[alancalverd]

By the movement of ions. In the case of alternating current, ions of both polarities move, on average, equally in both directions so any associated magnetic field alternates in sign at any point, with a net average of zero.

[hamdani yusuf (OP)]

By the movement of ions. In the case of alternating current, ions of both polarities move, on average, equally in both directions so any associated magnetic field alternates in sign at any point, with a net average of zero.

In the case if metal wire, if the positively charged particles (crystal lattice) and negatively charged particles (free electrons) move at the same speed but opposite direction, the force exerted to the stationary charged test particle is not zero. What makes you think it would be different for ions?

[hamdani yusuf (OP)]

Let me simplify the question. A long straight metal wire moves to the right at speed 1 m/s relative to the lab. Electrons in the wire move to the left 1 m/s relative to the lab. A positively charged test particle is stationary relative to the lab, 1 cm below the moving wire. Will it experience a force by the wire? Which way?

A few seconds later, the motions of the wire as well as the electrons in it have their directions reversed. Will the force to the test particle change?

[alancalverd]

In the case if metal wire, if the positively charged particles (crystal lattice) and negatively charged particles (free electrons) move at the same speed but opposite direction,

Obviously, they don't. The nuclei oscillate a bit (thermal energy) and the conduction electrons drift. We do see cases of electrostriction but AFAIK wires don't distort along their axis when carrying a current.

[hamdani yusuf (OP)]

In the case if metal wire, if the positively charged particles (crystal lattice) and negatively charged particles (free electrons) move at the same speed but opposite direction,

Obviously, they don't. The nuclei oscillate a bit (thermal energy) and the conduction electrons drift. We do see cases of electrostriction but AFAIK wires don't distort along their axis when carrying a current.

Do you think it doesn't make a difference if the wire moves or not?

[alancalverd]

A difference to what? Moving in what direction?

The propagation speed of current in a wire is about 10⁷ m/s, and the drift speed of the electrons is around 10^-5 m/s in a potential gradient of 1 V/m.

[Eternal Student]

Hi.

Let me simplify the question. A long straight metal wire moves to the right at speed 1 m/s relative to the lab. Electrons in the wire move to the left 1 m/s relative to the lab. A positively charged test particle is stationary relative to the lab, 1 cm below the moving wire. Will it experience a force by the wire? Which way?

   Yes, it should do.  It will be a force pushing the charged test particle directly away from the wire.

   Why?   Translate to the more typical frame of reference.    So we'll have the wire stationary, so we can assume the positively charged atoms of it are stationary.   Adjust the velocities of everything else appropriately.
    The free electrons of the metal now move with velocity 2 m/s to the left   ( my left direction is    < ---   ).   Actually slightly less if we use the the velocity addition of special relativity but  2 m/s will be OK.
     By conventional electromagnetism stuff, we have a magnetic field around the wire.   Conventional current is going  to the right  ( ---> that way ),   right hand rule shows the B field circles around the wire.
    The test charge, in this frame, is now moving at  1 m/s   (  <-- that way), or more importantly in the oppositie direction to the current.   A moving charge does feel force in a magnetic field,  we use the other right hand rule   for  F  = q  v x B.    We see the test charge is accelerated downwards,  i.e. directly away from the wire   (if I twisted my hands around in all the right directions).

    In the atypical frame where your situation applies   ( the wire moves ---> that way),   the same force and overall effect should appear.   However, it doesn't have to appear as a magnetic force and in this case it will be partly a force due to an electric field.
    We could draw the diagrams and do the maths - to show that due to relativisitic effects,  the average distance between charges has altered and the wire is no longer of overall neutral charge as far your test particle is concerned.

    The smart person would then ask a follow up question:   Aren't the electrons and positively charged ions affected equally?   Aren't the distances between +ve charges and between -ve charges equally contracted and the charge does remain overall neutral?    Indeed, in your original situation where the test paticle was at rest, it does seem that both the positively charged ions and the electrons are moving with equal speed, albeit in opposing directions.
    That's a good question.   The answer tends to be that the free electrons are not like the positively charged atoms in the metal wire.   The free electrons are, as in their name, "free",  while the metal atoms are much more locked in place in the crytsal lattice of the overall structure.
      An approximation is made here.   Here's the step-by-step reasoning:
1.   It does seem that when a current is passed along a wire, the free electrons do start drifting. 
2.   So there would be a corresponding length contraction for the distance between them, which would make the density of negative charges exceed the density of positive charges in the lab frame.
3.    However, the electrons are free to move.   This is the important approximation step.    By electrostatics, they will re-distribute themselves and the wire remains overall neutral in the lab frame, to a good approximation.    Note that the metal atoms could NOT do this, they are not like the free electrons.   We know that a static test charge placed next to this wire which has a current in it, doesn't seem to experience any force, that's the experimental evidence that we've got.  So it seems that the wire has remained overall neutral in charge, even when a current is made to pass through it.     

    One consequence of this is that the wire cannot be of overall neutral charge in other frames of reference.   In other frames, there must be length contraction as usual.   So, the easiest one to consider is where we move to a frame where the electrons are stationary and the metal atoms were the things moving with (the negative or direction reversal of)  the electron drift velocity.   In this frame, the electrons are not moving, while they were in the first frame, so the average distance between electrons is now bigger,   -ve charge density is therefore lower.   Meanwhile, the metal atoms are now moving whereas they were stationary, so the average distance between them has decreased,  hence positive charge density has increased.   So overall,  the positive charge density now exceeds the -ve charge density and the wire has a net positive charge.

    So the irritating and relentless person may then ask - but the electrons are still "free" in this new frame, why don't they just re-distribute themselves again?   Why isn't a current carrying wire of overall neutral charge in any arbitrary frame I wish to use?
One half of that we can answer well:   Since the electrons do drift when there's a current in the wire, it is just not possible for the wire to be of overall neutral charge in every frame.   Do the maths carefully, there is only one frame where that can apply.    The second half... well I don't have a satisfactory answer for it.... 
    For some reason the frame of reference where the wire is overall neutrally charged is the usual one, the one where the wire (the positive metal atoms in it rather than the free electrons) is stationary.  You can try and dig up a few reasons, for example a current just does not flow unless there's a cell (a battery) or something that is maintaining a potential difference across the wire.  You can attempt to argue that as part of that setting up procedure,  just the right number of electrons are initially pulled out of the wire at the +ve terminal of this battery,  so the slight over-density of -ve charges that should appear in the wire when the electrons do start drifting along does not emerge.  However, we can't make too much speculation.   Let's just leave it as  "for some reason" the frame where the current carrying wire is overall neutrally charged is the one where the wire is stationary.

Best Wishes.

[hamdani yusuf (OP)]

In the case if metal wire, if the positively charged particles (crystal lattice) and negatively charged particles (free electrons) move at the same speed but opposite direction,

Obviously, they don't. The nuclei oscillate a bit (thermal energy) and the conduction electrons drift. We do see cases of electrostriction but AFAIK wires don't distort along their axis when carrying a current.

For simplicity, let's take only average velocities for positively charged particles, and average velocities for negatively charged particles.

[hamdani yusuf (OP)]

In the atypical frame where your situation applies   ( the wire moves ---> that way),   the same force and overall effect should appear.   However, it doesn't have to appear as a magnetic force and in this case it will be partly a force due to an electric field.

In the reference frame where the test particle is stationary, v is 0. Thus special theory of relativity interpretes that the force is purely electric.

[alancalverd]

For simplicity, let's take only average velocities for positively charged particles, and average velocities for negatively charged particles.

With an alternating current, the average drift velocity is zero for both. And you will note from reply #405 above that the drift velocity is irrelevant to the induced magnetic field.

[paul cotter]

Hi ES, I have a problem with your proposition on #406 and I cannot figure out a resolution. Whether the wire, which is carrying a current, moves right or left the only effect it will have on the B field will be on it's magnitude, ie the B field will look exactly the same to the charged particle as it did in a static case although it's magnitude will be different. As we know a static B field will have no effect on a static charge. Yet if we change frames of reference as you have done we get a force - what am I missing? PS: my brain is not in gear, the "boss", my eldest daughter and myself are all currently virused, nothing significant but cognitively impaired.   Late addition: The B field remains normal to the page, regardless of the lateral movement of the wire so there is no differential movement between the charge and the B field. Looked at from the frame of the moving wire the test charge and the B field will move in unison.

[Eternal Student]

Hi.

Sorry to hear you're poorly @paul cotter .

As we know a static B field will have no effect on a static charge.

   Yes.   Stationary test charge implies no force.

Yet if we change frames of reference as you have done we get a force - what am I missing?

    Well, the "static" test charge won't be static (stationary) in both frames.   It's the motion of the test charge that undergoes an important change when you switch frames.   The B field doesn't need to change, the test particle now has motion.

    Just to be clear,  @hamdani yusuf   was discussing a situation with this sort of diagram:



      ------------------------------------------------------------ this is the wire


                                   *                this is the test charge

The test charge is something outside the wire,  we can move the wire while leaving the test charge static in the lab frame.    Of course, the moment you switch frames, so that the wire is stationary, then you see the test charge will be moving.       

Well that's a short answer anyway.
----------

I've read what you've written a couple of times now.   It seems that you think it would be important to know if a B field is moving or not.   However, it is NOT important to know.

You said:

Looked at from the frame of the moving wire the test charge and the B field will move in unison.

It's not meaningful to talk about the B field as if it can have motion and therefore move in unison with the test particle.   All that matters is that the test charge has motion in this frame.
 
The Magnetic force law (Lorentz force law where E=0 ) is often mis-interpreted in some fashion similar to the following:
   There's a force when a charge moves through a magnetic field   OR  when the magnetic field moves past the charge.

But that's just pop-sci and sloppy speech.   The Lorentz force law (or just Magnetic force law) actually says nothing at all about a magnetic field moving past a static charge because it's not even clear if a magnetic field is some sort of "stuff" that can move through space.   All we need is that the magnetic field, B, is something that has a value (this one has magnitude and direction) at every point in space and at every time.

     Much as you seem to suggest, quite often we just couldn't tell if the B field was moving or staying still.   For convenience we'll align our straight wire with the x-axis, then the current carrying wire has a B field that depends only on y and z co-ordinates of a point in space.   At any time, the B field at x=a is no different to the B field at x=b.   We can't tell if the B field at x=b,t=now   has just come from position x=a,  a second ago, or if the B field just wasn't moving.
   Fortunately, the Lorentz force law does not ask about how the test charge was moving relative to the B field.  It just asks how the test charge is moving   (the implication being this is relative to the frame you have chosen).

    Read perhaps just the first section of the Wikipedia article about the Lorentz force law to yourself:   https://en.wikipedia.org/wiki/Lorentz_force
    This time pay carefull attention to the LACK of something:   The lack of any need to know about the movement of the B field.

      The Lorentz force law states that a particle of charge q moving with a velocity v in an electric field E and a magnetic field B experiences a force (in SI units) of    F = q ( E + v x B ) .

    The velocity, v, that appears in these formulae is never stated as being the velocity of the test charge relative to the B field.   It is just the velocity of the test charge  (in whatever frame you have chosen to use).   Specifically, we start by picking a frame of reference,  then the test particle has a velocity v in that frame.... that's the only v we're interested in.... we don't care if the B field is something that might also be moving, it doesn't matter, we don't ever need its velocity. 

Best Wishes.

[paul cotter]

Hi ES, thank you but I am not really feeling poorly, just sort of thick-headed or stoopid!. That was a really stupid mistake to add that "unison" concept as I know and fully accept ALL you have said about the B field and have accepted such for decades. I think one of the best examples of the non-moving B field is, in my opinion, the homopolar generator with the magnet glued to the disk. All that said, I still have two scenarios with conflicting real world outcomes in mind. I will have a further "internal" discussion and get back to you. Thanks again and season's greetings to you and your family in the case I do not return to this over the next few days.  PS: I must learn not to make assertions when in a brain fog state!

[paul cotter]

Hi ES, I have thought about this question further and I now attempt to explain the apparent discrepancy, hopefully less ham-fisted than the first attempt. (1) in the frame of the moving wire we have a magnetic field due to the current and as the charge is moving in this frame it also produces a magnetic field and hence there is a force. (2) in the frame of the charge let us assume first the wire is not moving: then we get a B field with no effect on charge. Now move the wire along the x axis: now we have a B field again with a change in magnitude due to the change in effective current only and in this scenario I see no force on the charge- any form of charge moving along the x axis will produce a B field with no effect on the charge. I can only assume a relativistic effect of the movement of + and- charges in the conductor produces an electric field with force on the charge(WAG- wild ass guess). I am not happy when I cannot reconcile two perfectly valid approaches.

[Eternal Student]

Hi.

    I think I can see what you ( @paul cotter ) are saying and yes, you do seem to have the right idea.

I can only assume a relativistic effect of the movement of + and- charges in the conductor produces an electric field with force on the charge

Yes.
In post #406 I said the following:

....the same force and overall effect should appear.   However, it doesn't have to appear as a magnetic force and in this case it will be partly a force due to an electric field....

  In this situation described by @hamdani yusuf ,  the velocity of the test particle became 0 exactly and so we have the most extreme case.    Using the Lorentz force law and putting v=0 we would have that   F = q(E + v x B)  =  qE  =  a force entirely due to an Electric field.
This was mentioned in reply #408 by Hamdani:

In the reference frame where the test particle is stationary, v is 0. Thus special theory of relativity interpretes that the force is purely electric.

     In this frame of reference, the distance between the -ve charges (the free electrons) and between the +ve charges (the metal atoms in the lattice) has been affected (by length contraction under SR) such that the wire had an overall +ve charge of exactly the right magnitude to repel the positive test charge as required.   Although a B field would still be there circling the wire, it plays no part in exerting a force on the particle (in this frame).

---------
Minor notes:    You said the following (with red text added by me):

and as the (test)  charge is moving in this frame it also produces a magnetic field...

.
    Yes that's actually completely true  BUT  you are over-thinking this. 
     We were just considering it as a test charge.   They do not create or alter the field(s) around them, they just experience the field(s) created by other things    https://en.wikipedia.org/wiki/Test_particle .       
     In the real world, any moving charge would create a magnetic field in the space around it and this would combine with the magnetic field from the wire  -  but  we're talking about 1 small charge  (maybe just a proton) compared to billions  (~10²⁴ atoms per cm³)  of free electrons and metal ions in the wire.   The alteration in the magnetic (or electric) field caused by our little postive charge is utterly insignificant and we can consider it as an idealised test charge.
   
Best Wishes.

[hamdani yusuf (OP)]

For simplicity, let's take only average velocities for positively charged particles, and average velocities for negatively charged particles.

With an alternating current, the average drift velocity is zero for both. And you will note from reply #405 above that the drift velocity is irrelevant to the induced magnetic field.

The Lorentz force is not simply determined by the average velocity.
F= B. q. v
You need to multiply the velocity with the charges before averaging them.

[alancalverd]

So you might expect some ionic stratification within a liquid electrolyte in a magnetic field if you apply a direct current through the liquid. But if you reverse the current, you will reverse the stratification, so the net effect in your AC experiment is zero.

[hamdani yusuf (OP)]

So you might expect some ionic stratification within a liquid electrolyte in a magnetic field if you apply a direct current through the liquid. But if you reverse the current, you will reverse the stratification, so the net effect in your AC experiment is zero.

What do you mean by stratification?

The third table shows the force experienced by test particle, which is simply the multiplication of each cell in both tables above.
   v+   -4   -3   -2   -1    0    1    2     3     4
v-                              
-4       0    -3   -4   -3    0    5   12   21   32
-3       4     0   -2   -2    0    4   10   18   28
-2       8     3    0   -1    0    3     8   15   24
-1      12    6    2    0    0    2     6   12   20
0       16    9    4    1    0    1     4     9   16
1       20   12   6    2    0    0     2     6   12
2       24   15   8    3    0   -1     0    3     8
3       28   18   10   4   0   -2    -2    0     4
4       32   21   12   5   0   -3    -4   -3     0

For simplicity, let's just consider the anti-diagonal in the table above, which referred to the symmetrical case where positive charges are moving in the opposite direction but equal in magnitude as the negative charges. The numbers there are all positive, as long as v is not zero.

[alancalverd]

Now reverse the voltage gradient and add the two matrices. If you don't get a null matrix, you have made a mistake!

[hamdani yusuf (OP)]

Now reverse the voltage gradient and add the two matrices. If you don't get a null matrix, you have made a mistake!

Why should I do that? Do you know about Lorentz force?
The tables said nothing about voltage gradient.