A generator or mains supply does not provide an EM field at its terminals it provides voltage and or current. These come first.

I find this misleading, and utterly devoid of meaning without a detailed explanation or reference to support it.

Which voltage and which current comes first? Are you saying voltage and current comes before the fields as if they are a source of the fields?

A generator works via Faraday's Law. That law says the the emf is equated with negative rate of change of magnetic flux. One is not the source of the other, but they are fields that exist hand in hand. The emf is the line integral of the electric field. There is no generation of emf without fields.

Or maybe you are talking about where the magnetic field came from? That could be due to a current. A permanent magnet is something like a field due to current at the atomic level. Or a generator with field windings needs current. Or, an induction generator needed an initial rotor current to get the induction process started.

There is just too much detail washed over by your statement and you just refuse to provide any external reference to support your view. If it is your own view, then at least state that, and provide a detailed explanation of your viewpoint.

The closest thing I've seen to what you are saying is that many field books describe charges and currents as the sources of fields. I think this is a useful viewpoint for anyone trying to understand fields, but even this view can be criticized. For example at a more fundamental level, electrons and other particles have charge and field and magnetic moments. Also, the photon is the force exchange carrier for the electric force. How do you separate any of this and say that voltage and current come first?

There is also another field approach where the vector potential A and scalar potential V are used in place of the magnetic and electric fields. This means that they are different viewpoints of the same thing. So again, how do you say voltage comes before electric field? It's too simplistic to think that way, or at least the statement begs for support.

 

That was a suprisingly unmild reply for the SteveB I recognise.

The OP asked for confirmation the wavelength of a 60 hz signal in copper wire.

Unfortunately he used the wrong velocity in copper wire in his calculations`.

I posted the correct version in post31.

Meanwhile several respondents pointed out (correctly) the most AC mains power is actually via the (edit: external) EM fields that are generated by the passage of current the the wires.
The velocity of signals propagating in these fields is indeed c, or pretty close to it.

If you would like the calculations that led to post 31 I will try to find the time to write them out.

go well

 

If you would like the calculations that led to post 31 I will try to find the time to write them out.

I don't intend my "unmild" sounding post to be disrespectful, and i apologize if it comes across that way. I do feel some (friendly) frustration though because I'm genuinely trying to understand your points. But, for whatever reason (perhaps my own fault) I can't follow exactly what you are trying to say.

I don't ask for the calculations, I'm just curious about the justification and logic for some of your statements. Perhaps I'm misinterpreting some of them.

For example, on this point of the copper wave speed, why do we care about the wave propagation speed in copper? I don't see where the OP brought up the copper at all. Perhaps I missed it, but I don't feel like trying to go back and find it. Please identify the post if you think this is important for understanding your comments. I know you brought it up in #31, but at the time it wasn't clear to me why you brought it up.

For a propagating wave, the wave is not trying to propagate into the copper but is traveling along the copper with the copper providing a boundary condition.

The propagating wave has a speed that depends on all materials that the wave is traveling in. In an approximate sense, it is a weighted sum relating to the proportion of power within that medium. Since only a small exponential tail of the wave exists due to the boundary condition at the copper surface, the wave speed and propagation constant are closer to that of the air and insulation of the cable. For a coax, it is primarily the dielectric that controls the speed. For a twin wire, the air and insulator combine to produce an effective wave speed. For a single wire that uses a ground plane, the air may be much more important than the insulator. In all cases, the fact that the conductor provides a conductive boundary condition is the most important consideration. The field distribution and propagation constant are often solved assuming the conductivity is infinite, simply because the it's easier and gives the same answer, and then a small loss term can be added to the propagation constant as needed.

The losses due to the copper significantly affect the wave attenuation much more so than the effect on wave propagation speed. Since the air and/or dielectric are good insulators with low loss, the losses of in the copper are a major contributor to the wave loss.

 

It is funny, during my sophomore class on antennas they used 96%c. No math was given.

 

It is funny, during my sophomore class on antennas they used 96%c. No math was given.

Yeah, it's one of those things that is easier to measure, than to calculate. That goes for both attenuation loss and for wave speed. I remember doing the wave speed measurement on RF coax, in a junior level class. The transmission line was quite short, and a 100 MHz analog O-scope was more than sufficient to pick up the reflection off of the shorted load end. A pulse with a very fast rising edge was put in and the return pulse was quite obvious. In this case, it was a coax with dielectric such that the wave speed was 2/3 the speed of light which gives a 5 nm/m delay (obviously, propagation length is twice the line length with a reflection).

The loss measurement is quite easy too, although I've only done that one on fiber optic cable over many kilometers of length.

One can do measurements with high confidence. A calculation will leave you wondering if you made a mistake, and even without a mistake, you will not be very accurate.