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?

@ Jensen Paull But in your question (perhaps after editing), "magnet's" means "of the magnet". The apostrophe is being used correctly to form the possessive case of "magnet". I wonder if you are thinking of the special rule for "its and "it's". The rule is that "its" needs no apostrophe if it means "belonging to it". [Technically "its" is a possessive adjective and not the (non-existent) possessive case of the pronoun "it".] But if you want to abbreviate "it is", then you write "it's". "Magnet's" can also mean "magnet is" but this is rather informal and would be unusual usage on this site. — Philip Wood, Dec 15 '21 at 18:18

score 6 · Accepted Answer · edited Dec 13 '21 at 15:09

6

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.

edited Dec 13 '21 at 15:09

Nihar Karve

8,445

answered Dec 13 '21 at 14:41

Philip Wood

35,641

1

Yeah I can see what you mean, the fields the result from either scenario would produce the same effect GIVEN charges WERE there, which is ultimately what Maxwell's equations are useful for. EMF is work done PER unit charge, thus i would assume that Hypothetically if charges were there, there WOULD be in EMF so defining emf even the presence of no charges IS usefull,, But in the case where there are no charges EMF * 0 = 0 work. – jensen paull Dec 13 '21 at 15:13
@Nihar Karve Thank you, Nihar, for doing the link. How did you do it? – Philip Wood Dec 13 '21 at 15:14
in the case where there is a changing magnetic field, there IS an electric EMF, but no charges thus the work done would be zero, Likewise for defining a "imaginary" magnetic emf, there are no charges so the work done in BOTH cases will be the same, Which is ultimately the quantity EMF is trying to describe – jensen paull Dec 13 '21 at 15:15
@Jensen Paull Note that in my question I quoted from Page and Adams, who disagree(d) completely. – Philip Wood Dec 13 '21 at 15:17
I Would completely dissagree with page and adams then. Defining EMF in the absence of charges and current IS NECCESSARY for the consistancy of faradays law. This also doesnt violate anything as the actual WORK done would be the same for both cases, given a particle Were there. which is what EMF is describing Emf = Work/charge Emf * charge = work, Emf * 0 = 0, defining Emf to be zero would be the same physical result, but allowing the "Emf" to exist " without a charge is just more self consistent. that is neccessary for faradays law to hold. – jensen paull Dec 13 '21 at 15:25
I would also argue that it doesnt matter either way, as motional EMF isnt even apart of the 4 maxwells equations, So solving for the fields are the same regardless on how you define Motional EMF. The "motional emf" part comes into play when using the fields with the lorentz force law to find forces on PARTICLES ( you can use lorentz force law to also split up faradays law into (Motional Emf + faraday-maxwell law) – jensen paull Dec 13 '21 at 15:28
@ Jensen Paull Good! And Ján Lalinský seems to agree – which is also good. – Philip Wood Dec 13 '21 at 15:45
for the link there are 2 ways, firstly you can use the syntax like this "Title to be displayed" or secondly on the box where you are typing, you can just click the "hairpin" image toward the right side of the option for bolding and italicising. I think the rest would be easy to figure out – Adil Mohammed Dec 13 '21 at 15:55
@ Adil Mohammed Many thanks for trying. I don't understand about the hairpin image: on my Mac I just use command B and command I to embolden and italicise. – Philip Wood Dec 13 '21 at 16:12

score 6 · Answer 2 · answered Dec 13 '21 at 15:31

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 expression here anyway, for a given moving loop, whether it corresponds to some real circuit or not. 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 the motional EMF experienced by the circuit. If there is no real circuit, there is no real motional EMF present, but 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 loop or circuital conductor in magnetic field.

At first glance, The LHS of 1st equation looked "Emotional". :D — Nilay Ghosh, Dec 17 '21 at 08:40

score 1 · Answer 3 · edited Dec 17 '21 at 10:29

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

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)

I would also like to mention that jefimenkos equation is a PARTICULAR solution to the inhomogenous wave equation, Such that the initital condition sets the homogenous solution to be zero. Infact the general solution to maxwells equation is jefimenkos equation + the homogenous solution. which means the E and B fields are not uniquely determined the the charges and current. but by the charges and currents, and the initial conditions of the universe itself. maxwells equations allow for a background EM field to be present. which is amazing — jensen paull, Dec 13 '21 at 16:28

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

3 Answers3