[varsigma (OP)]

What does anyone here think of the idea, commonly found, that electrons flow along wires and that's what you pay for, when the bill is due?

I learned that electrons barely move in a typical low frequency circuit; that power supply for your computer is not a thing that "delivers" electrons to the motherboard. The electrons in the motherboard and in all those chips and other discrete devices, are already there. So what happens?

Apparently what happens is all outside of the wires where the fields have free space and all the charges appear on the surface; so in a semiconductor this surface effect and external fields must also be the case. It's all about changes in the fields, some are internal but most of the action is in free space.

Electron flow and induction explain how transformers work, but there is no flow of electrons between two windings. Induction is a field effect.

[alancalverd]

There is movement of electrons, transmitted at the speed of light in the medium, but the net flow (drift) is very slow. This from Wikipeda

Assume a current I = 1 ampere, and a wire of 2 mm diameter (radius = 0.001 m). This wire has a cross sectional area A of π ? (0.001 m)2 = 3.14?10−6 m2 = 3.14 mm2. The charge of one electron is q = −1.6?10−19 C. The drift velocity therefore can be calculated as.....
.....2.3 x 10^-5 m/s

If the difference is difficult to comprehend, imagine pushing a stick into soft tar. The far end of the stick moves almost as soon as you start to push (depending on the speed of sound in the stick) , but the rate at which the stick moves through the tar is very slow.

[varsigma (OP)]

There is movement of electrons, transmitted at the speed of light in the medium, but the net flow (drift) is very slow.

You mean the movement is transmitted, not the electrons?

Another reason the electrons don't move much is, they don't have to, because electrons pack a lot of energy into a small volume.

[Eternal Student]

Hi.

What does anyone here think of the idea, commonly found, that electrons flow along wires and that's what you pay for, when the bill is due?

    Advice:   If you plan on making a significant change in topic from what you started in this thread, then it may be best to start a new thread.   I'm not staff and it doesn't bother me,  it's just that you might get more replies and/or more relevant replies if people know they don't have to read the last 11 pages to join the current discussion.

    Anyway, in answer to that question:    It's a good enough idea for school level physics.   For higher level physics it's understood that E and B fields cause a flow of energy in the direction of the Poynting vector.  So the energy is not really flowing along the wires.   As @Alancaverd mentioned, some electrons are slowly moving along the wires but you aren't getting super-energy electrons delivered and low-energy tired old electrons taken away at the other side or anything like that.   Energy is flowing out to space from the battery and then in from space to the rest of the wires and the device that is consuming power,   the flow of electrons just provides some magnetic field to make  the Poynting vector  E x B   do what is needed in the right places.
    None-the-less a model or conceptualisation based on electrons* charge carriers flowing and somehow carrying the energy with them, dumping this energy into the power drawing device and returning to the battery to be "re-energised" again,  is still a good enough model to explain a lot of things.

(LATE EDITING:   *postive charge carriers and conventional current may be discussed instead of electrons).

    There are several PopSci videos on YouTube explaining the situation.    Veritasium did a video about a year ago called "Energy doesn't flow in wires" which most people will quote but I think this video (below) actually beat him to it by 2 years.

"Circuit Energy doesn't flow the way you think",  Science Asylum,   (duration under 8 minutes).

Best Wishes.

[alancalverd]

You mean the movement is transmitted, not the electrons?

effectively, yes

Another reason the electrons don't move much is, they don't have to, because electrons pack a lot of energy into a small volume.

Er, no. There are just an awful lot of free electrons in a conductor! Current is the quantity of charge passing through a plane per unit time.

 In 1 m³ of copper, there are about 8.5x10²⁸ atoms. Copper has one free electron per atom, so n is equal to 8.5x10²⁸ electrons per cubic metre.

[varsigma (OP)]

In 1 m3 of copper, there are about 8.5x1028 atoms. Copper has one free electron per atom, so n is equal to 8.5x1028 electrons per cubic metre.

I was trying to make the point that a small fraction of that number is what moves under an applied electric field.
So the energies aren't because a lot of electrons are moving, it's because electrons are charged. You only have to disturb a small number to see a proportionally large current. Or see a spark jump across a gap.

And you have Fermi levels involved in the dynamics. The applied electric field changes the Fermi levels, but around defects in a metal lattice electron flow is . . . different.

[alancalverd]

I was trying to make the point that a small fraction of that number is what moves under an applied electric field.

No, all the conduction electrons  move. Same as my "poking tar with a stick" example.

[varsigma (OP)]

No, all the conduction electrons  move. Same as my "poking tar with a stick" example.

All the electrons move, but only a fraction move in the same direction at the drift velocity. This is much lower than the random velocities, with no net flow.

[alancalverd]

If there is no net flow, why do we measure a current? Why does the battery discharge? Why is there a magnetic field around the conductors? All these phenomena are associated with unidiredtional movement of charge.

[varsigma (OP)]

If there is no net flow, why do we measure a current?

The "no net flow" condition is when there is no applied field (no voltage or current source). An applied field gives some of the valence electrons a net flow, with a corresponding flow of holes in the opposite direction. The bulk of the electrons remains dynamically thermal though, so I say there is no net flow for those.

I guess at the fundamental level a picture is needed of how electrons "flow" in conductors. Or in semiconductors.

[alancalverd]

In the absence of an applied voltage, there is no drift velocity. You keep changing the question if the answer doesn't agree with your preconception!

Valence electrons are irrelevant: we are talking about electrons in the conduction band. The picture is simple - just think of the stick model, or if that's too complicated, imagine a crowd leaving a stadium. When the final whistle blows they all move. The ones nearest the gate leave immediately, the number leaving the stadium per unit time (the current) depends on the ratio of the width of the gate to the width of a person, and the drift velocity of those inside the stadium may be very slow indeed.

There is no necessary reverse flow of holes. The Hall effect shows that the moving entities are EITHER electrons (in most metals and all n-type semiconductors) OR holes (some metals, all p-type semiconductors).

[varsigma (OP)]

The picture is simple - just think of the stick model, or if that's too complicated, imagine a crowd leaving a stadium. When the final whistle blows they all move. The ones nearest the gate leave immediately, the number leaving the stadium per unit time (the current) depends on the ratio of the width of the gate to the width of a person, and the drift velocity of those inside the stadium may be very slow indeed.

Ok. There's always a problem with analogies but I get the gist of what you say. The big difference is when electrons do this, the changes in the fields move too, at the speed of light, and most of the field is outside the conductor. In free space.

It isn't the flow of electrons or holes, although that's what explains semiconductor v-i characteristics, but the flow of energy in the field around the wire. In semiconductors you have junctions between p and n type crystals in say a diode. There's a region--the space charge layer--which is part of the lattice. There isn't any real benefit in analysing the fields outside the solid state, but you could.

[alancalverd]

The field outside the wire is mostly irrelevant to the transfer of charge. We use bi-filar winding or twisted pairs to cancel the external field when making precision resistors or AC power transfer cabling, and the current flows just the same. That said, an inductive delay line can be used to make a temperature-stable sinusoidal oscillator.

Not sure what you mean by changes in the external field. The external field around a straight wire at distance r is always μi/2πr for a current i, regardless of the drift velocity of the carriers.

[varsigma (OP)]

Not sure what you mean by changes in the external field.

I mean in the sense electromagnetic signals travel at c.
Whether you ignore the fields external to the conductor or not, they're still there.

[alancalverd]

Electromagnetic signals are only generated when charges accelerate. A d.c. flow of electrons produces a static magnetic field around the conductor but no external electric field.

You can't actually measure the speed of propagation of a magnetic or electric field because varying one would generate the other, and produice a selfpropagating electromagnetic pulse.

[Eternal Student]

Hi.

    While I agree with a fair amount of what you've said @alancalverd ,  I don't think you are being entirely accurate and consequentially you may be dismissing a few ideas or comments from @varsigma that actually aren't entirely wrong.   I'll address one of your points first because you probably won't be too offended. 
    @varsigma, I don't think you're getting or presenting the arguments based energy transfer with E and B fields in the best way.   I think that @alancalverd has correctly found some faults.
    The main thing is:   I'm no expert or final arbitrator.  It's a forum, this is a discussion, that's all.   
   Now, Alancalverd seems to be denying E and B fields and their importance in delivering power in electrical circuits, that's something I might be able to change with some discussion.

A d.c. flow of electrons produces a static magnetic field around the conductor but no external electric field.

    That's only a simplified truth.   A more complete truth is that there will also be a static electric field around the conductor.   
     Wires have some small resistance.  As such a small electrical potential gradient must be established along them in order to maintain a steady d.c. flow.
   How is that done?  It is thought that charges re-distribute themselves slightly, such that when you take a cross-section through the wire you would have some surface charges on that conducting wire.    At steady state, the surface charges are going to be positive (abbreviated +ve) and fairly dense near the +ve end of the battery,  becoming less dense as you move away from the +ve battery terminal toward the power drawing component of the circuit.   After reaching approximately 0 surface charge about half-way around the circuit, the surface charge will now start to become -ve as you move toward the -ve end of the battery.
   With this arrangement of surface charge, an electrical field is maintained inside the wire or equivalently an electrical potential gradient exists along the wire.

     
    Image taken from  "Projects and Practices in Physics",   https://msuperl.org/wikis/pcubed/doku.php?id=184_notes:resistors,    mainly because they display a "non-commercial share-alike license" on their website so I don't think they'll mind me using the diagram.
   A gradient of  surface charges is shown in the circuit,  with the steepest gradient existing over the resistor (the narrow piece of wire).

   Anyway, this distribution of surface charges means that there will inevitably be an electrical field outside the wire in addition to there being an Electrical field inside the wire in the direction of the conventional current.

    I'm keeping this post short so won't say much more now.   To start with, we just need to establish that electric fields do exist OUTSIDE of a current carrying wire.   It's already commonly known we will have magnetic fields outside the wire, so in another post we'd be ready to start looking Poynting vectors  E x B both inside and outside of the wire.

1.  There is work done by Jefimenko (et.al.) that show the existence of an electrical field in the space outside a current carrying conductor    [ there's a copy of a 1961 article here:   https://zjui.intl.zju.edu.cn/course/ece329/Secure/LectureNotesforCalendar/optional/Jefimenko62.pdf  with pictures on the second page (numbered page 20) for those who prefer pictures ].   

2.    This paper:   The Electric Field Outside a Stationary Resistive Wire
Carrying a Constant Current,   by  Assis, Rodrigues  and  Mania,  Foundations of Physics, Vol . 29, No. 5, 1999            is also commonly cited.   Although it does fall back on some numerical approximations,  formulae for the force exerted on a point charge located some distance away from a current carrying wire and purely due to surface charges on the current carrying wire are exhibited   (see section 4 of the article).

3.   More modern treatments exist.   Using the basic ideas of continuity conditions at a boundary between two media for electromagnetic fields we can actually show that there is an E field outside the wire with very little work   (see https://en.wikipedia.org/wiki/Interface_conditions_for_electromagnetic_fields ).   But until or unless LaTeX support for mathematical symbols is restored I'm not going to do that today.

Best Wishes.

[paul cotter]

I agree, ES. If power is being delivered there has to be a non-zero Poynting, hence both a B and E field.

[alancalverd]

 Wires have some small resistance.

Interesting. I've  just spent the morning measuring the magnetic field outside a closed superconducting ring where this statement is clearly not true!

[paul cotter]

About a half hour after I posted superconductivity came to mind. Say you have a superconducting wire delivering power to a load what would the Poynting vector be? zero?( on the superconducting wire )

[Eternal Student]

Hi.

   For a superconductor,  you're (@alancaverd) right.   I'm going to take for granted that for a more ordinary wire, you'd agree that it does have some resistivity.

Say you have a superconducting wire delivering power to a load what would the Poynting vector be?

   
   That's actually a very interesting question.  You just won't be able to establish a potential gradient across a length of superconductor.   It has precisely 0 resistance and will not support an infinite current in steady state  (So  V=IR always gives V=0 since R=0  etc.).   So there is no need for surface charges to accumulate on the surface of the superconducting wire (similar to the diagram a few posts ago).  There is zero E field inside the conductor (in steady state), any current that was there will just always keep flowing, it does not need any E field to overcome any resistivity.  In practice, if you tried connecting a battery then you'll have a problem, the current will escalate rapidly and the material will get warm fast,  so in practice the superconductivity will be lost or something else in the circuit (like the wires to the battery) will break  etc.

   For one thing, you'll need to be very specific about what the load is and how it was attached without itself being a superconductor:  For example, if it's a resistor (or lamp filament or something) which is also supercooled and superconducting then you just can't establish a potential across the resistor (or filament) either,  so no work can be done when charge passes through that resistor.   Specifically, it's not sufficient to say  "you have a superconductor delivering power to..(whatever)....", in some cases you just would not deliver power or have power consumed by what you thought was the load.     If you dunk a superconducting circuit complete with a lamp filament that can also become superconducting into a tub of liquid Nitrogen then you may find that when you connect a battery, the lamp filament just would not light up.

   To connect a battery and have it all work (hopefully),  there must be some regions where you don't have a superconductor.  For example the battery and some wires to it are NOT in liquid nitrogen, they are at room temperature,  similarly the load  (lamp filament or whatever) is not superconducting and it has some wires to/from it which are also not in the liquid nitrogen.    Now you're in a position where there are some regions of the circuit where E and B fields can be set up fairly conventionally.   So there are some E fields outside the superconducting piece of wire just from the bits of circuit that were not superconducting.   
    That's my best estimate anyway:
1.   There will still be a magnetic field outside the superconducting piece of wire because there were moving charges in it.
2.    there will still be E fields outside the superconducting piece of wire but they were sourced from the non-superconducting bits of the circuit rather than directly from surface charges on the superconducting piece.
3.    If you prevent (2) and don't have E fields produced in space from anything or anywhere in the circuit  (e.g. dunk the whole circuit into liquid nitrogen) then the bulb does not light and NO power is delivered anywhere or consumed by any part of the circuit.

Best Wishes.