As far as I know,

  1. Electric and magnetic field inside a perfect conductor plane are zero (it can be proved by combination of Ohm's law and Faraday's law of induction).
  2. When there is a magnetic field, eddy currents must exist inside a superconductor to make the net magnetic field zero inside the plane.
  3. The eddy current flows in nonlinear paths so a force must be applied to the charges to make this movement.
  4. Lorentz force formula includes all possible forces to an electric charge ($F=qE+qvB$)
  5. In our case, $E$ and $B$ are zero, so the Lorentz force must be zero.

So which force is applied to the electric charges and causes their movement (eddy current) in a perfect conductor plane?


1 Answer 1


It is still the electromagnetic (Lorentz) force. A couple of your assumptions are wrong.

First, in a superconductor no force is required for a steady current to exist, even a circular or more complicated "bent" current. You should not think of the electrons forming a superconducting current as little classical point particles racing around and accelerating as they turn. They are not spatially localized very strongly and they come in pairs which "balance" each other out. There is no force involved in a steady superconducting current.

Second, magnetic fields do penetrate a short way (the London depth) into a superconductor. This is called the skin effect in normal conductors, and it is slightly different for a superconductor than a normal conductor but it still exists.

So, in short, your assumptions 1 and 3 are not correct. There is a magnetic field down to the London depth, and no continuous force is needed to maintain a steady current in a superconductor. The magnetic field down to the London depth allows the surface currents to form, and once formed they need no further force to continue.

  • $\begingroup$ Thank you. But even if magnetic field penetrates into the perfect conductor, in a situation which a magnet levitates above it, the field doesn't change so it can not make an electric field to cause an electric force; Also the force can not be qvB because v is zero. You mentioned that the presence of a force is not necessary but in the curved paths I am not still convinced. $\endgroup$ Sep 11, 2021 at 6:50
  • $\begingroup$ The DC resistance of a superconductor is 0, this by definition means that no E field is required to maintain a current. I don’t know how you can be “not still convinced” when that is the definition of a superconductor. If any E field were required to maintain a steady current then the resistance would be non zero. You are thinking of the electrons as little classical point particles, but they are quantum mechanical. They don’t follow classical rules. If they did then superconductivity could not exist. Until you can let go of your classical preconceptions you will not understand superconductors $\endgroup$
    – Dale
    Sep 11, 2021 at 11:12
  • $\begingroup$ I apologize if I am making mistake about the matter. Let me explain my problem better in another way. It is definitely a quantum phenomenon but if some parts of my following reasoning contradict with quantum principles please let me know. The electric charges of a superconductor which is in a magnetic field, move in nonlinear paths. (at least some of them don't move linearly). Having no resistance does not mean that for nonlinear movement we don't need force. It means there is no need to do work or pay energy for movement and normal force does not do work. Accelerated motion requires force. $\endgroup$ Sep 12, 2021 at 10:58
  • $\begingroup$ I am just repeating myself here now. As I already said you are thinking of the electrons as little classical point particles, with a definite location and velocity. They don’t follow classical rules. They don’t have a definite location and velocity that needs to accelerate. Instead they have a state where they exist as a Cooper pair composite boson with a long range interaction with lattice phonons that is not localized. They are not little classical balls zipping around and accelerating. Sorry, I cannot teach Quantum Mechanics in comments $\endgroup$
    – Dale
    Sep 12, 2021 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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