Electrons In A Circuit: CR4 Challenge (07/08/08)

Finger-in-the-wind, about half an hour.

In a solid the drift speed is in the order of millimeters per second ... so to estimate say five millimeters a second.

10 * 1000 / 5 = 2000 seconds

so ... 33 minutes and 20 seconds .... ish

More than one hour, probably closer to a day. Off the top of my head.

I checked on diffusion - sort of:

In an hour electrons in silver might diffuse about 10-cm, so not as wildly far away from the challenge's distance as you might expect.

But I don't have the proper tools or data to hand, so it's using data that is very indirect - and may indeed be wildly out for some of the non-fundamental data.

Almost instantaneously.

The uniform drift theory is a swell theory, and the movement of the "charge cloud" has, I'm sure some valuable uses, but it is relative to averages. Some electrons are booking along, some will be stable, and some are going to be involved in traffic accidents.

Now, as was already mentioned by other, the time decreases proportionally to decrease in cross section of the wire. [In the uniform, or cloud theory]

So, Suppose we have a silver wire one atom in diameter [and also that we could keep it cool enough not to burst into flame instantaneously]. How fast then? Darn near instantaneous!

Now transpose this to [say] 18AWG silver wire. There is inside the 18 AWG wire a [an abstract] wire one atom thick that the luckiest lil electron is going to take to make it to the positive post on the battery in [relatively] no time at all.

Add a bit of lateral movement as the path is not likely to be straight, and a compensation for the under-defined resistor and we still get

Almost instantaneously.

tom

=-==

Looking at previous replies, can I ask everyone if we're allowed to tag each electron, so we know which is which ?

Quantum mechanics does not permit the labeling of individual electrons in a way that makes them all indistinguishable. This fact has a big influence on the statistical behaviour of multiple electron systems, but probably does not significantly influence the answer to this question or the observation by "Physicist?" that all the electrons take almost exactly the same time to travel along the wire.

Supplementing Elroch's answer: electrons are distinguishable only by virtue of their states. That includes their kinetic energy and their location. In principle you could track one all the way along the wire - albeit the energy required to observe and identify it and the frequency of the required observations would at best significantly modify its temperature and reduce its drift velocity (so in this case the watched electron is indeed the last to boil).

Hi BobD,

I see we are at least in the same ballpark with our figures.

I give you a GA as you pipped me to the post

I finally manage to see why our answers differ, although in the same ballpark.

You assume the wire is 0.8 mm² while I have assumed 1 mm² which is why you have got a lesser number of days 72.3/90.3 = 0.8

Equivalent to mine too (#4), once you account for the factor of 10 area. But I can't for the life of me see why the pair of you want to go to the velocity and back again; surely you only need to know when a charge corresponding to the total valence electrons in the wire has flowed?

Incidentally, with regard to BobD's comment about the wire resistance: in the hypothetical limit of zero-thickness wire (or if you short-circuit the resistor), the time would still be as long as 20.7 minutes (at 20^OC).

If Switch is not closed : Electrons will never reach.

If Switch is closed : Electrons will reach Instantaneously.

28 minutes

Current = 12/1000 = 12mA or Coulombs/second

Number of electrons per second=

(12 X 10^-3 C/s) / ( 1.6 X 10^-19 Coulombs/electron) = 7.5 x 10¹⁶ electrons/s

Now for Silver conductor the atomic weight is 107.8 grams/mole.

Divide this by the number of atoms in a mole (6 X 10²³ ) to get

17.97 X 10^-23 grams per atom

SG of silver is 10.5g/cm³ , therefore the number of atoms per cm³ =

(10.5 g/cm3 ) / ( 17.97 X 10^-23 grams per atom ) = 5.84 X 10²² atoms/cm³

Assuming wire is about 1 mm² ( 10^-2 cm² )in cross-section, one metre (100cm)

length has a volume of 1 cm³ .

Assuming 1 electron per atom involved in current flow, the average drift speed of the electrons will be:-

(7.5 X 10¹⁶ electrons/s) / (5.84 X 10²² atoms/m) = 1.28 X 10^-6 m/s

Time for 10 metre trip = 10m / (1.28 X 10-6 m/s) = 7.8 X 10⁶ seconds

= 2,166 hours = 90.3 days

Propagation Delay: 85% speed of light

speed of light in a vacuum = 300,000,000 meters/ second

Therefore Speed in Silver wire = 255,000,000 m/s

So at 10 Meters, Propagation from terminal to terminal = (1/255,000,000)x10

3.921568627450980392156862745098e-8 or 39.22nS ?

Hmmm I think something wrong here, calc, been too long..

I think you may be confusing the speed of propagation of the electromagnetic field with the speed of drift of the electrons in the wire.

For explaining the difference between propagation speed and speed of each constituent part alone, consider a non-deformable pipe full of water. The pipe may be as long as you desire, but as soon as you switch on the tap on the one end, water will almost instantaneously appear at the other end. Nevertheless, you can make water in the pipe flow as slow as you wish...

Maybe BobD and Maths_Physics_Maniac answers seem at absurd, but indeed, electrons need not travel that fast to produce some good amount of electric current (even enough to kill you), as there are sooo many of them. Imagine Amazon river for instance. Its waters may seem still, but nevertheless the flow of water is enormous.

less than a secnd

wow, good question.

My answer is electrons never arrive at positive terminal from negative terminal forever !!!

electron can move only very tiny distance at very lower speed, only electric field can transmit instantially all circuit. even if the length is up to 10000km.

electric field speed is 300000km, as we all know.

They'll drift at about walking pace...say 4miles per hour, so about 6000m/hour. So 100m/min.

That's 10m in 6 seconds

Del

Does it really matter, they will get there & hopefully in time for dinner.

I always thought cats stopped for regular rests - that might get you closer to the minimum possible time.
Presumably the drift rate comes from some published figure that covers a different special case?

• If you would be able to "mark" a single electron, close the switch and wait until it arrives at the positive terminal, you will have to wait a very long time. as it may never arrive...Electrical conductivity is a series of electron movements within the molecular frame. They "bang" into each other in a polarized direction, but actually never migrate. I believe the question should be asked a little differently.
• How long would it take from the first "bang" to the last? Sort of, how long will it take from the first Domino to move until the last one falls down...

My 5 cents...

Wangito.

Where do you people come from? Electron flow has been the theory for many years:

http://en.wikipedia.org/wiki/Electric_current

what is the diameter of the wire?

I think we should assume that the information provided in the challenge defines the parameters; in this case, that the wire resistance is sufficiently low that the 1-k resistor sets the current to within about 30%.

Does the resistor limit the number of electrons or their speed?

I asked a few but they just drifted on by humming to themselves...blue shiny bas*ards

Del

The answer must be the speed, as the number of free electrons in the wire stays essentially the same, regardless of whether there is a resistance added to the circuit. The way it achieves this is by decreasing the voltage gradient in the wire (most of the voltage difference is across the resistor).

Your answer doesn't make sense to me, maybe it's badly explained.. or I'm thick (which I know isn't true)

Any mention of the wire is irrelevant, we can consider it a perfect conductor with infinite free electrons if you like.

I was asking about the resistor....thus your statement
' as the number of free electrons in the wire stays essentially the same,' is no help.

...the thing about voltage gradient is obscure in the extreme...it's just a re-phrasing of ohms law and doesn't answer anything.

Forget the wire!

Back to my question.....
Does the resistor slow 'em down or limit their number?

Del

I too didn't understand your question originally. But: the answer could be both - it depends how you look at things; but: if we assume linear resistors and conductors, the answer that the resistor determines the velocity at every point is certainly appropriate. On the other hand, what we are concerned about is the number crossing the boundary between the wire and the battery terminals - so you could say that the resistor also determines the number/second - but mainly by way of controlling the velocity in the wire.

So far as the original question is concerned, we are of course only interested in the time for all the electrons ahead of our chosen few to reach the battery terminal. So we don't actually need to know the velocity. All we need to know is the current (Coulombs/sec) and the total charge (Coulombs) available to be transferred ahead of the ones we are interested in.

The relevance is that since there is no build-up of charge anywhere, the same charge per unit time (i.e. the same number of electrons) passes through the resistor and the wire. You are right that we cannot infer from this that the number of charge carriers in the resistor remains constant as the voltage across it changes: this depends on the precise type of resistor and the range of voltages. For example in semiconductors, a rise in temperature (which could result from an increase in the current due resulting from an increase in the voltage) will increase the number of charge carriers. Note that your original question is not well defined. With chosen conditions there are a certain number of charge carriers in the resistor and they move at a certain speed. Which of them is "limited" by the resistor only makes sense if you compare two different scenarios (which you did not specify). This is why I was implicitly comparing the situation with a circuit with and without a resistor, to discuss its "limiting" effect on the current in the wire.

I hope this is a little clearer!

(i.e. the same number of electrons) passes through the resistor and the wire.

Exactly the point I was trying to make earlier!... The implication of this is that they must travel slower in the resistor (else it is no different from the wire and doesn't 'resist')

The whole question is about the transit time ...therefore the resistor is the important thing...thus I wish everyone would stop rabbiting on about the silver wire

Del

Yes, it's about time, not velocity. The only necessary assumption is that the total number of conducting electrons in the resistor is small compared with the total number in the wire. The electron velocity in the resistor may be fast or slow compared to the wire - that depends only on the relationship between the number of electrons per unit path length in the two media. If they are slow in the resistor and the total number of conducting electrons in the resistor is small, it simply follows that the path-length in the resistor is very small.

Think water flowing through a pipe under pressure, and the resistor as a constriction (or a vortex-generator) - you will easily see that total volume (charge) and flow rate (current) tell you all you need to know.

P.S. a water analogy for slow movement through the resistor would be if the high-resistance section of the was a wide porous filter. (That corresponds to lots of electrons in a medium with very low mobility).

However, if we take the example of a metal-film resistor, the effective cross-sectional area of the conductor would likely be substantially below 100-sq-um, so the constricted pipe is probably a better analogy - fewer-electrons per length, and also lower mobility than in the wire.

Back to my question.....

Does the resistor slow 'em down or limit their number?

And, would it depend on a chip resistor, or a wire wound, or metal film, as the distance for each component is different.

A chip resistor could be about 1mm long, and a wirewound resistor could be a meter long (unwound)

I wrote to Del: "You'll be sorry you asked". Nevertheless, I'll do my best:

As stated elsewhere, so far as the free electrons in the silver wire is concerned, the construction of the resistor is irrelevant - it leaves their number unchanged and limits the field (and hence the average velocity).

If you are concerned with the number and/or velocity in the resistor, that of course depends on the construction of the resistor. Most resistors (thick film, thin film, carbon film...) have rather small active cross-sections, so the average velocity of the active electrons within these resistors is relatively high.
Bulk carbon and wire-wound resistors have relatively large cross-sectional areas, and the average velocity of the active electrons within them is accordingly relatively low. For both nichrome and carbon film resistors there is another factor - the density of valence electrons that are involved in the conduction is much higher than for silver (about 6x for Nichrome, and 7x for graphite). Note that this is not in these cases identical to the number of free electrons**, which in Nichrome is about half the number for silver - but the free electrons interact with the lattice and the other valence electrons, and effectively all the valence electrons are at some point involved in the conduction. That means that the cross-sectional area of these resistors can be as small as 1/6th (1/7th) that of the silver wire, and still the average velocity within the resistor will be lower than in the wire.

**At least in Nichrome, the increase in the number of free electrons with increasing temperature is at least partially responsible for the anomalous temperature-characteristics.

So, the electrons in the silver wire would move faster than the ones in the (say wire wound) resistor?

If this was possible, then the electron flow would either spurt or the density would decrease once the free ones fell out the end before getting refilled by the ones fallung out of the resistor?

??? we are talking about the over all transit time of an electron..!!!

Imagine a continuous column of soldiers marching, they come to a steep hill (resistor) and have to slow down. The hill slows down the transit time...it doesn't create any problem of varying soldier density*...missing soldiers or 'spurting' !

Alternatively imagine they are marching 5 abreast and come to a gate only wide enough for one...again this will slow down the transit time...

I don't see any problems... ok they aren't perfect analogies, maybe theres a gate and a hill ...with a bit of boggy ground too, and an army of marauding cats?

Del

* actually you would get bunching which is an inrease in density...but it's not a 'problem'

1)2*Pi*r*B=uo*I,(I=12mA)by ampere;E=v*B,(E=12volts/10m)by Lorentz.

2)v=(2*Pi*r*R)/(uo*L),L=10m,r=silver wire radius,uo=magn.perm.;by 1)

3)R=R1*L/Pi*r**2,by Ohm,so R=1000ohms by hypothesis,so you get "r".

4)t=L/v

Most of the 12-Volts is across the resistor, which is not the silver wire.
Even if it were the silver wire, the velocity is limited by factors other than B and μ.
What you need (to get started) is Ohm's law (for the current), plus
the Mole density of Silver, and Avogadro's number for the number of silver atoms, together with knowledge that (almost exactly) the single Valence electron is all that is involved in the conduction (for the number of electrons to be shifted before the relevant ones reach the terminal).

If those equation are wright and the hypo.too,the results must be wright too.I assume the resistor is the silver wire or may be have not solution for this problem.

Possibly i am wrong,but, i think you must point the wrong equation, hypothesis, calculation or etc.

I think most people are assuming that the silver wires are of a reasonable size and that the 1K resistance is an actual resistor of value 1k.

If you take the assumption that the 10 metres of silver wire provides the 1000 ohms, then the silver wire would need to be very, very, very, thin.

Resistivity of Silver is 1.59 X 10^-8 Ω.m = (Resistance*Area)/length = 1000.A/10

Thus the cross-sectional Area of the silver wire would need to be:-

(1.59 X 10^-8 X 10) / 1000 = 1.59 X 10^-10 m²

This corresponds to a wire diameter of 7.1 microns

I cannot imagine this carrying a current of 12mA without melting like a fuse.

Imagine the fun of trying to solder the connections. You would want a very steady hand.

BAB

Even if i use d=1mm,(r=0.5mm) in v=2*R1/uo*r ,R1=silver resistivity ,i get a speed ten times the Bobd's value.Why's that?

Never thought before that speed could be so slow,now, how to explain the fast charge of a capacitor "R*C" ?

When you "push" electrons in one end of a wire, different electrons immediately come out of the other end, so the time it takes the electrons to travel along the wire does not have any influence on how long it takes a capacitor to charge. The analogy of a pipe leading from a reservoir to a tap is a good one. Water starts to fill the sink immediately you turn the tap on, even though it takes a very long time to travel along the pipe from one end to the other.

Although I would agree that unsupported 7-um diameter silver wire would be very tricky to handle, I doubt it would get excessively hot.
If we assume the wire is uncoated and is lying an average of 10-cm above the ground in perfectly still air, (probably very worst case for such flimsy wire?), air conduction between the wire and the ground would give a thermal resistance in the order** of ln(0.1/(∏.r))/10/Κ_air, or 35^OC/W. With 144-mW dissipation, we would only be looking at a 5^OC temperature rise.

**The calculation is based on twice the thermal resistance between complete concentric cylinders

Wow, that is interesting.

Just shows you how intuition can let you down.

Please show how your equations can lead to a meaningful result - including presenting it. {I may be blind, but I can't see anything usable in your equations}.

So far as I am concerned, Fyz and jim35848 already answered the question if the resistor is the (uniform thickness) silver wire (i.e. about 20.7 minutes at 20^OC).

Instantly...

While we have Vsupply greater than total voltage drop We will have enough electrical field that force electrons to move.

Think of thermocouple applications. Even PtRh-Pt wires that has larger specific resistance than silver can send data instantly to receiver under a few millivolts potential (instead of 12V)!

I think so as Guest commented (#6).

By my knowledge the mobility factor is regarding semi-condustors and gases.

In our case medium is well conductive solid-metal-.

According to quantum mechanics if wavelength of an electron is greater than ion's radius it can move without collide any other ion.

Any charge will expands over the whole conductor acc. to:

ρ = ρ₀e^(-σ/ε)t => For silver the time will be ε₀/σ = 8,054.10^-12 seconds. (σ:conductivity, ε:dielectric constant = ε₀)

mean instantly... wrong?

okay,

Taking into account the resistor don't change my way. The resistor is neither a semiconductor nor gas. The parameter that will change is conductivity only!.

With the resistor the time will be 12,57.10^-4 seconds.

Still instantly...

I'm not certain how you are interpreting the question. I think the intention of the challenge is to imagine that we could tag an individual electron that started at the negative terminal and calculate how long it takes to reach the positive terminal. I also think that you are working on the basis that tagging is not possible, and that you have an electron cloud with no electron having any identity:
if we take that view the identity progresses from the switch to the positive terminal at the speed of light in the conductor - on the other hand, only a fraction of the electron that represents the proportion of the total conducting charge on that terminal will actually be at that terminal - which raises the question as to whether it was ever truly at the negative terminal - and equally whether it ever really reaches the positive terminal.

I think the best (still highly artificial) "way out" is to take the view that the silver wire is not a continuous silver crystal, so there are multiple electron clouds and the electrons transfer from one cloud to the next - still only probabilistic fractions, but we can assign a very high probability of individual transfer between crystallites at times that are short compared to the total transit time.

Dear Guest,

I'm aware what you mean, thank you.

According to classical conductivity theory it is even impossible one electron can reach to the positive terminal. Namely we wait forever.

On the other hand the quantum mechanics talks about one electron can move along the lattice without hurdled using wave properties of electron. If this is possible this electron can fly over the following lattice since the electrical field is continuous along the wire, why not?

(By the way, i would not like to dig into schrödinger equations mere to prove this idea.)

thanks again

Regardless of such considerations, doesn't the average speed of the electrons still have to be the answer derived from the known current, the number of free electrons per meter of wire, and the electron charge?

One question is whether electrons on the outside of the wire move at the same speed as those within the conductor. If the current was alternating, there would certainly be a difference, but I don't see why there should be much difference with direct current. In any case, I believe the diffusion of electrons would thoroughly mix those on the surface with those within the conductor over significant periods of time.

Voltage / current is transmitted near instantaneously, but electron migration much slower. I've never seen a calculation of the time for migration. . . "who cares"?

Wait a minute, one could easily calculate electrons per second equivalent of a certain number of amps. . . but that doesn't give us the answer. I guess we would have to know the number of outer shell electrons in the wire? ? A very heavy wire would have the same amps, but much slower migration?

"I guess we would have to know the number of outer shell electrons in the wire?"

Silver has a valency of one; in metallic silver these form the conductive electron cloud.

Please! So you just assume .1 cm/sec? How about calculating it from amperes? This is no answer at all!

Agreed - totally arbitrary apparent guesswork - no better than the first three answers (whose authors presented them as first-cut finger-in-air exercises)
IMHO any of the four "Good Answers" did a much better job - assuming you accept the validity of the question.