but some AC resistance. So AC resistance is present, but as the first section showed it is not necessary. Please re read the answer, you seemed to have misunderstood the first part. $\endgroup$
    – Dale
    Apr 17, 2021 at 10:58
  • $\begingroup$ Regarding “it is a different explanation than the other provided so I am a little bit lost now”. Indeed I wrote this answer specifically because I disagree with that answer. There is just the E field of Maxwell’s equations, not an E and e field. The same E goes into both Ohm’s law and Faraday’s law. $\endgroup$
    – Dale
    Apr 17, 2021 at 11:09
  • $\begingroup$ For your last comment. I agree with you. This is also what confused me in this other answer. $\endgroup$
    – StarBucK
    Apr 17, 2021 at 11:59
  • 2
    $\begingroup$ @StarBucK, I'll jump in here if it's OK. A ring of perfect conductor is a very different configuration from a coil of perfect conductor (ends open to be connected to a circuit or source). If a coil (of perfect conductor) is connected to an (ideal) AC current generator, the current through the coil is determined entirely by the generator. Dale's correct analysis above of a closed ring of perfect conductor cannot be applied to this configuration. The two configurations are not equivalent. $\endgroup$ Apr 17, 2021 at 13:24

Your confusion is caused because "EMF" has, similarly to "voltage", somewhat different meanings in different contexts. Some sources define total EMF for a path as integral of total electric field over that (possibly closed as in your question above) path. This sounds reasonable as total electric field is usually what drives mobile charges in real conductors, but unfortunately it is not how induced EMF is defined in the context of AC circuits.

When we talk about induced EMF in AC circuits, we really mean integral of induced electric field only. Induced field, in AC circuits, is only a part of the total electric field; there is also the Coulomb field due to electric charges distributed on the conductor surface and sources.

In case of a perfect core-less inductor with terminals A,B the induced EMF is defined based on induced field $\mathbf E_i$ only:

$$ emf_{AB} = \int_A^B \mathbf E_i \cdot d\mathbf s. $$

This can be expressed as $$ emf_{AB} = - L\frac{dI}{dt}. $$

One can't define induced EMF on the inductor based on total electric field and get this result. "Total field" EMF for a path is much lower than induced EMF for the same path, because the conductor counteracts the induced field with its own field. In a perfectly conducting wire segment, the Coulomb field and the induced field completely cancel each other.

The induced EMF can, however, be calculated using the Faraday law

$$ \oint_\gamma \mathbf E \cdot d\mathbf s = - \frac{d\Phi}{dt} $$ where $\gamma$ is a closed curve, $\mathbf E$ is total electric field and $\Phi$ is magnetic flux through that closed curve.

This is accomplished by completing the path that goes inside the conductor from terminal A to terminal B by a path from B to A that does not follow the coils (it can be straight line from B to A or it can be curved somewhat in order to avoid the conductor but must not follow the coils closely).

The left-hand side is approximately equal to emf as defined, because we can re-express it using the induced field and the Coulomb field. Contribution from the Coulomb field to the integral is zero, so the Faraday law can be also stated using induced field only:

$$ \oint_\gamma \mathbf E_i \cdot d\mathbf s = - \frac{d\Phi}{dt}. $$

Since contribution from the added path is negligible (the added path is short and not following the lines of induced electric field), this integral is approximately equal to emf as defined:

$$ emf_{AB} \approx \oint_\gamma \mathbf E_i \cdot d\mathbf s. $$

The right hand side can be expressed in terms of magnetic field inside or current that flows in the inductor. Magnetic flux $\Phi$ is approximately equal to magnetic field inside the inductor $B$ times the area of one coil $A$ times the number of coils in the inductor $n$:

$$ \Phi \approx n A B. $$

So we arrive at $$ emf_{AB} \approx -nA\frac{dB}{dt} $$ and this can be further re-expressed as $-L\frac{dI}{dt}$.

The case of a closed ring is different in that the Coulomb field, while it can be present, can't cancel the induced field everywhere in the conductor - the Coulomb field can't have non-zero circulation (line integral over closed loop). So total induced field itself has to be zero. How is that possible? Because even if external bodies produce their induced field $\mathbf E_{i,ext}$, the induced current distribution in the ring, howsoever weak, will produce counteracting induced electric field $\mathbf E_{i,ring}$ and these two cancel each other in the perfect conductor.

So in case of a ring, it doesn't matter whether we integrate the induced field or the total field over the closed loop. The result is the same - zero.

A ring made of perfect conductor will indeed have zero total electric field inside and thus both the first ("total field") and the second ("induced") kind of EMF is zero.

This does not mean electromagnetic induction does not happen, only that induced electric field due to external sources is exactly cancelled by induced electric field due to electric charges on the ring itself.

  • $\begingroup$ Thank you for your answer. I am not sure to understand however. If the emf is defined with only the induced field, then I don't know how you can relate it to the flux, and then get $emf=-L \frac{d I}{dt}$. To get to this equation you need to use maxwell faraday which is valid for the total electric field. $\endgroup$
    – StarBucK
    Apr 17, 2021 at 14:31
  • $\begingroup$ I've addressed this in comment here physics.stackexchange.com/questions/629305/… . In short, you can re-express the integral in definition of emf for path from A to B in conductor, as line integral of total field over a closed path that is a result of completion of the A->B path, the additional segment being any imaginary path that goes from B to A outside the conductor in region of space where induced field is negligible. $\endgroup$ Apr 17, 2021 at 20:13
  • $\begingroup$ While I give you a thumbs up for using the divergence-free, or as you put it "induced" E field, I disagree with your assertion that "In a perfectly conducting wire segment, the Coulomb field and the induced field completely cancel each other." This is only true if current density is constant, per [the London Equations] (en.wikipedia.org/wiki/London_equations) . $\endgroup$ Apr 23, 2021 at 2:49
  • 1
    $\begingroup$ @MathKeepsMeBusy We are talking about the simplest model of ideal conductor (zero Ohmic resistance, neglecting momentum/energy of the charge carriers), not about superconductors. In this model, total electric field in ideal conductor is zero. In London's model, momentum and energy of the charge carriers are not neglected, it is a more detailed model. Electric field in this model then isn't zero when current increases or decreases. But this is outside the scope of the question. $\endgroup$ Apr 23, 2021 at 11:09
  • $\begingroup$ Zero total field in perfect conductor (and neglect of electron kinetic energy/momentum) is a standard useful "lie" that helps us understand electrostatics and AC circuits of normal (non-perfect) conductors. It is insufficient only in more exotic cases like behaviour of superconductors where electron's mass and inertia have to be taken into account. $\endgroup$ Apr 23, 2021 at 11:13

The reason why electric field is zero inside a conductor is the existence of plenty mobile charges. After a short transition, they are displaced by the applied E field in order to neutralize it.

But we can not ignore that short transition when inducing an emf by an variable magnetic field. The very displacement of charges is the variable current that is proportional to the emf. $$V = L\frac{dI}{dt}$$

While we can assume zero ohmic resistance in a circuit, it is not possible to suppose zero inductance. What can be done is to represent the inductance in a part of the circuit, joined by no inductance wire.

It is like mechanical problems were pulleys or strings are supposed massless. It is valid since there is a mass somewhere, and their masses are negligible compared to it.

  • $\begingroup$ Thank you for your answer. However I am not sure to understand how it solves my problem. To say that the emf is equal to $L \frac{d I}{d t}$ you had to relate $\int \mathbf{E}.\mathbf{dl}$ to the magnetic flux through Faraday law. But if the electric field inside the conductor is zero, you would get a zero emf. Which is the thing I don't get. $\endgroup$
    – StarBucK
    Apr 17, 2021 at 17:00
  • $\begingroup$ "The reason why electric field is zero inside a conductor is the existence of plenty mobile charges. After a short transition, they are displaced by the applied E field in order to neutralize it." E fields are normally present in normal conductors conducting current. J=σE. $\endgroup$ Apr 23, 2021 at 13:24

The EMF induced by a magnetic field is not $\oint_{\text{coils}}\mathbf{E}.\mathbf{dl}$ but

$$ \mathscr E_{induced} = {\int_a^b}_C \vec{E_{rot}} \cdot d\vec{\ell}$$


  • $\vec{E_{rot}}$ is the divergence free (or rotational) component of the $\vec{E}$ field, and is a solution to the equation

$$\vec{\nabla} \times \vec{E_{rot}} = \frac{\partial\vec{B}}{\partial t}$$

  • the integral is taken from a point $a$ on the wire to a point $b$ on the wire along the wire path $C$

By Helmholtz's decomposition theorem, a well-behaved 3D vector field can be decomposed into two components, one curl-free and the other, divergence-free. The divergence-free component of the $E$ field, namely $E_{rot}$, may be non-zero in some places even when the total $E$ is 0 in those places.

How to precisely solve this confusion ?

Edit: This answer may be deleted soon, because I agree with the answer of @JánLalinský. I am leaving part of my original answer here for the moment to address concerns raised by @StarBucK.


You have in general $E=E_{rot}+ΔE$ where $ΔE$ is conservative. Then, to relate the emf to the magnetic flux (which you need to have the law $emf=\frac{LdI}{dt}, \Phi=LI$) you must integrate over a closed surface. In this case integrating with $E$ or $E_{rot}$ will not change the emf. It means that your integral with $E_{rot}$ will vanish as well.

Per my understanding of how the term "perfect conductor" is being used, is that it is shorthand for taking the limit of an Ohmic conductor, as the conductivity increases to infinity, or as the resistivity decreases to 0.

The simple answer is that a real inductor has both an induced EMF based upon $\frac{\partial \Phi}{\partial t}$ and the number of turns, and it also has a resistance. The EMF doesn't change as the resitance is changed, and, since we are taking a limit, the EMF is still present as the resistance goes to 0. If we should run into a mathematical problem such as dividing by 0, we shall ignore that, because a "perfect conductor" is really just an approximation of a conductor with a really, really, small resistance, i.e. a limit.

That is not to say that one actually does run into such a mathematical problem. You say one "must integrate over a closed surface". I'm not sure what you mean. With the formula

$$ \mathscr E_{induced} = {\int_a^b}_C \vec{E_{rot}} \cdot d\vec{\ell}$$

one is integrating over a line, not a surface. Perhaps you mean one must integrate over a closed path, but that is only for a conductor which forms a closed loop or ring. If there is a matghematical difficulty, I can, at the moment, only forsee it occurring in that case. Give me some time, or if you want to expand the explanationi of your concern it may help.

  • $\begingroup$ About your second point, I am not sure to agree. You have in general $E=E_{\text{rot}}+\Delta E$ where $\Delta E$ is conservative. Then, to relate the emf to the magnetic flux (which you need to have the law $emf=L dI/dt$, $\Phi=LI$) you must integrate over a closed surface. In this case integrating with $E$ or $E_{\text{rot}}$ will not change the emf. It means that your integral with $E_{\text{rot}}$ will vanish as well. $\endgroup$
    – StarBucK
    Apr 23, 2021 at 8:47
  • $\begingroup$ @StarBucK Did my edit explain things more clearly for you? $\endgroup$ Apr 24, 2021 at 22:55

Your Answer

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

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