Current: the speed of charges vs the number of charges past a point

[artis]

Current is determined as the amount of charge moving past a certain point of reference (like the point in a wire with current meter clamps around it) in 1 second.
This got me thinking , so in a classical resistive load situation with Ohm's law there would be two ways to increase the amount of charge flowing past a point, either to increase the voltage keeping the resistance fixed or decrease resistance keeping the voltage fixed.

But what about if we have a case where the speed of current past a point (or relative speed between current and measuring point is changed) we have a copper wire of certain length with fixed resistance and voltage and fixed DC current running through it.
Assume the wire is of some considerable length much like a transmission line. I put an ammeter clamp around the wire while standing still and the clamp meter registers a current of say 1A.

What would change if I put the ammeter clamps around the wire but I was traveling at the same speed that the electrons in the wire travel, would I still measure the 1A of current as when standing still? The same goes for traveling in the opposite direction of electrons?
The same question just in a different setting could be asked about a solenoid with a coil around a core,for this purpose assume the coil is a spiral type , like a corkscrew and it isn't wound back and forth with multiple layers like ordinary coils, again a fixed DC current flows through this coil but in addition to that the coil is rotated around it's longitudinal axis, does this increase the strength of the B field in the core as if more current would be flowing in an ordinary coil that doesn't rotate?

 

[anorlunda]

Your question is tricky. It leads to special relativity.

If you do the equations correctly, you'll even find that electric fields appear to be magnetic and magnetic fields appear to be electric depending on the speed.

You will also find that electron drift speeds are very slow. The important parameter is not the speed of electrons, but the speed of propagation of the EM field.

Electromagnetism can not be correctly visualized imagining electrons to be like marbles in a pipeline. If that's what you're trying to do, cut it out.

 

[Mister T]

Assume the wire is of some considerable length much like a transmission line. I put an ammeter clamp around the wire while standing still and the clamp meter registers a current of say 1A.

What would change if I put the ammeter clamps around the wire but I was traveling at the same speed that the electrons in the wire travel, would I still measure the 1A of current as when standing still?

Are the clamps moving with you or are they stationary relative to the wire? If the latter then the meter still reads 1.0 A. If you keep the clamps stationary relative to you then you wouldn't have to move very fast to get the meter to read zero.

 

[jartsa]

What would change if I put the ammeter clamps around the wire but I was traveling at the same speed that the electrons in the wire travel, would I still measure the 1A of current as when standing still? The same goes for traveling in the opposite direction of electrons?

When electrons started moving at speed v, the density of electrons did not change.

When electrons stopped moving and protons started moving at speed -v the density of protons increased. That is the difference.

The same question just in a different setting could be asked about a solenoid with a coil around a core,for this purpose assume the coil is a spiral type , like a corkscrew and it isn't wound back and forth with multiple layers like ordinary coils, again a fixed DC current flows through this coil but in addition to that the coil is rotated around it's longitudinal axis, does this increase the strength of the B field in the core as if more current would be flowing in an ordinary coil that doesn't rotate?

If there is any change of density, it's same for electrons and protons. But rotation of the coil can cause the protons to gain more speed than the electron lose, if the speed of the electrons is relativistic. That would mean more current.(Well in classical physics none of this happens)

 

[hutchphd]

Are the clamps moving with you or are they stationary relative to the wire? If the latter then the meter still reads 1.0 A. If you keep the clamps stationary relative to you then you wouldn't have to move very fast to get the meter to read zero.

If I have an uncharged wire and the electrons are moving relative to the protons, my (v<<c) frame of measurement will not change the current! If I am moving at the electron drift velocity, the protons will be "carrying the current" but it will still be 1 Amp. If we are at relativistic speeds then there are contraction issues that create charge imbalance.
Am I missing something here?

 

[artis]

@anorlunda thanks for the video very simple and nicely put, I had read this before , now one clarification, when I use points of reference here with respect to the ammeter for all intents and purposes assume that the ammeter itself is the point of reference and the "conscious" observer.

@hutchphd well so you say that if I the ammeter move along the wire in the direction of electron current and with the same speed I still get current reading from the wire but now due to protons? It seems to me this would go against what I know and what is also said in the video @anorlunda posted, because if my speed is the same as the electrons then I can't get any current measurement from the electrons, so then we are left with a bunch of protons that are seemingly closer together and are said to be "length contracted" but why would that register as current? I think it's the same as passing along a positively charged metal strip, you just feel the E field but there is no current, because a sign of current is magnetic field, but passing along a positively (or negatively for that matter) charged metal strip or wire produces no B field in the moving reference point but also in the stationary one. There is still attraction from the E field and also in the video it was said this to be the reason why parallel current carrying conductors attract.
Am I right or wrong with this?
Seems like @Mister T also said this in a much shorter way, ?
Well as for my rotating spiral coil around a core question, to which @jartsa replied, but here is the thing, my electron current is 1A with my coil being stationary and my ammeter being stationary around the coil (even though ammeters with current transformers don't work for DC, but for the sake of argument here let's suppose they do )
Now I start rotating my spiral shaped coil so that the wire moves in the same direction as the electron current in it, my ammeter is still held stationary, what the ammeter now registers is...? An increase in current ?
If it does register an increase in current then the B field through the coil core has also gone stronger ?

Isn't this much like having a train moving on a track with speed 100, and then suddenly the track/rails beneath the train also start moving with a speed of 100, so now a stationary observer would register the train moving at a speed of 200. So in 1 second twice as many railcars would pass the observer as when the train went at 100, so current has doubled?

 

[artis]

PS. one more thing about trains, a DC electric train moving from A to B. The overhead wire power supply is such that electron current also is in the direction from A to B.
The train starts traveling from A to B then stops at B and travels back to A. Would it then be the case that as the train was traveling from A to B it got less current available than when it was traveling from B to A? because traveling back to A from B the train was moving in the opposite direction than the electrons so it's pantograph/current collector was able to sweep up more current per the same speed and time?

 

[Ibix]

Am I missing something here?

Well - relativistic effects always apply. But you would certainly need a ridiculously precise ammeter to detect a difference at electron drift velocities, I agree. The ratio of currents measured in the moving frame compared to the rest frame is $\gamma$, which is approximately $1+v^2/2c^2$ ($v\ll c$) so at drift velocity would differ from 1 on around the fifteenth or sixteenth decimal place.

 

[Ibix]

Would it then be the case that as the train was traveling from A to B it got less current available than when it was traveling from B to A?

No. Current density is a four vector, and if it is $(0,j,0,0)$ in the wire frame then at velocities $\pm v$ it Lorentz transforms to $(\mp\gamma jv/c^2,\gamma j,0,0)$. So the sign on the charge of the wire flips but the magnitudes of the charge and current densities are the same.

Edit: note that if you consider only the electrons then the current density four vector in the wire frame for these is $(-\rho,j,0,0)$ and transforms to $(\gamma(-\rho\mp vj/c^2),\gamma(j\pm v\rho),0,0)$. You can consider the protons separately too, which have wire frame current density of $(\rho,0,0,0)$ and transforms to $(\gamma\rho,\mp\gamma v\rho,0,0)$. So the train does pass more/less electrons in one direction than the other, but ditto the protons. When you add the electron and proton current densities together to get the actual density then the direction dependent terms cancel each other out, leaving only the direction independent ones as in my first paragraph.

 

[jartsa]

The train starts traveling from A to B then stops at B and travels back to A. Would it then be the case that as the train was traveling from A to B it got less current available than when it was traveling from B to A?

Yes. (@Ibix I disagree with you.)

Train observer aims a telescope at the power plant, sees a robot putting electrons on a wire one by one. Those are the electrons that form the "available current".

When the observer changes his velocity he observes a change of the working rate of the robot. At the same time he notices that the available current changes accordingly. In the new frame the robot has always worked that fast.

So interestingly the speed of the electrons does not matter.
(Perhaps the train sees a current in the wire that is not an available current for the train)

 

[Ibix]

Those are the electrons that form the "available current".

You've forgotten the proton current, which is non-zero in the train frame.

 

[Nugatory]

The train starts traveling from A to B then stops at B and travels back to A. Would it then be the case that as the train was traveling from A to B it got less current available than when it was traveling from B to A?

Unless I’m misunderstanding what you’re asking, the current is going to be determined by the impedance of the train’s motor and the voltage difference between wire and ground.

 

[artis]

let's just forget the train analogy for now as there is already some confusion here without it.

@Ibix, unless I have something wrong why would there be a "proton current" if I as an observer travel along a wire at the same exact speed as the electrons travel? I would just see a net positive charged wire with an E field.there are still unanswered questions in my post #6 but let me try to rephrase, some posters here said that if my ammeter would travel along a DC current in the direction of the current at the same speed the ammeter would register no current, which seems kind of odd because the electron drift velocity is rather low.
So when the ammeter would register no current , when it would travel along a DC current wire at the speed of the electrons or at the speed of c through that particular medium? I wonder which one is it. I kind of feel it will be the light speed.

 

[hutchphd]

Are your questions about relativistic speeds or about common everyday speeds? You seem to be conflating them: yes relativity always is true but usually it is practically irrelevant. The "drift speed" of electrons is slower than a snail's pace. If we move with the electrons at any speed, the background positive charge will move in the opposite direction. As the speed approaches c we will additionally see "relativistic effects". There is no "either" here: both exist but one is usually very small.

 

[Nugatory]

he "drift speed" of electrons is slower than a snail's pace.

The drift velocity is indeed very small, but remarkably the relativistic effects are still significant - Purcell's first year E&M textbook uses them to derive the velocity-dependent transformation betwen electrical and magnetic fields around a current-carrying wire. Neglecting these effects may introduce errors of the same magnitude as the magnetic field itself.

 

[artis]

well in that case moving the ammeter with any practical speed doesn't result in any change of measured current both for a stationary observer looking at the meter or one sitting and "riding" along the meter.If this is the case then I still am curious about what happens to a magnetic field in a core surrounded by a coil that is shaped like a spiral and can be rotated, because current is the amount of charge past a given point in a given time, if electrons move at low pace then current past a given point should theoretically double by doubling the amount of electrons passing that said measuring point.
so if I have a cylindrical core with length 0.5m and a much longer corkscrew type coil then i can just rotate the coil so that it "screws" past the core and the core would behave as if there is double the current in the wire around it aka resulting in a stronger B field?

 

[hutchphd]

The drift velocity is indeed very small, but remarkably the relativistic effects are still significant - Purcell's first year E&M textbook uses them to derive the velocity-dependent transformation betwen electrical and magnetic fields around a current-carrying wire. Neglecting these effects may introduce errors of the same magnitude as the magnetic field itself.

Is this the same issue as ignoring the magnetic field of a moving point charge? I am not sure what the measurable "errors" are...seems to me they are still small (and quadratic in v/c), Can you be a little more specific?
In particular suppose I wish to know the force (per length) between parallel conductors as calculated in moving frames.

 

[Ibix]

@Ibix, unless I have something wrong why would there be a "proton current" if I as an observer travel along a wire at the same exact speed as the electrons travel?

Because in this frame the protons are moving past you. The electrons are (or may be, depending on your velocity) stationary.

Far and away the easiest thing to do is what I did in post #9 - write the current density (either the total as in my first paragraph or the electrons and protons treated separately as in my edit) as a four vector and Lorentz transform it. If, however, you want to go with an electron-counting argument as @jartsa does in #10 then you need to count protons passing you too.

I must say I assumed this was in the relativity forum, so I used relativistic tools without comment. However, I now see that it's in classical. Nevertheless, this whole topic is a lot easier with relativistic tools, and I recommend looking into them.

 

[Ibix]

So when the ammeter would register no current , when it would travel along a DC current wire at the speed of the electrons or at the speed of c through that particular medium? I wonder which one is it. I kind of feel it will be the light speed.

You can read that off from my post #9. Current density measured when you travel at speed $v$ along the wire is $\gamma j=j/\sqrt{1-v^2/c^2}$ which is never zero. In fact, your measured current density increases as you accelerate towards $c$ relative to the wire.

 

[Ibix]

if electrons move at low pace then current past a given point should theoretically double by doubling the amount of electrons passing that said measuring point.

Again, you cannot forget the protons in the wire. Any bulk movement of the electrons that you induce by moving the wire also induces the same bulk movement of protons, with an equal and opposite charge and hence an opposite electrical current. So you only get a tiny change in the total current.

 

[Ibix]

Is this the same issue as ignoring the magnetic field of a moving point charge?

Kind of. The microscopic cause of the magnetic field of a wire is the magnetic field of the many moving electrons, after all.

The problem emerges when you consider, for example, a free charge moving parallel to a current carrying wire. The charge feels a force $q\vec v\times\vec B$. Now (naively) transform to the rest frame of the charge and, since $\vec v=0$, the force vanishes! That's inconsistent, since the charge can't be accelerating in one inertial frame and not another. Presumably that kind of thing is what @Nugatory is referring to.

This can only be resolved by using relativity (edit: Einstein cites a similar scenario in the introduction to his 1905 paper as an open problem, before going on to close it decisively). Properly transforming the current in the wire shows that it acquires some charge in this frame, and an E field emerges which turns out to give you the force $q\vec E'$ that you need for a consistent picture in both frames. But where did the wire's charge come from? Purcell traces it to differing length contraction of the spaces between the initially moving electrons and the initially stationary protons - the electron density in the wire does not equal the proton density in this frame.

So that's why you need a relativistic analysis of electromagnetic fields even for absurdly low velocities.

 

[sophiecentaur]

PS. one more thing about trains, a DC electric train moving from A to B. The overhead wire power supply is such that electron current also is in the direction from A to B.
The train starts traveling from A to B then stops at B and travels back to A. Would it then be the case that as the train was traveling from A to B it got less current available than when it was traveling from B to A? because traveling back to A from B the train was moving in the opposite direction than the electrons so it's pantograph/current collector was able to sweep up more current per the same speed and time?

For every electron entering the locomotive via the overhead wire, there is an electron leaving via the ground or the other overhead wire so how would that affect your model? The motion of the train away from or towards the power source has the effect off increasing or decreasing the area of the 'loop' which will alter the magnetic flux and the Magnetic Energy. So the power supplied by the generator would be greater with the train traveling away (assuming you could actually measure it, of course).

 

[anorlunda]

because current is the amount of charge past a given point in a given time, if electrons move at low pace then current past a given point should theoretically double by doubling the amount of electrons passing that said measuring point.

I call BS on all this stuff about trying to analyze electricity by tracking electrons one at a time.

Visualizing electricity as similar to marbles moving in a pipeline is wrong. We should not be indulging it.

The definition of current as being the flow of charge per unit time is an average. It does not imply a column of electrons marching in single step.

@artis , please cut it out.

 

[Ibix]

The definition of current as being the flow of charge per unit time is an average. It does not imply a column of electrons marching in single step.

I agree that electrons don't move along in neat single file, but current is the flux of charge across a surface. Interpreting that as the count of individual charge carriers times the charge they carry isn't wrong, I think, although pushing electrons one at a time into the pipe probably is.

It seems to me that where the charge carrier count analysis is going wrong in this thread is that the surface you are considering needs to be at a fixed position. In a frame where the wire is moving that means that the wire's protons (and their charge) are crossing the surface too, and this is being forgotten. That's compounded by fun with relativistic velocity addition/length contraction, which leads to a changed current and a charged wire - which is why I said the easiest thing to do is learn what the current density four vector is and learn how to Lorentz transform it (it's just a matrix multiplication). Then you can worry about why that works in a microscopic picture, IMO.

 

[artis]

I feel I am being wrongly criticized here, @anorlunda nowhere did I say or speak about current as individual electrons moving past somewhere or marbles and pipes, I do realize the flaws in those analogies yet I never brought them up in the first place.
I indeed was talking about current "in bulk" as the flow of charge past a given point in a given time interval , when the amount of charge that passes that point changes we say current changes, I don't see anything wrong with this.

 

[artis]

@sophiecentaur I already said forget the train analogy as I realized myself it's not the best example.

Let me give this one specific example of the homopolar aka Faraday linear motor. Which we know as two conducting rails on which a perpendicular conducting rod is placed, then a DC potential is attached between both of the rails, the rod starts moving forward or backward depending on polarity.
I made a drawing of this model in the attached picture.
Now to make this clear at least for me I want to ask , assume the DC potential is capable of unlimited current and there is no voltage drop, also the rails and the rod that connects them have 0 resistance and the rails strecth away for as long as needed. So if there is no resistive voltage drop and no current/voltage limitation from the power source , when the rod has moved away from the power source for a considerable length is the current through the loop still the same as at start when the rod was right at the power source?

For a real world situation the further the rod moves the higher the resistance becomes and current decreases , but what happens in my example?

 

[sophiecentaur]

the wire's protons (and their charge) are crossing the surface too, and this is being forgotten

That is only true during a change in PD across the conductor when the lattice distorts slightly. But the mass of the nucleus, relative to its free electron, is huge so the displacement will be negligible. I have proposed (but not carried out) an experiment with a tray of mercury, carrying a hefty current in which the force on the nuclei should cause the liquid to lean towards the negative end of the tray. I would expect this to be a very small slope but it could be magnified using AC of an appropriate frequency to build up a longitudinal wave in the liquid. (A Q factor of several tens would be reasonable, I guess)

Movement of heavy +ions happens in electrolysis too.

 

[sophiecentaur]

So if there is no resistive voltage drop and no current/voltage limitation from the power source

It is probably not valid to impose conditions like that. It's very likely that the Energy that is put into the motion of the rod, appears as a Resistance like the Radiation Resistance that's measured at the feed point of a radiating RF antenna.

 

[Ibix]

That is only true during a change in PD across the conductor when the lattice distorts slightly.

I was talking about the rest frame of the electrons drifting along the wire. In this case the electrons are stationary and the protons are moving. In fact, the movement of protons is the source of the current in this frame.

 

[sophiecentaur]

I was talking about the rest frame of the electrons drifting along the wire. In this case the electrons are stationary and the protons are moving. In fact, the movement of protons is the source of the current in this frame.

Is there a lot of point in the 'frame game' here? The electrons have a range of velocities extending to fractions of c, I believe. The actual drift velocity of electrons in a circuit, once the current is known, will vary with the thickness of the wire (remember that horrid water analogy?). Any analysis like the one you suggest would have to deal with that - or to assume that the wire gauge was uniform. It seems a very hard way of looking at things.

 

[hutchphd]

The drift velocity is indeed very small, but remarkably the relativistic effects are still significant - Purcell's first year E&M textbook uses them to derive the velocity-dependent transformation betwen electrical and magnetic fields around a current-carrying wire. Neglecting these effects may introduce errors of the same magnitude as the magnetic field itself.

Yes. Consider two identical charged particles moving past each other in their CM frame. Their magnetic interaction force is smaller than the electric by a ratio of (v/c)² where v is their relative speed.
In a wire we have eliminated the strongest electric interaction because the conductor is not charged.
It is therefore not surprising the leading order correction to the electric field and the magnetic field itself are of equivalent magnitude.

 

[sophiecentaur]

I was well impressed by that alternative idea. The electrons in two current carrying wires ‘see’ a different density of + and - charges due to Relativistic effects for even a very slow mean velocity. The force is the same (calculation) as if you use Ampere’s law and magnetic fields.
So much for ‘what really happens’

 

[vanhees71]

I was talking about the rest frame of the electrons drifting along the wire. In this case the electrons are stationary and the protons are moving. In fact, the movement of protons is the source of the current in this frame.

I've just written an FAQ article about this case. I've to still proof-read it tomorrow once more. Then I'll put it up as an Insights article too:

https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf

 

[Ibix]

Is there a lot of point in the 'frame game' here?

The OP specifically asked (in #7 and again, more clearly, in #13) about the current measured by a moving ammeter. In that case, changing frames is the easiest way of answering. The answer is in #9. The discussion since has been about the various traps in naive electron counting and when and why you have to use relativity.

 

[jartsa]

So the wire looks c

You've forgotten the proton current, which is non-zero in the train frame.

Oh. The train moving in the same direction as the electrons sees a positively charged wire.

I guess both trains see the same current passing by then.

If we assume that the wire is neutral in its rest frame, then the train moving against the electron current sees a negatively charged wire, from which the electrons do not feel any particular need to leave. So this kind of wire can not be transmitting any energy anywhere.