# How Fast Does an Electrical Impulse Travel in a Copper Wire?

• Byron Forbes
In summary, the speed of an electrical impulse in a copper wire is just under c. The speed is well known to be just under c, but often when this topic is spoken about, the term "EM field" is used. There is no EM field or force in play here - it is simply an initial "electric" field, electrons and what is basically a longitudinal wave acting through them. Maxwell was able to do his math, make measurements and derive the magnitude of c. But that was derived with an electric field and a magnetic field, not just with an electric field, and it certainly never factored in any particles (electrons), only their fields. So the question is, is it reasonable to see this "electrical
Byron Forbes
The topic here is about the speed that the "electrical" signal (impulse) travels at in a copper wire.

The speed is well known to be just under c.

Often when this topic is spoken about, the term "EM field" is used and I don't understand why. We have electrons with a power supply that supplies an "electric" field - there is no need to introduce magnetism here whatsoever as far as I can see. And so the assumption, and the wrong assumption I believe, is that because EM fields are in play here that it is no surprise that the electrical impulse in a wire is approx c.

My view here is that we have an electric field at the power supply that acts on the local electrons in the wire - this is the beginning of the "pulse". From there, the electrons in that vicinity begin to move, thus pushing the ones in front of them, and so on and so forth all around the wire and back to the power supply. There is no EM field or force in play here - it is simply an initial "electric" field, electrons and what is basically a longitudinal wave acting through them.

As we know, Maxwell was able to do his math, make measurements and derive the magnitude of c. But that was derived with an electric field and a magnetic field, not just with an electric field, and it certainly never factored in any particles (electrons), only their fields.

So the question is, is it reasonable to see this "electrical impulse" is traveling at c and think that makes perfect sense? I see no EM here, just pure electric fields, particles with mass, and energy.

aniseed and Delta2
A pure electric field is static. It does not change with time. If you have an electric field that changes in time then you necessarily also have a magnetic field.

nasu, DaveE, vanhees71 and 2 others
What one observer sees as a pure electric field, an observer in motion with respect to the first will see as an electromagnetic field. Similarly with a magnetic field. There is no invariant sense in which there is "only" an electric field in any circumstances. There is always an electromagnetic field.

aniseed, Rcastro, hutchphd and 1 other person
Dale said:
A pure electric field is static. It does not change with time. If you have an electric field that changes in time then you necessarily also have a magnetic field.

Ok, so the electrons move, their fields move with them, and so they generate a consequent magnet field.

So how does that play any role in determining the speed of the longitudinal wave acting through the electrons that is purely electric?

This is a longitudinal wave acting through particles with mass - what is the relationship to light?

A "field" refers to a collection of points versus looking at the singular dimension. Ignore the term for now.

What you are proposing is appropriate because otherwise you would be discussing whether action at a distance occurs [like how can the ends of a wire just connected to some battery, 'know' the difference of potential without being informed through the wire.

In this case, the electrons DO only operate locally and work their way to the end as a kind of 'pulse'. But for whatever reason, the way it is taught often doesn't respect that such questions need to be asked. But then again, most teachers (or texts) ignore the original approach to begin from questioning from motivational philosophy to the reason science has adopted some view. Rather, the philosophy now is reversed by expecting the students to 'trust' up front the general overall topic and then gradually gets deeper and deeper until you get to see the possible step-by-step rationale that gave us these theories AFTER you have invested in the courses. It was a practical politic that favors those who could adopt a faith-in the authorities up front because most people would not take up science as a possible career when it required better emphasis on understanding foundational ways of thinking. [I believe the switch in the West came in the 1960s when the competetition to the Soviet Unions' space program led many to fear the Western kids were turned off from science. So they shortened the route educationally to get people to become practicioners of science before they are eligible to the philosophy. Note that the Philosophical Degree (PhD) comes at the end rather than the beginning. To properly understand, one has to take a 'foundational' approach if one wants to truly understand the science.

Delta2, davenn, etotheipi and 4 others
Byron Forbes said:
So how does that play any role in determining the speed of the longitudinal wave acting through the electrons that is purely electric?
Well, that is a little more complicated. You don’t really get longitudinal waves in a transmission line. Longitudinal waves are a characteristic of a waveguide, and they are typically used for substantially higher frequencies than normal transmission lines.

Here is a good page on the topic: https://www.ntnu.no/wiki/download/a...veguide_propagation_06.pdf?version=1&amp .; p

Edit: note that here we are talking of transverse and longitudinal fields, but I believe that the “longitudinal” wave you are thinking of is the current itself. So this is basically the circumferential component of the B field, which would be considered a transverse wave in the usual language.

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Byron Forbes said:
Ok, so the electrons move, their fields move with them, and so they generate a consequent magnet field.

So how does that play any role in determining the speed of the longitudinal wave acting through the electrons that is purely electric?

This is a longitudinal wave acting through particles with mass - what is the relationship to light?
The electron speed in a usual household current is tiny, on the order of about ##1 \;\text{mm}/\text{s}##. It's not this drift velocity of the electrons that determines the speed. If this would be the case, it'll take some time until the light goes on when you flip the switch. What transports the signal and the electromagnetic energy is in fact the electromagnetic field whose velocity is the speed of light. That's why your light goes on (almost) instantaneously when switching it on.

Rcastro, Daanh, Delta2 and 1 other person
Byron Forbes said:
From there, the electrons in that vicinity begin to move, thus pushing the ones in front of them, and so on and so forth all around the wire and back to the power supply.

It sounds like the OP is visualizing the EM wavefront to be analogous to a sound wavefront. Sound propagates only via particle collisions. But when an electron moves, its field changes with infinite extent and the change propagates at speed c in a vacuum. So when the first electron moves, that pushes all the other electrons in the wire (after propagating the EM field), not just the ones immediately in front. That makes electrons unlike gaseous neutral particles.

aniseed, Delta2, hutchphd and 5 others
Byron Forbes said:
This is a longitudinal wave acting through particles with mass - what is the relationship to light?
Since the electrons have mass, it makes sense that the signal propagates at less than c. But the changes in their fields propagate at c. And light is just that, propagating changes in EM-fields.

vanhees71 and davenn
Byron Forbes said:
So the question is, is it reasonable to see this "electrical impulse" is traveling at c and think that makes perfect sense? I see no EM here, just pure electric fields, particles with mass, and energy.

No it isn't reasonable.
A moving electric field generates a moving magnetic field and on and on ... so there is your EM field

Also, another misunderstanding you have ... the EM field doesn't travel in the wire (other conductor),
it travels along the outside of the wire/conductor with the wire acting as a waveguide . It is this reason that the EM field travels less than the speed of light (in a vacuum = c)
For a bare wire, it's around 90-95% of c. adding insulators to the wire decrease the propagation speed even more.

In coax cables, for example, the propagation of the EM field/wave can be as low as 0.60 of c.
A lot of work is done with designing the insulating material between the inner and outer conductors so as to keep the
velocity factor as high as possible 0.7 - 0.8 or so. This reduces losses that get worse as the frequency gets higher

Rcastro, vanhees71 and DaveE
Unfortunately, it's just really hard to explain why electricity moves as an EM wave (as opposed to electrons bumping into each other) unless you study the underlying physics. I, for one, don't know how to do it. As with many concepts in physics that are a bit beyond our current level of education, you may need to just accept that someday you will study this and you will understand it better; also, that people that do understand it can help with analogies, which are good, but somehow not complete.

Notwithstanding the conspiracy theories about how academia is ruining scientific inquiry, since 1960, anyway.

Delta2
A metal is sometimes considered to contain a dense electron plasma. I think it is reasonable to have plasma oscillations, and transmission line effects resembling longitudinal waves. The motion of an electron is dictated by the surrounding permittivity and permeability in the same way as elasticity and inertia for any other transmission line. The electric and magnetic fields are the energy stores of the vibrating electron.
For a wire in isolation we see a transmission line mode where the electric field lines jump between points on the wire and the magnetic field surrounds the wire. These fields seem to be the energy stores of the vibrating electrons.

Dale said:
You don’t really get longitudinal waves in a transmission line.
Yes you get, because as you also say the current wave is longitudinal , because the electric field inside the wires that sustains the current is in the same direction as the direction of propagation of current (according to ohm's law ##\mathbf{J}=\sigma\mathbf{E})##.

However this longitudinal electric field is small in good conductors (because ##\sigma## is really big), and it transfers only the energy that it is lost as heat in the ohmic resistance of the transmission line. The main energy transferred by the transmission line is by the transverse fields outside of the wires.

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Delta2 said:
Yes you get, because as you also say the current wave is longitudinal , because the electric field inside the wires that sustains the current is in the same direction as the direction of propagation of current.

However this longitudinal electric field is small in good conductors, and it transfers only the energy that it is lost as heat in the ohmic resistance of the transmission line. The main energy transferred by the transmission line is by the transverse fields.
I think the majority of the electric field in the longitudinal direction will be across the inductance per unit length. For a single wire in isolation we do not have a transverse component.

Delta2
It look like the confusion arise by imagining the conduction electrons moving because they push each other. There are many images and descriptions in texts for children showing this idea of electrons like balls in a tube pushing each other while they move from one end to the other.
However this is far from the situation in a metal carrying electric current. Actually the interaction between electrons plays practically no role in the electric current. Most of the properties of electric currents can be described by the model of free, independent electrons. The electrons move with the drift velocity due to the external field created by the power supply and not due to the pushing of other electrons. They all "start" at about the same time so there is no longitudinal wave of electrons inetracting with other electrons. It may be an electromagnetic longitudinal wave in some situations but this is not what the OP seem to describe.

etotheipi and Delta2
nasu said:
They all "start" at about the same time so there is no longitudinal wave of electrons inetracting with other electrons. It may be an electromagnetic longitudinal wave in some situations but this is not what the OP seem to describe.
The OP is asking about signal speed, which would be infinite, if they all started at the same time. The propagation speed of the "wave" seems exactly what the question is about.

Byron Forbes

"From there, the electrons in that vicinity begin to move, thus pushing the ones in front of them, and so on and so forth all around the wire and back to the power supply. There is no EM field or force in play here - it is simply an initial "electric" field, electrons and what is basically a longitudinal wave acting through them. "

What he decribes here does not happen. I did not say that they start at the same time, I said "about the same time". Which is taking into account that the drift velocity is so small compared with the speed of the signal that the electric field already propagated all around the circuit by the time it would take one elctron to travel one "inter-electron" distance at the drift elocity. Assuming that there is such a thing as "inter-electron distance".

Also, another wrong concept is that the signal moves from one terminal of the battery to te other. Like the elctrons at one end of the circuit are waiting for the kick from the electrons starting to move at the the other end.

nasu said:
...the drift velocity is so small compared with the speed of the signal that the electric field already propagated all around the circuit...
Doesn't this depend on the length of the circuit?

nasu said:
...by the time it would take one electron to travel one "inter-electron" distance at the drift velocity.
Why does the electron have to travel one "inter-electron" distance? Only the change of its field has to propagate to the next electron.

No, the difference between the two velocities does not depend on the length of the circuit.
There is no "next electron", this is part of the problem. The electrons are not aligned along the circuit and they are not static but move with speed of the order of 10^5 m/s. I suppose you could imagine the equivalent of a sound wave in the free electron gas but I don't think this (assuming that is possible) has anything to do with the signal propagation through the circuit.

https://www.bpa.gov/news/newsroom/Pages/Hitch-a-slow-ride-on-the-Electron-Express.aspx
Wouldn't it be fantastic if we could travel at the speed of an electron? Imagine a trip from The Dalles, Ore., to Los Angeles in the blink of an eye on the 846-mile Electron Express known as the Pacific DC Intertie, which just marked its 40th anniversary.
Unfortunately, you would be in for a disappointingly long ride.

Incredible as it may sound, the electrons themselves make a rather leisurely journey. For one to travel on the DC line from Celilo Converter Station near The Dalles to Sylmar Converter Station near Los Angeles would require hundreds of years. Assuming the line operates at full power, had you hitched a ride with the first group of electrons that left Celilo in 1970, when the line was energized, your Electron "Express" would reach southern California about the year 2912 – at the earliest!

How can this be? When we flip a light switch, the bulb turns on immediately. Doesn't electricity travel at the speed of light?

Delta2
Delta2 said:
as you also say the current wave is longitudinal
Part of our role here is to teach the standard terminology. When RF engineers talk about longitudinal waves they are not talking about the currents or the minuscule Ohmic field. They are talking about the dominant modes of energy and information transfer.

So although there is a current and an there is an Ohmic field, I would recommend against calling that a longitudinal wave. They point in the longitudinal direction, but they actually transport energy radially into the conductor dissipating the signal. This is not what the standard terminology refers to.

sophiecentaur, vanhees71, DaveE and 3 others
Dale said:
but they actually transport energy radially into the conductor dissipating the signal
Yes I guess you are right here, inside the conductor the Poynting vector and the propagation of the EM wave changes direction and follows the radial direction so the EM wave is still transverse, while outside the conductor the Poynting Vector is along the direction of the wire (tangential to the wire).

nsaspook
Delta2 said:
Yes I guess you are right here, inside the conductor the Poynting vector and the propagation of the EM wave changes direction and follows the radial direction so the EM wave is still transverse, while outside the conductor the Poynting Vector is along the direction of the wire (tangential to the wire).
Yes, exactly. There is a surface charge on the wires that causes the sudden shift in the direction of the E field. The B field doesn’t change similarly so the sudden change in the E field at the surface produces a sudden change in the direction of energy transport at the surface.

vanhees71 and Delta2
Byron Forbes said:
So the question is, is it reasonable to see this "electrical impulse" is traveling at c and think that makes perfect sense? I see no EM here, just pure electric fields, particles with mass, and energy.
Here we can see "magnetic impulse" traveling at slow speed. We see no EM here, just pure magnetic fields, particles with mass, and energy.

Delta2
jartsa said:
Here we can see "magnetic impulse" traveling at slow speed. We see no EM here, just pure magnetic fields, particles with mass, and energy.
Actually it is the speed of magnets that is slow. The speed of their interaction is equal to the speed of light.

We have EM waves produced by this arrangement of magnets too, only that the frequency of those waves is really small, approximately equal to the frequency of the mechanical oscillation of the magnets (they oscillate like a system of coupled simple pendulums ), that is a few Hz I believe.

Dale
nasu said:
...that the electric field already propagated all around the circuit by the time it would take one electron to travel one "inter-electron" distance at the drift velocity.
A.T. said:
Doesn't this depend on the length of the circuit?
nasu said:
No, the difference between the two velocities does not depend on the length of the circuit.
My question refers to your comparison of durations not velocities: Doesn't the time it takes the electric field to propagate all around the circuit, depend on the length of the circuit?

What is the capacitance of a long thin wire?
Because when electricity propagates in a wire, it is basically an electromagnetic wave, not through free space but along the wire which serves both as a capacitance and inductance. And the inductance of the wire is the magnetic part, right?
It´ s easy to estimate the capacitance of an isolated sphere, because the electric field diverges as 1/r2, which nicely integrates. But the field of a long wire would diverge as 1/r, which gets ugly when integrating...

Delta2
snorkack said:
But the field of a long wire would diverge as 1/r, which gets ugly when integrating...
It isn't that ugly, it integrates as ##\ln|r|## but the electric field outside a straight thin wire it falls as 1/r (and it also has additional terms 1/r^2 and 1/r^3) when the thin wire is actually a dipole antenna.

Say you have a wire 20 mm diametre and 5000 km long. And switch on 110 kV at one end. How do you estimate the total amount of charge that goes into charging the whole length of wire? How long before the other end is charged to 10 kV? To 100 kV? Will the other end overshoot past 110 kV, or approach with pure damping and without overshoot?

Delta2
snorkack said:
Say you have a wire 20 mm diametre and 5000 km long. And switch on 110 kV at one end. How do you estimate the total amount of charge that goes into charging the whole length of wire? How long before the other end is charged to 10 kV? To 100 kV? Will the other end overshoot past 110 kV, or approach with pure damping and without overshoot?
hard to answer your question exactly as it depends on many parameters (is the wire part of a transmission line like a coaxial cable, is it over good conducting ground e.t.c), but i believe a good approximation is that the other end will be at 110kv in time approximately equal to ##\frac{5\cdot 10^6}{3\cdot 10^8}## because we know that in wires the voltage waves travel at about the speed of light.

Delta2 said:
hard to answer your question exactly as it depends on many parameters (is the wire part of a transmission line like a coaxial cable, is it over good conducting ground e.t.c), but i believe a good approximation is that the other end will be at 110kv in time approximately equal to ##\frac{5\cdot 10^6}{3\cdot 10^8}## because we know that in wires the voltage waves travel at about the speed of light.
It´ s a bad approximation precisely because the waves necessarily travel slower. The question is how much slower? And does it travel as a "wave"? If it starts at one end of wire as a jump when switch is closed, does it arrive at other end in the same shape, or as a completely different shaped disturbance? In which case you cannot even define a specific speed.

yes ok, if it is part of a coax with dielectric then it travels slower, you also start talking about dispersion, ok.

There might not be a dielectric in the coaxial cable itself. But because of the 1/r divergence of electric field which only integrates as ln r, to which place does the capacitance of the wire go? To infinity 5000 km away (order of magnitude of total wire length) or just to dielectric ground 10 m beneath the wire?

snorkack said:
It´ s a bad approximation precisely because the waves necessarily travel slower. The question is how much slower? And does it travel as a "wave"? If it starts at one end of wire as a jump when switch is closed, does it arrive at other end in the same shape, or as a completely different shaped disturbance? In which case you cannot even define a specific speed.
We study that in power systems where the wires are very long. Switching surge is one case. Lightning strikes are another. Every study depends critically on the design parameters of the line.

The speed of propagation depends on the line parameters.

The shape of the pulses are greatly distorted as they travel.

There are many nonlinear effects, one of which is lightning arrestors.

It is hard to be more specific because every real case depends on the parameters.

Delta2 said:
Actually it is the speed of magnets that is slow. The speed of their interaction is equal to the speed of light.

We have EM waves produced by this arrangement of magnets too, only that the frequency of those waves is really small, approximately equal to the frequency of the mechanical oscillation of the magnets (they oscillate like a system of coupled simple pendulums ), that is a few Hz I believe.
It seems to me that those magnets move quite fast, much faster than electrons in a wire.

If we made the magnets trillion times stronger, and also gravity would be that much stronger, then the signal would move much faster through the line of magnets. I trust that J.C. Maxwell could calculate that speed, and it would be close to c.

If we wanted to make electric wire to be more like that series of magnets, we should make all segments of wire to have less charge. Which we can do by making the 'wire' to be a vacuum tube.

If we make the amount of charges in the 'wire' small enough, then a normal power source can make the charges move like the magnets in the video, or faster. The mass of electron would be important in that case.

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