What Are the Forces Behind Electrical Conduction and Drift Speed?

[Gokul43201]

Hi all,

If we state that low (drift) speed free electrons are responsible for the transport of power than it follows we need big forces. Regardless the explanation of what happens inside a wire i.e. classical / semi / qm etc. as far as I’m concerned this wire is a box in which I push small marbles with speed and force.

For example say we generate a power of 2 Watt in a piece of wire area 2 mm2 and length 1 m. Normally electronic drift speed caused by voltage in good conductors is ~0.001 m/s, therefore it follows that this force needs to be ~2000 N ! The pressure becomes even more impressive!

Why the exclamation? The average force on a single electron due a small field of say 1000V/m is F = eE ~ 10^{-16} N. In a wire with 10^{20} electrons, that's a total force of about 10,000N from the field. Naturally, the net average force on the electrons is zero, since they are not accelerating. There's nothing bizarre about this.

One way we could check whether any such forces are generated is by sticking a couple of probes of a power supply horizontally in some liquid mercury. Since we have roughley as many +ve ions as free electrons there should be a drift of liquid towards the –ve probe. I’ve got a feeling there will be no shift of liquid.

So did you actually do the experiment before going back to the drawing board?

 

[erickalle]

How does what I explained causes this? All I said what that the electrons are highly mobile and are very efficient at moving. I didn't say there's any accumulation of charges of any kind! In fact, I even said that on average, the mercury ions "... NEVER become electrically charged..." How did you arrive as what you said above based on what I said? You'd think I would know enough about electrical conduction to not make that silly conclusion.Zz.

If the mercury ions never become charged then we are left with the charge of the –ve electrons. So the total sum is: zero charge on the ion and 1 negative electronic charge per atom. This makes for a lot of –ve charge on the whole wire.

But you ARE! Your premises started from the erroneous assertion that you would have electrons in a SOLID CONDUCTOR gaining such a high speed that it can attain ridiculously high temperature! Would you like me to point out where you said such a thing, or did you forget? This is what I am countering with the Drude model, and the fact that in most application, a wire simply doesn't melt that easily. So I am attacking your starting premise, which if show to be false, makes the rest of your scenario utterly moot.

If you got objections to me melting the wire with a high current I can always melt it in another way and then send a high current through. My point here was that the ions are not moving length wise even in the liquid state.

You may not believe it, but I happen to be working on a similar problem. I'm trying to investigate breakdown in copper and other metallic structure under very high field gradients, especially when there are "sharp point" or field enhancement regions. And we're working with RF fields up to 90 MV/m! The initiation of breakdown effects is thought of to be preceeded by the melting of a localized region of Cu due to either heating effects of the surface currents that these field enhancement regions, or due to the field-emitted current though through very small surface area and thus generating huge current density and Ohmic heating. But these things do not occur THAT easily, and it certainly does not occur in non-field enhanced region! It isn't THAT easily to melt copper under such a condition using charge transport in itself.

You’re very privileged.

 

[erickalle]

Why the exclamation? The average force on a single electron due a small field of say 1000V/m is F = eE ~ 10^{-16} N. In a wire with 10^{20} electrons, that's a total force of about 10,000N from the field. Naturally, the net average force on the electrons is zero, since they are not accelerating. There's nothing bizarre about this.

Hi Gokul.
1000V/m is a huge field for the conductor we are considering!
What do you think is providing a counter force of 10,000 N per atom so that there’s no net acceleration? What is the total pressure exerted by this counter force?

edit: 10,000 N in total. Not per atom

So did you actually do the experiment before going back to the drawing board?

I do this experiment each time I connect a mercury switch in a circuit. If the liquid was moving, it and the current would soon start to oscillate.

 

[ZapperZ]

If the mercury ions never become charged then we are left with the charge of the –ve electrons. So the total sum is: zero charge on the ion and 1 negative electronic charge per atom. This makes for a lot of –ve charge on the whole wire.

Er... say what?

A "liquid metal" is still a metal. Stick your electrodes in liquid mercury and tell me if you have a charge liquid after 5 minutes.

If you got objections to me melting the wire with a high current I can always melt it in another way and then send a high current through. My point here was that the ions are not moving length wise even in the liquid state.

And my question is, why should they? The whole lump of metallic liquid is neutral. A liquid metal is STILL a metal!

You’re very privileged.

And I don't just talk the talk without having to get to DO it also, which means I don't just make it up as I go along.

Zz.

 

[erickalle]

Stick your electrodes in liquid mercury and tell me if you have a charge liquid after 5 minutes.

It’s probably me but I don’t understand this sentence.

You have stated (#28) that on average the mercury ions are neutral. That still leaves a non neutral net charge per atom because of the –ve electron. Are you saying that the electrons are on average neutral as well?

 

[ZapperZ]

It’s probably me but I don’t understand this sentence.

You have stated (#28) that on average the mercury ions are neutral. That still leaves a non neutral net charge per atom because of the –ve electron. Are you saying that the electrons are on average neutral as well?

It's my turn to say that I don't understand what you are saying.

The Hg liquid is ELECTRICALLY NEUTRAL. You could FREEZE the Hg and turn it into a solid and it will behave the same way. A liquid metal is STILL a metal (by definition!) and the electrons are still as mobile. There's always the same number of electrons in the Hg liquid as necessary to keep it electrically neutral... just like a solid metal!

Now is that any clearer?

Zz.

 

[erickalle]

However, under normal conditions, the electrons are MORE mobile than the ions. It means that if you apply any reasonable potential across mercury, the electrons are zipping in and out of the mercury so fast, the mercury ions, on average, NEVER become electrically charged! So these so-called "ions" are never ions in the first place!
Zz.

According to all electrical conduction theories:
Say a conduction electron under the influence of an external electrical field leaves an atom. This atom now becomes an electrically charged ion. There’s absolutely no doubt about that! There are electrically charged +ve ions for as long as the current flows.

 

[ZapperZ]

According to all electrical conduction theories:
Say a conduction electron under the influence of an external electrical field leaves an atom. This atom now becomes an electrically charged ion. There’s absolutely no doubt about that! There are electrically charged +ve ions for as long as the current flows.

Think again. The conduction electrons are in BANDS already - they no longer belong to ANY ions, meaning they are non-localized! Their energy level no where near resemble that of an atom!

What does it mean? It means that even without an applied field, the conduction electrons are not a part of any atom. Yet, do you see the ions here reacting to any external field? Nope! Why? Look up "skin depth" or a metal, or why the conduction electrons can be such an effective shielding of any external field!

The metal remains neutral. The ions do not see appreciable field. It is why when you do a Gauss's law on a metal for a static field, there's zero E-field inside a metal!

Zz.

 

[erickalle]

Think again. The conduction electrons are in BANDS already - they no longer belong to ANY ions, meaning they are non-localized! Their energy level no where near resemble that of an atom!

What does it mean? It means that even without an applied field, the conduction electrons are not a part of any atom. Yet, do you see the ions here reacting to any external field? Nope! Why? Look up "skin depth" or a metal, or why the conduction electrons can be such an effective shielding of any external field!

The metal remains neutral. The ions do not see appreciable field. It is why when you do a Gauss's law on a metal for a static field, there's zero E-field inside a metal!

Zz.

Now put a pd across a conductor. What will be the net result of forces by the field exerted on the conduction electrons and forces exerted on “on average” neutral ions?

 

[ZapperZ]

Now put a pd across a conductor. What will be the net result of forces by the field exerted on the conduction electrons and forces exerted on “on average” neutral ions?

What forces? Under STATIC equilibrium, there are not NET fields in the conductor itself! This is what I've been trying to get across to you, that ON AVERAGE, due to the high mobility of the electrons, they zip into the anode and come out of cathode so fast as to shield the inner ions from any net electric field! Go do Gauss's Law on a conductor in a static field! What's the E-field inside the conductor? You continue to ignore my question on this.

At NO POINT in a simple static equilibrium electrical transport is there a net charge on the conductor. The conduction electrons make sure of that.

Besides, what IF there is a net force on those anyway? Are you saying that they SHOULD move due to your calculated "large" force? Or are you still arguing that you have calculated enough energy that a copper wire SHOULD melt?

Zz.

 

[erickalle]

What forces? Under STATIC equilibrium, there are not NET fields in the conductor itself! This is what I've been trying to get across to you, that ON AVERAGE, due to the high mobility of the electrons, they zip into the anode and come out of cathode so fast as to shield the inner ions from any net electric field! Go do Gauss's Law on a conductor in a static field! What's the E-field inside the conductor? You continue to ignore my question on this.
Zz.

Quote of Ashcroft / Mermin solid state physics near the top of page 7:

The resistivity ρ is defined to be the proportionality constant between the electric field E at a point in the metal and the current density j that induces: E=ρj.

It quite clearly means to say there’s an E field in the metal.

 

[Dr Transport]

If there is an E-field in a metal, please produce the experimental evidence and I'll be more than happy to email the people in Stockholm to nominate you for the Nobel because you would be putting the entire physics community on it ear...

 

[ZapperZ]

Quote of Ashcroft / Mermin solid state physics near the top of page 7:

It quite clearly means to say there’s an E field in the metal.

Again, what was the MODEL being used there? It is the free-electron gas based on a CLASSICAL IDEAL GAS!

Jeeeze! It's like talking to a WALL!

Zz.

 

[erickalle]

If there is an E-field in a metal, please produce the experimental evidence and I'll be more than happy to email the people in Stockholm to nominate you for the Nobel because you would be putting the entire physics community on it ear...

Hi Dr.Transport. Thanks for joining in. If you had joined earlier, it would have saved just about 25 or so replies in this thread.
So there’s no E-field in a metal and therefore the relation F=E*q in a metal is not valid anymore. This is new to me but not entirely unexpected. My edition of the book I mentioned is from 1976, it has a whole chapter called “Failures of the free electron model”. In there is no mention of the above fact, also throughout the book formula’s are used using an E-field existing in the metal.
I don’t think I’m the only one caught out here.
Is there an E-field just outside a conductor? Can you recommend a textbook?
Can you put that email on hold?

 

[Dr Transport]

If you can find an edition of Ashcroft and Mermin that isn't from 1976, please pass it along, they never wrote a second edition.

As for the electric field just outside a conductor, sure there is one. Look at Jackson's E&M book for the expression of a conductor in an electric field. If there isn't an electric field applied, there won't be one unless there is excess charge.

I'll be happy to hold off on that email...

 

[Farsight]

Perhaps "wire explosion" is relevant to what you're talking about, Eric. Search Google.

http://flux.aps.org/meetings/YR00/DPP00/abs/S1360079.html

"The Joule energy deposition in the initial stages of exploding wires has been investigated. A Maxwell 40151-B trigger generator, producing a maximum current of ~3kA and voltage of ~120kV, with current rise times of ~170A/ns in a fast regime and 22A/ns in a slow regime, was used in the explosion of different fine wires. The wires were 2 cm in length and 4-40 microns in diameter. Current, voltage, current and voltage derivatives, wire boundary evolution, interferometry, shadowgraphy, absorbography, time-resolved self-luminosity imaging, emitted light intensity were monitored in the experiment. The evolution of wire diameter was determined using a diode laser back lighter and streak camera. Two significantly different modes of wire explosion have been found: fast and slow. It has been found that electronic emission from wire surface plays important role in the Joule energy deposition inside the wire..."

Does anybody know why my cheap little DC welder exhibits a strong attraction between the rod and the subject?