Recognitions:
Gold Member

 Quote by Per Oni So the general consensus is that the conduction discussed here is an energy transport somehow carried out by very slow particles. Using this type of conduction, power up say a hairdryer P=100W: U=240V so that I=0.4A It is possible to transport this current using a soft copper wire having an extremely small surface area, say 4/100 of a square millimetre. Have you ever worked out the force which needs to be pumped through this small wire? In this example for drift a drift speed of 0.006m/s, F=P/V=100/0.006=17KN which is equivalent to a mass of 1700kg or the weight of 2 small cars. Why does this wire not explode / implode? I mean if you would again use that stick (or those sweeties) of such a surface area would it not buckle (pulverise)? I have used here the standard UK mains voltage but instead I could use a voltage hundreds of times higher which would allow for a much smaller surface area or alternatively a much higher power / force could be transported. Have generations of students been miss let?
I think you are using free electron theory here.If the current were carried by a single electron the average force might be of the order that you calculated.In a typical metal there are about 10^28 electrons per cubic metre carrying the current so the average force is very much smaller.

 Quote by Dadface I think you are using free electron theory here.If the current were carried by a single electron the average force might be of the order that you calculated.In a typical metal there are about 10^28 electrons per cubic metre carrying the current so the average force is very much smaller.
More significant here is the amount of electrons per square meter ~ 10^19.

In my example the force is 17KN and the area is 4X10-8 m2 so that the pressure is F/A=4x10^11Pa. Now whether there’s one particle or 10^11 particles is for this calculation immaterial. The small copper wire has to deal with this pressure!

To mehslep:
You are correct about the power losses. But keep in mind that those electrons also have to transport power to the point where its needed. How in your opinion does the power reach its destination far away from the source? (Remember we're still dealing with the stick/sweetie/water analogy)

Recognitions:
Gold Member
 Quote by Per Oni How in your opinion does the power reach its destination far away from the source? (Remember we're still dealing with the stick/sweetie/water analogy)
The electromagnetic field does the work.

 Quote by mheslep The electromagnetic field does the work.
Well yes but now you are diverting away from our original analogy.

So it looks like indeed many generations of students have been taught the wrong concept. Its impossible to transport any kind of everyday meaningful amount of power using such a low velocity. The stick theory just doesn’t work. How many physics books are still going to be written with this wrong concept?

We could go into how the electromagnetic field does the work but that is too far away from the OP and could do with a new post.

Recognitions:
Gold Member
 Quote by Per Oni Well yes but now you are diverting away from our original analogy. So it looks like indeed many generations of students have been taught the wrong concept. Its impossible to transport any kind of everyday meaningful amount of power using such a low velocity. The stick theory just doesn’t work. How many physics books are still going to be written with this wrong concept? We could go into how the electromagnetic field does the work but that is too far away from the OP and could do with a new post.
You can transport a meaningful amount of power because the actual equation is as below:

P=N times Fv

N= number of electrons in wire
F=force acting on each electron=electron charge * electric field strength
Because N is large the power is large.Please note that A cancels and that the force originally calculated would be the force if the power was carried by one electron only.

 Quote by Dadface You can transport a meaningful amount of power because the actual equation is as below: P=N times Fv N= number of electrons in wire F=force acting on each electron=electron charge * electric field strength Because N is large the power is large.Please note that A cancels and that the force originally calculated would be the force if the power was carried by one electron only.
In order to proceed with your explanation you have to be more case specific. Eg you state N=number of electrons in wire, but how long is this wire? Is the function of this wire to connect source with load? How much is E etc.? If you do that we can perhaps come to a conclusion whether the stick theory works or not.

Recognitions:
Gold Member
 Quote by Per Oni In order to proceed with your explanation you have to be more case specific. Eg you state N=number of electrons in wire, but how long is this wire? Is the function of this wire to connect source with load? How much is E etc.? If you do that we can perhaps come to a conclusion whether the stick theory works or not.
Let wire length=l and wire cross sectional area=A

I=nAve..........................................1.
P=VI..............................................2.
n=no of free electrons/unit volume=N/Al
Substituting and tidying up we can write:
P=N*Ve/l*v=N*F*v

 Quote by Dadface Let wire length=l and wire cross sectional area=A I=nAve..........................................1. P=VI..............................................2. n=no of free electrons/unit volume=N/Al Substituting and tidying up we can write: P=N*Ve/l*v=N*F*v
I’m not going to argue about what you stated here. We are talking about different things.

First we have to define what the “stick” theory means. It means the transport of power from source to load in a longitudinal (length way) manner. Much like when, holding a stick, you can push and pull a heavy stone across a rough surface.
This way of power transport is similar to hydraulic, water, chain, etc in fact any mechanical way of power transport.

Do you go along with that explanation?

Then my claim is that to have any meaning full transport of power this way we need some kind of reasonable speed of medium otherwise forces and pressures become too high for any everyday material (and size) let alone soft copper.

 Mentor Per Oni, your calculation of the stress in a current carrying wire is WAY off. The stress due to an arbitrary EM field is given by the Maxwell stress tensor which is: $$\sigma _{ij} = \varepsilon_0 E_i E_j + \frac{1} {{\mu _0 }}B_i B_j - \frac{1}{2}\bigl( {\varepsilon_0 E^2 + \tfrac{1} {{\mu _0 }}B^2 } \bigr)\delta _{ij}$$ For a typical maximum 15 A circuit on 14 gauge copper wire used in home wiring the diameter is about 1.6 mm and the resistivity is 1.72E-8 Ohm m. The magnetic field due to a long isolated current carrying wire is perpendicular to the wire (right hand rule) and of magnitude: $$B=\frac{\mu_0 I}{2 \pi r}=0.00375 \: T$$ The electric field in a resistive wire is parallel to the wire and is of magnitude: $$E=\rho J = 0.128 \: V/m$$ For the tensile stress on an isolated current carrying wire on the z axis you would be interested in the z,z component of the stress tensor. $$\sigma _{zz} = \varepsilon_0 E_z E_z + \frac{1} {{\mu _0 }}B_z B_z - \frac{1}{2}\bigl( {\varepsilon_0 E^2 + \tfrac{1} {{\mu _0 }}B^2 } \bigr)\delta _{zz} = -5.60 \: Pa$$ Which is more than 10 million times smaller than copper's yield stress of 70 MPa. To get the kinds of stress you are imagining would require something on the order of 4 MA. Yes, 4 MA will indeed destroy 14 gauge copper wire. In any case, the density of the charge carriers can be easily measured with the Hall effect. Once the density of charge carriers is known then there is no ambiguity whatsoever about the drift velocity for a given current density. The density of charge carriers is enormous, leading to very slow drift velocities.

 Quote by DaleSpam In any case, the density of the charge carriers can be easily measured with the Hall effect. Once the density of charge carriers is known then there is no ambiguity whatsoever about the drift velocity for a given current density. The density of charge carriers is enormous, leading to very slow drift velocities.
I've never claimed that the density/drift velocity was anything other then what previous posters claimed it was. No argument there. Its just because of this low velocity that I doubt that any meaning full amount of power can be transported that way.

I will have a closer look at your calculations when I've got some time. In the mean time thanks for taking part.

 Dalespam: I’ll try to explain as carefully as pos what my objections to this thread are, not that I want to talk down to anybody but I am plainly bad at explaining myself. The suggestion in this thread is that the signal speed of current in a wire can be explained using the analogy of a tube of sweeties where one enters, side pushes all the adjacent sweeties so that the last one pops out of the tube at the far end. This picture is wrong for a number of reasons. I wanted to show what the implications are, considering the forces involved transporting any average power that way, knowing that the medium (sweeties) travel at an extremely slow speed. Because that way since F=P/v, force and therefore pressure become very high indeed. A second objection is that it doesn’t even explain the electrical signal speed. Use the analogy of say a metal bar. When we push this bar a small distance length ways the pressure wave travels at approx the speed of sound for that metal, far short of the signal speed which is close to the speed of light. There are more points but I hope you get why I objected. As for your solution, mheslep and I already indicated that electro-magnetic theory can explain the ins and outs. I am a little puzzled why you reacted to my correct post and not to some of the previous wrong posts.
 Recognitions: Gold Member Hello Per Oni.I think all analogies have their weaknesses and shortcomings and you have pointed out some of these with the stick and sweeties analogies.Despite these weaknesses these analogies can be useful provided they are presented with care.I suppose it could be argued that free electron theory is itself an analogy.

Recognitions:
Gold Member
 Quote by Per Oni .... Because that way since F=P/v, force and therefore pressure become very high indeed.
The power expended in moving the electrons and delivered to the load (the 100W in your example above) are two different things. The output or load power delivered by the conductor is not the force on the charge carriers (electrons) x their drift velocity.

Recognitions:
Gold Member
 Quote by mheslep The power expended in moving the electrons and delivered to the load (the 100W in your example above) are two different things. The output or load power delivered by the conductor is not the force on the charge carriers (electrons) x their drift velocity.
But its interesting that when you equate P=VI with I=nAve you do get a P=Fv type equation the full equation being P=N*Fv(see my post above).I know it seems odd.

 Quote by mheslep The power expended in moving the electrons and delivered to the load (the 100W in your example above) are two different things. The output or load power delivered by the conductor is not the force on the charge carriers (electrons) x their drift velocity.
Correct for the case of real electrons.
However now use sweeties / ball bearings etc. in a tube instead of electrons and you will find that for 100W to be supplied with v= 0.006 m/s there’s a big force present on this medium.

I suppose the real difference is that electrons carry (are carriers of) electro magnetic fields. Any disturbance in these fields are propagated close to the speed of light, depending on what kind of cable and insulation.

Mentor
 Quote by Per Oni I am a little puzzled why you reacted to my correct post and not to some of the previous wrong posts.
Thanks for the clarification as to your intent. I simply reacted to your post because the number you gave for stress was off by 11 orders of magnitude.
 Quote by Per Oni The suggestion in this thread is that the signal speed of current in a wire can be explained using the analogy of a tube of sweeties where one enters, side pushes all the adjacent sweeties so that the last one pops out of the tube at the far end. This picture is wrong for a number of reasons. I wanted to show what the implications are, considering the forces involved transporting any average power that way, knowing that the medium (sweeties) travel at an extremely slow speed. Because that way since F=P/v, force and therefore pressure become very high indeed.
I think you are overreacting here. The analogy is adequate for its intended purpose which is to demonstrate that a low drift velocity is consistent with the observed fact that the lights come on as soon as we flip the switch. It is just an example used to help students get comfortable in an intuitive mechanical way with the idea that the drift velocity does not limit the propagation of the electrical current.

Also, the analogy is not so far from the truth. When you push one sweetie in on one end one pops out on the other end (conservation of charge). Moving the sweeties heats up the tube (resistance) and reduces the force that the last sweetie can push with (voltage drop). The last sweetie doesn't quite start moving instantaneously (finite speed of light). Etc.

As with all analogies it fails at some point, and it is good to point that out to students. In this case it fails to explain the stress of the EM field. But if we were to insist that all analogies be completely perfect in all situations then we could never use any analogies at all. The analogy is good (although I prefer the plumbing analogy to the sweeties one) and is useful to students in learning a challenging subject. It correctly addresses the OP's question which did not extend to EM stress or other topics outside of the proper scope of the analogy.

 Quote by DaleSpam Thanks for the clarification as to your intent. I simply reacted to your post because the number you gave for stress was off by 11 orders of magnitude.
This amount of stress I calculated is correct for the case of sweeties in a tube. I wanted to indicate just how ridicules this analogy is in point of view of forces involved. And yes compared with real conduction its way way off, but that’s just the point I’m making. I don’t need a house rewire each time something like a hairdryer gets switched on.

As to your feeling that the plumbing theory it is sort of ok, some time ago I read an article (it could be wiki) which proposed 2 different theory’s of electrical conduction one based on plumbing and one following the theory of the Poynting vector. (Perhaps it has been removed?) If we don’t point out discrepancies eventually it will end up all the way up there.
Just reading some wiki pages I found this: http://en.wikipedia.org/wiki/Poynting_vector:
 DC Power flow in a concentric cable Application of Poynting's Theorem to a concentric cable carrying DC current leads to the correct power transfer equation P = VI, where V is the potential difference between the cable and ground, I is the current carried by the cable. This power flows through the surrounding dielectric, and not through the cable itself.[7] However, it is also known that power cannot be radiated without accelerated charges, i.e. time varying currents. Since we are considering DC (time invariant) currents here, radiation is not possible. This has led to speculation that Poynting Vector may not represent the power flow in certain systems.[8][9
So it looks like there’s still some controversy.

There’s another aspect of electrical conduction theory I like to mention, namely: energy leaves the source at near light speed towards the surrounding dielectric. Therefore it follows that the source should receive a kickback in the opposite direction. Has anyone ever seen a paper dealing with that fact, or anyone any thoughts?

 Tags drift speed