Faraday's law paradox during relative motion between conductor and magnet

Question

Faraday's law states that the induced EMF about a closed path is equal to the negative rate of change of magnetic flux enclosed by that path.

Consider the classic magnet and conductor ring problem.

If the ring is stationary and the magnet moves, there is an induced EMF caused by the electric field

However in the moving magnet's frame of reference, $B$ is constant and there is a moving ring that experiences a magnetic Lorentz force, causing an EMF.

Now here's where the problems start.

As I am currently aware, the Maxwell-Faraday law, that relates the electric field induced EMF, holds for any ring that I choose, even if there is no physical ring present.

Thus if I pick an "imaginary" ring instead of a physical conducting ring, then if a magnet were to move, there would be an induced electric field.

Now, if the ring I choose is not a physical ring, then in the magnet's frame of reference, there is a moving imaginary ring... there are no charged particles in this ring, thus there is no magnetic Lorentz force present, thus I would conclude that there is no EMF.

Isn't there a contradiction between the two scenarios? What is going on?

Answer 1

Your paradox is not quite as different as you think. I asked a very similar question early in 2020 under the title "Conductors and induced emfs: an inconsistency?".

For what it's worth, the conclusion I came to is this. We're prepared to define an electric field in terms of the force $\mathbf F =q\mathbf E$, that would be experienced by a stationary test charge, if one were present. I think that we must do the same for a magnetic field, defining it from $ \mathbf F = q\mathbf v \times \mathbf B$ in which $\mathbf F$ is the force that would be experienced by a charge $q$ moving at velocity $\mathbf v$, if one were there.

In this way we can talk about emfs of both kinds in non-material loops, because we imagine them being replaced by material loops with charge carriers in them.

I'm far from satisfied with this answer, but at least I'm showing solidarity.

Answer 2

If the ring I choose is not a physical ring, then in the magnets frame of reference, there is a moving imaginary ring... there are no charged particles in this ring, thus there is no magnetic lorentz force present, thus I would conclude that there is no EMF

You are correct that physically there is no motional EMF present and acting on current.

But we can define motional EMF here anyway, for a given moving loop, whether it corresponds to some real circuit or not (and is merely imaginary). The expression is $$ \mathscr{E}_{motional} = \oint_{loop} \mathbf v \times \mathbf B \cdot d\mathbf s, $$ where $\mathbf v$ is velocity of the loop element, and $d\mathbf s$ is infinitesimal vector of the loop element.

In case the loop is inside and follows a real circuit made of conductor, this fictitious EMF gives correctly value of the motional EMF experienced by the current in the circuit. If there is no real circuit, there is no real motional EMF present, but we can still talk about motional EMF that would be present if conductor was there.

So the situation is analogous to induced EMF, where we define induced EMF for a loop (whether it corresponds to a real conductor or not) as

$$ \mathscr{E}_{induced} = \oint_{loop} \mathbf E_i \cdot d\mathbf s, $$ where $\mathbf E_i$ is induced electric field.

The EMF in the first case is called motional EMF, as opposed to induced EMF, because it is not due to induced electric field, but due to motion of the conductor in magnetic field.

Answer 3

A quote from here¹:

...neither Maxwell's equations nor their solutions indicate an existence of causal links between electric and magnetic fields. Therefore, we must conclude that an electromagnetic field is a dual entity always having an electric and a magnetic component simultaneously created by their common sources: time-variable electric charges and currents.

In other words, at any point in space $\vec r$ and time $t$, both the electric and magnetic fields there, $\vec E(\vec r, t)$ and $\vec B(\vec r, t)$ can be entirely attributed to charges $q(\vec r',t_r)$ and currents $\vec j(\vec r', t_r)$, and their time derivatives on the past light cone, as defined by the retarded time:

$$ t_r = t-\frac{|\vec r -\vec r'|}c $$

The idea that the time derivative or curl of one is inducing a curl or time derivative on the other is not supported by causality, but it is a great way to design motors, generators, and various sensors.

Reference

1. Oleg D. Jefimenko, Causality Electromagnetic Induction and Gravitation, 2nd ed.: Electret Scientific (Star City - 2000) Chapter 1, Sec. 1-5, page 16 ISBN 0-917406-23-0. (pdf link)