newtonrulez
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 traveling so fast in the conducting wires with such a low drift speed?

Thanks!

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

newtonrulez
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?

Repainted
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.

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.

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.

DrMik
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?

pallidin
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.

DrMik
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?

pallidin
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.

pallidin
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.

pallidin
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.

Last edited:
DrMik
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)

Per Oni
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?

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.

Per Oni
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)

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.

Per Oni
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.

Per Oni
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

Last edited:
Per Oni
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.

Last edited:
Per Oni
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.

Per Oni
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.

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.

Last edited:
Gold Member
... 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.

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.

Per Oni
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
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.
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.

Last edited:
Per Oni
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: [Broken]
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?

Last edited by a moderator:
akinokoe
Here is the simple explanation. The switch-on time is not dependent on the drift velocity. After the circuit is switched on, the electric field is applied on the appliance immediately (the speed of the propagation of the field is about 1/10 of the speed of light). At the same time the electrons start to drift and thus current is formed. We can also say within the time of the length of the circuit divided by 1/10 of the speed of light, the drift velocity along the wire can reach the same value. The time that is consumed is not dependent on the drift velocity but the speed of the propagation of the electric field.