Hi!

I was just reading about drift speed and I read that the speed is about 10^-4m/s. It then struck me that if electricity is carried by electrons, then in a given circuit, how is the light bulb lighting up so quickly when according to my calculations the time taken to travel 0.5 metres for electrons (current carriers) is :-

T = D/S = 0.5m / 10^-4m/s which is a very large value for the time.

How are electrons supposedly travelling so fast in the conducting wires with such a low drift speed?

Thanks!

The electrons themselves move at a low average drift velocity, but the current in the wire moves at close to the speed of light.

Thank you for your reply! I however, still am not completely clear.

If the electrons' drift velocity is low, how is the current moving so fast?

Aren't electrons supposed to carry charge and therefore current themselves?

I'm not exactly sure on this since I've only read a very layman explanation for it.

Basically, from what I read, using your light bulb example, the charges near the bulb 'feel' the acceleration due to the emf from the power source almost immediately as the E-field changes due to the source takes the same time as light to travel from the source to the point in the circuit to be 'felt'. So the charges near the bulb start moving almost instantaneously.

Ugh I'm not sure if I phrased that very well.

Matterwave
Gold Member
Imagine electrons traveling in a circuit like when I push a very long pencil. I push on one side of the pencil, and the other side moves (almost instantaneously)! The current, then, is not determined by how fast I can push the pencil, but by how much of the pencil moves past a certain point within a certain amount of time.

All of the electrons within the wire start moving almost instantly after the circuit is switched on. This is very much like how all of my pencil starts moving almost instantly after I push it.

mheslep
Gold Member
Or: Take two current flow scenarios. In the first imagine only one electron moves past a 'point' as Matterwave mentioned, and assume it is moving fast, say, several meters per second. Imagine further that only one such electron moves past the point every, say, 10 minutes. In the second scenario imagine an electron crawling past the same point at only several microns per second, and again only one such electron happens along every ten minutes. What is the current with respect to the point in each case? The same. The drift velocity is high in the first case, low in the second.

Imagine a cardboard tube full of sweets. If you push in another sweet at one end another drops out of the opposite end, no matter how long the tube is. That shows that the sweet put in only travels a couple of millimetres yet its action can be seen instantly at the other end of the tube which could be a metre away! How is that?

Imagine a cardboard tube full of sweets. If you push in another sweet at one end another drops out of the opposite end, no matter how long the tube is. That shows that the sweet put in only travels a couple of millimetres yet its action can be seen instantly at the other end of the tube which could be a metre away! How is that?

Umm... that's not true at all.

Umm... that's not true at all.

It was the same 'laymans' terminology as the pencil analogy given earlier. I don't think this question is above the level of 'high school' so that was the level of answer I gave. If you have a better analogy why don't you just give it to the questioner then?

The reason I did not expound is that the OP is asking about electron drift.
Your analogy is dealing with pressure waves; a wholly different subject with extremely different processes and outcomes.

Just trying to keep the focus on the OP's question.

Still, DrMik, in your scenario(or any classical scenario) there is no such thing as "instantaneous"
When you push the candy in the tube, the "exit" candy does NOT respond instantaneously. Far from it. You just think it does because it's only 1-ft long(or several meters) and your natural senses can not perceive the delay.

I think DrMiks analogy is pretty good.The sweets are analogous to the electrons,pushing in an extra sweet is analogous to closing the switch,and the resulting movement of the sweets is analogous to the resulting drift of the electrons.Analogies are never perfect but they can be helpful.

I think DrMiks analogy is pretty good.The sweets are analogous to the electrons,pushing in an extra sweet is analogous to closing the switch,and the resulting movement of the sweets is analogous to the resulting drift of the electrons.Analogies are never perfect but they can be helpful.

He portends that this event is "instantaneous" It is not and in fact MUCH lower than c. That needs to be firmly understood. I heard somewhere that electron drift velocity is on the order of about 100,000 miles per HOUR in copper. That is far lower than 186,000 miles + per SECOND with regards to c. That's about a 3,600 difference.

And, c is most certainly not instantaneous. Not even close.

He portends that this event is "instantaneous" It is not and in fact MUCH lower than c. That needs to be firmly understood. I heard somewhere that electron drift velocity is on the order of about 100,000 miles per HOUR in copper. That is far lower than 186,000 miles + per SECOND with regards to c. That's about a 3,600 difference.

And, c is most certainly not instantaneous. Not even close.

I think drift velocity is of the order of 0.006m/s compared to 3*10^8m/s for light.Quite a big difference.Yes I agree that there are imperfections with DrMiks analogy but isn't this the case with all analogies?The water circuit analogy is the one most commonly used for electricity and generations of students have found this to be useful despite its imperfections.I think the usefulness or otherwise of the analogy depends on the present level and understanding of the O.P.

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Still, DrMik, in your scenario(or any classical scenario) there is no such thing as "instantaneous"
When you push the candy in the tube, the "exit" candy does NOT respond instantaneously. Far from it. You just think it does because it's only 1-ft long(or several meters) and your natural senses can not perceive the delay.

Err yes. I was not using the term instantaneous so literally, I meant it as if you actually did it using the tube of sweets. Why was this so different from the pencil analogy? But you are correct anyway (apart from drift velocity in copper I though it was about tenths of a mm per sec). I must try harder to keep it real, (bows head in shame and hides in wardrobe)

I think drift velocity is of the order of 0.006m/s compared to 3*10^8m/s for light.Quite a big difference.Yes I agree that there are imperfections with DrMiks analogy but isn't this the case with all analogies?The water circuit analogy is the one most commonly used for electricity and generations of students have found this to be useful despite its imperfections.I think the usefulness or otherwise of the analogy depends on the present level and understanding of the O.P.

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?

mheslep
Gold Member
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.
The 100W of output power in your example is not related F=P/V to the drift velocity. The drift is connected to the I^2 x R resistive power losses in the wire.

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.

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)

mheslep
Gold Member
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.

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.

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.

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.

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

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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.