Current of AC transmission lines and electricity in general

[jim hardy]

Working with the gas analogy, I get that pressure and kinetic energy (heat) are related to the potential of the gas to move to a more expanded state. (the expansion exerts a Force outwards)

again we have to be careful about applying analogies
i used pv because that's analogous to electrostatic force between charges=electric field intensity and it's potential energy, unlike temperature which is mv²/2 of individual particles and is kinetic energy.
As anorlunda points out electrons are so near massless they have zip for kinetic energy at drift velocities When they're hauling butt through a vacuum toward a pentode tube plate it's a different matter altogether. Look up secondary emission.

However, if there's a mass*acceleration of steam water moving over the fins of a turbine per second, so for the same energy transfer with a cooler mass there would have to be less moving over the fins? (over the same amount of time)

Why ? Less what - mass or volume ?
Impulse turbine works solely on the momentum change, moving blade absorbs kinetic energy of the steam . Moving blades have same Inlet and exit pressure . Reaction turbine's moving blade works on expansion too, allowing some enthalpy (U + pv) to turn into kinetic ½mv² which it then extracts just like an impulse stage. Its outlet pressure is lower than its inlet pressure because of the expansion that went on. . (competent mechanical engineers kindly correct me ?)
http://www.eolss.net/Sample-Chapters/C08/E3-10-03-02.pdf

Regarding Coulombs, I suppose the charge of approximately 6.242×1018 electrons is like mass, so Volts would have to be like the acceleration component, to effect the amount you can get through in the time interval?

i don't think that way at all.
Electrons are so near massless that at drift velocities they have ~zero momentum and ~zero kinetic energy. Think of them as dandelion puffs unaffected by gravity or momentum, but highly charged so they repel one another. That repulsion potential energy just as is a compressed spring.
Potential is how hard they're pushing against one another and against the insulation of the wire they're contained in.
Voltage is the difference between that potential and potential somewhere else like other end of a resistor in the circuit.
Introducing Time is a diversion.

 

[tim9000]

If the AC is a high frequency, then the modes described become important.
Just for completeness, regarding the radiated fields, of course these will travel away for ever as lost energy. The electric radiated component is at right angles to the longitudinal E field near the wire and so it can have a different value. These are the radiated and induction E fields encountered with antennas. However, it seems that the magnetic radiated component is wrapped around the wire and cannot be distinguished from the transmission line field, with which it is identical. So close into the wire, we cannot distinguish radiated and induction components of the magnetic field. The magnetic field from the part of the wire behind the impulse, where the current is steady, initially falls off with 1/D, but when the distance is about greater then the length of the wire, it will fall off very rapidly. The radiated component will continue to fall off slowly as 1/D for ever. This action of waves on wires seems to be very interesting.

Ah, so for a low frequency electricity isn't a TM, interesting.
How is there electric components in the direction of propagation and at right angles to propagation? Radiated EM has always interested me, I ignorantly assumed the HF radiated the magnetic component and as it switched there just had to be an electric component by default as the magnetic field radiated away. Probably childish of me...actually is this something like what you mean:

Is that what you mean by radiated E component perpendicular to the longditudinal E field, because that's what I did envisioned.
When you say the transmission line field, is that which energy snaps back into the wire after you stop exciting the wire? (as in, inductance)
But what did you mean by the "magnetic field from the part of the wire behind the impulse, where the current is steady"?

I really don't think you will ever do better than the best analogy ever invented - and that is Mathematics.

Very true!

What do you mean by “wires”. Use the term conductor rather than wire. The insulation on a wire is not conductive. The current moves on the surface of the conductive material. The energy is propagated through the air and insulation, guided by the surface of the conductors that make the full circuit.

For AC you must stop thinking about currents and fields inside conductors. Think only about the current as flowing on the surface of the conductor. Any AC current, or current induced EM field that enters the body of the conductor will be delayed and lost. Luckily, only a very small proportion of the induced surface current penetrates a good conductor because the conductive surface is like a highly reflective mirror to EM fields. Any magnetic field incident on a conductive surface induces a perpendicular current on the surface. That current generates a perpendicular counter magnetic field that comes very close to cancelling the incident magnetic field. Hence the reflection.

Sorry, conductor. I always imagine the thickness of the conductor being that, or less than, the skin-depth. Ah, so it's the electrons ONLY on the surface, that are responsible for radiation leaking out, that dose make sense. The power of free charges on a surface is one of the few things I did take away from my EM unit. Ah, so basically if EMR is incident on a surface, some counter EMR in according to Lenz's law?

Thanks all, I'll get to the rest of this thread tomorrow (hopefully).

 

[Baluncore]

Ah, so it's the electrons ONLY on the surface, that are responsible for radiation leaking out, that dose make sense.

No. The electrons on the surface stop the vast majority of the external EM fields getting into the conductor.
The electrical energy does not travel in or on the conductors. It travels between the conductors, guided by the surface of the conductors.

 

[tim9000]

well would that be mv2 , and does that X 0.5 give kinetic energy ?

Umm...to be honest, I don't know, I wouldn't have thought so, I was talking about voltage. at any rate, there was no /2, not that that is too big a deal I suppose.

again we have to be careful about applying analogies
i used pv because that's analogous to electrostatic force between charges=electric field intensity and it's potential energy, unlike temperature which is mv²/2 of individual particles and is kinetic energy.
As anorlunda points out electrons are so near massless they have zip for kinetic energy at drift velocities When they're hauling butt through a vacuum toward a pentode tube plate it's a different matter altogether. Look up secondary emission.Why ? Less what - mass or volume ?
Impulse turbine works solely on the momentum change, moving blade absorbs kinetic energy of the steam . Moving blades have same Inlet and exit pressure . Reaction turbine's moving blade works on expansion too, allowing some enthalpy (U + pv) to turn into kinetic ½mv² which it then extracts just like an impulse stage. Its outlet pressure is lower than its inlet pressure because of the expansion that went on. . (competent mechanical engineers kindly correct me ?)
http://www.eolss.net/Sample-Chapters/C08/E3-10-03-02.pdf

i don't think that way at all.
Electrons are so near massless that at drift velocities they have ~zero momentum and ~zero kinetic energy. Think of them as dandelion puffs unaffected by gravity or momentum, but highly charged so they repel one another. That repulsion potential energy just as is a compressed spring.
Potential is how hard they're pushing against one another and against the insulation of the wire they're contained in.
Voltage is the difference between that potential and potential somewhere else like other end of a resistor in the circuit.
Introducing Time is a diversion.

Interesting, so even though a reaction turbine isn't pretty to look at, it's more efficient than an impulse turbine?
Yeah well I suppose that makes sense, and is fair enough considering that they (electrons) are just probabilistic standing waves.
Hmm, I never thought of potential as being how hard they're pushing on each other, I always just thought of them like suspended in a gravitational field from an height. So on this train of thought, we could say that C*V is like holding C electrons at a height of V.
Anyway, so say P = I*V = (C*V / s) * V, what is that Columbs*Volts² per second, mathematical analogy/model actually telling us about the real world?

No. The electrons on the surface stop the vast majority of the external EM fields getting into the conductor.
The electrical energy does not travel in or on the conductors. It travels between the conductors, guided by the surface of the conductors.

Yeah so the free electrons are like an extremely shiny mirror to EMR, but I'm a bit confused. Are you saying my diagram of EM being emitted is incorrect? Because clearly there is a B-field around a live conductor, however I understand that radiation incident on a conductor would have a very hard time going in far, or at all.
I find that statement about it traveling only between conductors interesting because it's not something that I've ever considered to be the norm for run of the mill conductors. So what if I had an infinitely long conductor, that is, there is an infinite amount of space between the return path conductor, and the positive terminal of the source? Does that mean no energy or current would start to flow? Actually I'll give a different example:

Say I have an infinity long distance the return path between two charged parallel plates (I don't know if this is an appropriate example of an electric source), does the electric field have to radiate off forever before charge can flow?

Sorry, Probably a silly thought experiment but I'm trying to get my head around the fact that there is no E-field in the wire, yet a B-filed surrounding it.

Cheers

 

[Baluncore]

Say I have an infinity long distance the return path between two charged parallel plates (I don't know if this is an appropriate example of an electric source), does the electric field have to radiate off forever before charge can flow?

No. When a two wire line is connected to a signal source, a differential wave travels away from the source guided by the two conductors. That transient wave will roll on down the transmission line, being attenuated by resistive losses and radiation until what energy remains reaches the load. There can be many transients traveling in a train down the line at one time. There will also be transients traveling back up the line due to impedance mismatches on the line and at the load.
If the conductors are far apart the energy density will be sparse and some “antenna” type radiation will take place. If they are close together, the energy will be dense between the conductors. By keeping the conductors close the energy radiated can be reduced. A coaxial cable internalises the fields and reduces cross coupling of stray signals.

I have no idea what you are getting at with that diagram.
Were you of the opinion that energy could not flow until the signal had run all the way down one wire, then all the way back up the return wire ?

 

[tim9000]

Hi Baluncore,
[I meant to write 'Finite' not "Fite"; how embarrassing]

No. When a two wire line is connected to a signal source, a differential wave travels away from the source guided by the two conductors. That transient wave will roll on down the transmission line, being attenuated by resistive losses and radiation until what energy remains reaches the load. There can be many transients traveling in a train down the line at one time. There will also be transients traveling back up the line due to impedance mismatches on the line and at the load.
If the conductors are far apart the energy density will be sparse and some “antenna” type radiation will take place. If they are close together, the energy will be dense between the conductors. By keeping the conductors close the energy radiated can be reduced. A coaxial cable internalises the fields and reduces cross coupling of stray signals.

I have no idea what you are getting at with that diagram.
Were you of the opinion that energy could not flow until the signal had run all the way down one wire, then all the way back up the return wire ?

In my last diagram I was trying to illustrate...nevermind. Yeah I was thinking maybe the electric field had to fully circulate before the current can flow because I was thinking about all the electron charge stays spaced evenly (doesn't bunch up).

So what would (by your definition) be the distance between three phase transmission? I assume you'd say the conductors are far apart? (What, like at least a foot between each aerial bundle) And the conductors of your extension cord are close together (because the conductors insulation is butted up against each other).

Energy propagation of three phase electricity:

Say that in the above figure the waveform is something like LHS is peaking, the middle is bottoming out and the RHS is climbing.
So I assume it doesn't matter other than increasing cross coupling or inductance if the conductors are close or far apart (it doesn't matter if the field is dense or sparse) the energy is still the same and the current will flow just as well.
So in principle this is still different from a micro-strip or wave-guide, neither the electric or magnetic component is in the direction of propagation?
How does the electric current propagate forward if there is no source electric field longitudinally?
Thanks

 

[jim hardy]

Interesting, so even though a reaction turbine isn't pretty to look at, it's more efficient than an impulse turbine?

I don't think i said more efficient
just the reaction stage let's steam expand.

Hmm, I never thought of potential as being how hard they're pushing on each other, I always just thought of them like suspended in a gravitational field from an height. So on this train of thought, we could say that C*V is like holding C electrons at a height of V.

Ahh the water analogy - it falls to ground as it exits the garden hose .
Bernoulli

I switched my water analogy long ago to use the pressure term because
within the constraints of a circuit
electrons have so little mass they're insensitive to gravity
and so little velocity the middle term disappears (exception when they're in a beam as in a CRT)

I encourage people to use pressure not height because to allow gravity invites misconceptions of Ground. It's not gravity that pulls lightning down.

So on this train of thought, we could say that C*V is like holding C electrons at a height of V.

see above objection.

A Coulomb X a Volt is to me more like holding a mole of charges at one atmosphere.

Anyway, so say P = I*V = (C*V / s) * V, what is that Columbs*Volts2 per second, mathematical analogy/model actually telling us about the real world?

from this question earlier ?

(So surely the Voltage part of power is related to how fast the electrons move?...or accelerate...actually that doesn't sound right, electrons have a velocity, not an acceleration...? and so power = C*V^2 / s)

by V do you mean Volts or Velocity ?

How about if you go back to energy = ½mass X velocity²
enter mass of an electron in kilograms
energy of one electron volt in joules
and figure what is velocity
and compare that to typical drift velocity.

Then we'll be on common ground about volts and velocity .

 

[Baluncore]

Say that in the above figure the waveform is something like LHS is peaking, the middle is bottoming out and the RHS is climbing.

That is not possible with balanced 3PH as there is a 120° angle between phases. One phase cannot be high at the same time that another is low. The sum of all three line voltages will be zero. The sum of all three line currents will be zero.

How does the electric current propagate forward if there is no source electric field longitudinally?

The biggest voltage changes along the line are due to the transient wave traveling along the line. Losses along the line due to resistance also result in longitudinal voltage drops, but only in proportion to the energy lost in the line resistance. That is because each section of the line represents a small load in series with the intended output load.

 

[dlgoff]

DriftVelocity = I /(n*Q*A)

that;s unfmailiar to me.

Image compliments of http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html
note: not a hot link image :)

 

[jim hardy]

Image compliments of

Thanks Don

as I've said so often i don't trust formulas until all the terms are defined

 

[tim9000]

That is not possible with balanced 3PH as there is a 120° angle between phases. One phase cannot be high at the same time that another is low. The sum of all three line voltages will be zero. The sum of all three line currents will be zero.

Woops sorry, well I was envisioning something like:
https://www.google.com.au/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjCvvPPwdzNAhVDFpQKHSccD6wQjRwIBw&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FThree-phase&psig=AFQjCNGKxz6s2ax0cXuSTmVcMCwYrpPyxw&ust=1467814595778898
https://www.google.com.au/url?sa=i&...6s2ax0cXuSTmVcMCwYrpPyxw&ust=1467814595778898
Anyway current direction aside the important part was were the electric fields what you meant?

The biggest voltage changes along the line are due to the transient wave traveling along the line. Losses along the line due to resistance also result in longitudinal voltage drops, but only in proportion to the energy lost in the line resistance. That is because each section of the line represents a small load in series with the intended output load.

I know start up, and faults, and adding and removing large load would cause transients, but is that all you meant? Or are transients caused by more things and more frequent than I understand?
My point with that diagram is that, from the description you gave (what I tried to illustrate) how is there a longitudinal voltage at all, if all the energy is between lines at right angles to the line itself?

I'll get to the rest of this thread hopefully tomorrow!

 

[tech99]

Electrons are not like billiard balls. The kinetic energy of electrons is negligible, even when electric power transmitted is huge.[/QUOTE]

When an electron moves, it builds a magnetic field, so that energy has to be supplied to get it moving. It is as if it has additional, or virtual, mass, and the inertia which results from this must appear as inductance.

 

[Baluncore]

how is there a longitudinal voltage at all, if all the energy is between lines at right angles to the line itself?

The energy travels along the line, not at right angles to the line. It is the electric field that is at right angles. The wave is traveling along the line. For any sinusoidal component, half a wavelength away the polarity is reversed. That gives a voltage gradient along the line but it cannot be measured because the propagation velocity will also occur in the meter leads.

There are two waves on the wire. One is the forward traveling wave, the other is a backward traveling wave reflected by the impedance mismatch at the load or along the line. Those traveling waves do not interact in the line, but where we make measurements we measure the sum of both which appears as what we describe and imagine to be a standing wave pattern.

The line is a good conductor so it takes very little voltage drop to drive the current against the copper resistance. Each small fragment of resistance along the line costs some energy, reduces the traveling wave voltage and generates another small backward traveling reflected wave component necessary to keep the directional ratios of v/i = Z₀, the characteristic impedance of the line.

 

[tech99]

The energy travels along the line, not at right angles to the line. It is the electric field that is at right angles. The wave is traveling along the line. For any sinusoidal component, half a wavelength away the polarity is reversed. That gives a voltage gradient along the line but it cannot be measured because the propagation velocity will also occur in the meter leads.

There are two waves on the wire. One is the forward traveling wave, the other is a backward traveling wave reflected by the impedance mismatch at the load or along the line. Those traveling waves do not interact in the line, but where we make measurements we measure the sum of both which appears as what we describe and imagine to be a standing wave pattern.

The line is a good conductor so it takes very little voltage drop to drive the current against the copper resistance. Each small fragment of resistance along the line costs some energy, reduces the traveling wave voltage and generates another small backward traveling reflected wave component necessary to keep the directional ratios of v/i = Z₀, the characteristic impedance of the line.

It appears to me that line has not only resistance but series inductance, so that a voltage (E-field) is required in the longitudinal direction in order for a current to flow.

 

[Baluncore]

It appears to me that line has not only resistance but series inductance, so that a voltage (E-field) is required in the longitudinal direction in order for a current to flow.

The characteristic impedance of a lossless line is Z₀ = √ ( L / C ) where L is the inductance per unit length and C is the capacitance per unit length.
A lossy line has series L, parallel C and also series resistance. That results in longitudinal energy losses and reflection of energy to balance the V and I equations along the line. The resistive losses occur for both the forward and backward waves. Longitudinal voltage is occurring in both directions at the same time.
https://en.wikipedia.org/wiki/Telegrapher's_equations#Distributed_components

 

[tim9000]

Hi Everyone,
This is one of many threads I've been trying hard to get back to.

It appears to me that line has not only resistance but series inductance, so that a voltage (E-field) is required in the longitudinal direction in order for a current to flow.

Yeah, I really don't understand how 'there isn't'?
Is the B field Outside the coax (below) anti-clockwise? And is the B field Inside partially canceled by the current in the other direction?
https://www.google.com.au/url?sa=i&...f1iqT5Bq4uHISC9aOSveT-ig&ust=1471353705804123

@Baluncore

The characteristic impedance of a lossless line is Z0 = √ ( L / C ) where L is the inductance per unit length and C is the capacitance per unit length.

Which is presumably = V(t)/I(t)
Is this diagram (frame at the top) subject to your description:

Anyone, Please answer the questions in the frames at your leisure.

Thanks very much!

 

[Baluncore]

Is the B field Outside the coax (below) anti-clockwise?

There are no fields outside the coax due to signals inside the coax. The current flow on the outside of the inner conductor is equal and opposite to the current on the inside of the outer conductor. Any currents on the outside of the coax will be related to currents in the external environment, they form a separate transmission line with a different velocity factor to the inside surfaces of the coax.

And is the B field Inside partially canceled by the current in the other direction?

You must stop confusing the two quite independent directions of energy propagation on the line from/with the two equal and opposite surface currents on the conducting surfaces of the one line. Your question is ambiguous unless you make it clear what you are referring to and in which direction you are considering it.

 

[tim9000]

You must stop confusing the two quite independent directions of energy propagation on the line from/with the two equal and opposite surface currents on the conducting surfaces of the one line. Your question is ambiguous unless you make it clear what you are referring to and in which direction you are considering it.

Sorry about the ambiguity, I'll try again. So if I was looking just at the outer conductor in the picture, with the current going out of the page, there would be a magnetic field anti-clockwise; however you said that the current is only flowing on the Inside of the outer conductor, so there is no magnetic field outside the outer conductor. So I am wondering why there is still a magnetic field inbetween the two conductors? Because the right hand rule for the inner conductor of the picture is shown, so I'm wondering why there is no magnetic field due to the outside conductor.
As for the confusion of direction of energy propagation, I wasn't aware I was doing that, but I concede that I may have been. Are you implying that if the magnetic field WAS canceled inside that there wouldn't be any energy propagation?

What about the top frame of the second picture, with E in blue and B in red, was that what you'd previously meant when we were talking about a two wire conductor (circuit)? [What you meant by the E field being in between, and at right angles?]

Cheers

 

[sophiecentaur]

You can't just look at the outer because there has to be a return path for current to flow. The 'corkscrew rule' tells you that the magnetic fields add in the inside and cancel on the outside. Look at the diagrams further back in this thread. E fields are zero outside too. A

 

[Baluncore]

Are you implying that if the magnetic field WAS canceled inside that there wouldn't be any energy propagation?

What about the top frame of the second picture, with E in blue and B in red, was that what you'd previously meant when we were talking about a two wire conductor (circuit)? [What you meant by the E field being in between, and at right angles?]

The inside of the outer conductor is a excellent conductor. That makes it a good EM mirror. Any magnetic field incident with the inside surface will induce a perpendicular longitudinal current that will create a reverse magnetic field to cancel the incident field. Those two fields cancel everywhere outside the inner surface of the outer conductor.
You are hypothesising that the magnetic field inside the transmission line is caused by a current flowing “in” the conductor. But that current was induced in that mirror surface by an incident internal magnetic field. You are forgetting that the EM field within a coaxial cable is constrained to the dielectric by the surface currents on the reflective “walls”.
When you connect a transmission line to a signal you are connecting the dielectrics and the conductive reflective walls. You do that by making sure that longitudinal surface currents and electric field between the surfaces will not be interrupted at the interface.

Hypothesising your own personal theory is a most inefficient way of approaching the subject. It wastes your time and the time of those who answer your questions. You question the paradigm by proposing a simpler model of reality that does not fit the paradigm. You then call for someone to say why your poorly specified model is wrong. That requires they understand your immature model, which is different to their reality. It is better to read and understand the physics than to learn in an ever changing feedback loop where communications in the English language gets two chances to be misunderstood per cycle.

 

[sophiecentaur]

The inside of the outer conductor is a excellent conductor. That makes it a good EM mirror. Any magnetic field incident with the inside surface will induce a perpendicular longitudinal current that will create a reverse magnetic field to cancel the incident field. Those two fields cancel everywhere outside the inner surface of the outer conductor.
You are hypothesising that the magnetic field inside the transmission line is caused by a current flowing “in” the conductor. But that current was induced in that mirror surface by an incident internal magnetic field. You are forgetting that the EM field within a coaxial cable is constrained to the dielectric by the surface currents on the reflective “walls”.
When you connect a transmission line to a signal you are connecting the dielectrics and the conductive reflective walls. You do that by making sure that longitudinal surface currents and electric field between the surfaces will not be interrupted at the interface.

Hypothesising your own personal theory is a most inefficient way of approaching the subject. It wastes your time and the time of those who answer your questions. You question the paradigm by proposing a simpler model of reality that does not fit the paradigm. You then call for someone to say why your poorly specified model is wrong. That requires they understand your immature model, which is different to their reality. It is better to read and understand the physics than to learn in an ever changing feedback loop where communications in the English language gets two chances to be misunderstood per cycle.

I'm not sure what you're implying here. Are you suggesting that the internal B field is eliminated due to reflections? (perhaps not) There is a B field inside the 'cavity, which is circumferential (Transverse). There are a load of links that show how this can be calculated using Ampere's Law and it is finite at the inner surface of the outer conductor and at the outer surface of the inner conductor. This link shows it in detail but there is a graph showing the results of the calculations and the way the field varies with the radius. It shows that the external field is zero and that it exists throughout the conductors (depending on the resistivity)
If tim looks at the link, above, he will see the actual situation and will not need to try an alternative personal approach.

 

[tim9000]

HI, thanks for the replies;

You can't just look at the outer because there has to be a return path for current to flow. The 'corkscrew rule' tells you that the magnetic fields add in the inside and cancel on the outside. Look at the diagrams further back in this thread. E fields are zero outside too. A

I am surprised that I'm getting the corkscrew rule wrong, could you please give me post number in the thread, to specifically refer to?

The inside of the outer conductor is a excellent conductor. That makes it a good EM mirror. Any magnetic field incident with the inside surface will induce a perpendicular longitudinal current that will create a reverse magnetic field to cancel the incident field. Those two fields cancel everywhere outside the inner surface of the outer conductor.
You are hypothesising that the magnetic field inside the transmission line is caused by a current flowing “in” the conductor. But that current was induced in that mirror surface by an incident internal magnetic field. You are forgetting that the EM field within a coaxial cable is constrained to the dielectric by the surface currents on the reflective “walls”.
When you connect a transmission line to a signal you are connecting the dielectrics and the conductive reflective walls. You do that by making sure that longitudinal surface currents and electric field between the surfaces will not be interrupted at the interface.

Hypothesising your own personal theory is a most inefficient way of approaching the subject. It wastes your time and the time of those who answer your questions. You question the paradigm by proposing a simpler model of reality that does not fit the paradigm. You then call for someone to say why your poorly specified model is wrong. That requires they understand your immature model, which is different to their reality. It is better to read and understand the physics than to learn in an ever changing feedback loop where communications in the English language gets two chances to be misunderstood per cycle.

That is harsh but pretty fair, but in my defense I didn't realize I was hypothesising my own personal theory, I thought I was just applying the corkscrew rule an illustration on the internet, I didn't realize that I was being overly reductionist and that it required a deeper level of understanding, hence I mistakenly expected a more simple answer than could be given [I do completely agree with the whole 'introducing two misunderstandings per cycle']. (also according to @sophiecentaur I'm applying it wrong, which doesn't help)
So I'm a little bit lost here "But that current was induced in that mirror surface by an incident internal magnetic field. You are forgetting that the EM field within a coaxial cable is constrained to the dielectric by the surface currents on the reflective “walls”.
When you connect a transmission line to a signal you are connecting the dielectrics and the conductive reflective walls. You do that by making sure that longitudinal surface currents and electric field between the surfaces will not be interrupted at the interface."
So are you saying that the current in the surface of the two conductors is caused by the magnetic field, and not the electric potential of the source? Okay, so the EM fields are contained within the dielectric, but what do you mean by 'interrupted at the interface'?

If tim looks at the link, above, he will see the actual situation and will not need to try an alternative personal approach.

Yeah I think I will need to look at that link to figure out what is going on with the B fields and how they are cancelling. Thanks

The bottom picture in my post #46 (particularly the top frame) I was trying to illustrate my interpretation of something(s) you said earlier:

It travels between the conductors, guided by the surface of the conductors.

And I was wondering if you could please comment on that.

There is a voltage difference between the conductors. So between the conductors we also have an electric field. The E and M fields are perpendicular. The cross product of those two fields gives the poynting vector which is the direction of energy flow. That direction is from the generator to the load.

When we measure the rate of energy transfer from generator to load by multiplying the current by the voltage to get power in watts, we are in effect (cross-)multiplying the electric and magnetic fields.
The energy travels through the insulation and air between the wires, guided by currents in the surface of the wires.

Yeah I think I more or less understand that, (and I am fascinated by the realisation that it is in essence the cross-product), and so (in the second half of that picture I drew in #46) I wanted to make a sort of hypothetical of well if the energy travels between them, what if they (the conductor and return conductor) were infinitely far spaced away from each other, but the source was a finite distance from the load, would energy ever actually be able to flow towards the load?Cheers

 

[sophiecentaur]

(also according to @sophiecentaur I'm applying it wrong, which doesn't help)

I have to backpedal here a bit. With a two wire feeder the B field between the two conductors will have contributions from both conductors but, of course, with a continuous outer, there is no contribution to the B field inside the cavity - if you consider the continuous outer as a number of parallel conductors with equal currents flowing in them, the 'corkscrew rule' will cancel the fields from diametrically opposite conductors. But Ampere's Law gives you the answer in a more pukka way.

what if they (the conductor and return conductor) were infinitely far spaced away from each other,

That's not an acceptable model to work with as you cannot connect a source or load in a conventional way. Your suggestion makes me think of radio communication between two antennae.

 

[Baluncore]

Are you suggesting that the internal B field is eliminated due to reflections? (perhaps not)

No, I'm saying there are no fields from inside the coaxial cable getting outside because the walls are mirrors. At a conductive surface the transmitted wave is canceled so it does not exist outside the coax.

 

[sophiecentaur]

No, I'm saying there are no fields from inside the coaxial cable getting outside because the walls are mirrors. At a conductive surface the transmitted wave is canceled so it does not exist outside the coax.

Yes, I see that now.
Also, Ampere's Law says what happens in any of these cases.

 

[tim9000]

I have to backpedal here a bit. With a two wire feeder the B field between the two conductors will have contributions from both conductors but, of course, with a continuous outer, there is no contribution to the B field inside the cavity - if you consider the continuous outer as a number of parallel conductors with equal currents flowing in them, the 'corkscrew rule' will cancel the fields from diametrically opposite conductors. But Ampere's Law gives you the answer in a more pukka way.

That's not an acceptable model to work with as you cannot connect a source or load in a conventional way. Your suggestion makes me think of radio communication between two antennae.

So with the conventional two wire feeder circuit (as I tried to illustrate in the top half of the bottom picture in #46) is that how the EM energy travels down the line? The B and E field vector orientations (if that is the right terminology) and so the cross product is down the line.

Ah yes, I see what you're both talking about with the continuous outer coax, thanks!

 

[sophiecentaur]

and so the cross product is down the line.

For a two wire feeder, there will have to be a finite radiation resistance, I think and this implies that the Power is mainly flowing down the line (TEM) but there will also be a slight divergence of the Pointing Vector field to account for the leaked power into free space. I guess that the E and B fields will have a small but finite in phase component.

 

[tim9000]

I was more interested in if the diagrammatic arrow representations of how you could think of the energy transfer component of the E and B actually looked, in the picture were accurate.
(also, what would this 'divergence' due to leaking power 'look like'?)

I think and this implies that the Power is mainly flowing down the line (TEM)

But I posed this before in about #12 and:

Why confuse things by using waveguide mode classification when phase velocity is not relevant because the distance between the wires is very very much less than a wavelength. EM radiation in space is TEM, just as are the fields between the good conductors that make up a circuit.

So are you saying that the conduction of electricity IS IN FACT a TEM model?

Cheers!

 

[sophiecentaur]

So are you saying that the conduction of electricity IS IN FACT a TEM model?

Why not? The E field is normal to the wire, the B field is in circles round the wire and the power flows along the wire. The fields are (at least nominally) transverse to the direction of power flow. (We ignore the forward component of the E field, which seems to get so many people steamed up because they think that field component is relevant)

(also, what would this 'divergence' due to leaking power 'look like'?)

Without bothering to actually draw it, I can describe it as having the Poynting Vector S parallel with the wires in the centre and very slightly spreading 'outwards' at locations near the wires. The divergence will be small as the radiation resistance is very low. If the wires were significantly resistive, part of the power would be 'into the wires' (more divergence, I guess) so the total flux (between the wires) would get gradually less and less. The explanation for this divergence is that the speed of propagation near the resistive wire (or the radiation leaves from sections of it) is a bit slower. The same effect can be seen when a Vertically Polarised Medium Frequency signal travels along the ground (a ground wave) and it is constantly 'dragged' and dissipated by the ground, giving a slight forward slope to the E field, directing Power down into the ground. It keeps the level of received signal higher than you'd expect because energy above the ground gets directed downwards. The 'shadow' of an MF signal, produced by the presence of metal buildings in a city, gets repaired by this mechanism and the received signal beyond the city gradually recovers.

 

[Baluncore]

So are you saying that the conduction of electricity IS IN FACT a TEM model?

How many different waveguide modes are possible on a coaxial cable?