Author Topic: Questions for those who know electromagnetism better than I do  (Read 7171 times)

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Offline T3sl4co1l

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Re: Questions for those who know electromagnetism better than I do
« Reply #25 on: August 27, 2021, 07:46:47 pm »
With that, you have the causal relationships all upside down. You'd be right if you only wanted to describe the historic sequence and linkage of discoveries, but as a causal chain of nature's features, it does not make sense.

Yes, my point was wholly historical, as is obvious from context, and as you so ably indicated.

I don't understand the confrontational tone, though.  I see a similar approach in other threads as well.  Have you had a stressful week?  Perhaps a vacation is in order?  They're just forum posts, you don't have to take every little thing personally. :-+

Tim
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Offline EPAIII

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Re: Questions for those who know electromagnetism better than I do
« Reply #26 on: August 28, 2021, 11:16:02 am »
I will have several points to comment on but have chosen only one previous post to quote. I hope that is not confusing. OK...

Re: Your paragraph which begins with "As with the Earth:". I think perhaps you confuse the three dimensional integration of gravitational force where a solid body (planet, moon, or star) is divided into layers and those layers into rings and those rings into segments and where each of those 3D segments is assumed to have the same density; with a situation where the density of such segments, of such rings is not uniform. When the density can be considered uniform, then YES, the integration does yield a result where the total mass in any one of those layers (or any two rings that are mirror images of each other about a plane passing through the center) can indeed be considered to act as if it were all concentrated at the center of the solid body. That is a classic problem in undergraduate mechanics and I well remember performing it, though many years ago. But if the density is NOT uniform, then the balance in the integration process is destroyed and the mass in that layer or mirrored pair of rings does NOT work out to be equal to the same mass being concentrated at the center. A simple and practical example of this is the detection of various ores and other deposits beneath the earth's surface by using a sensitive gravitational measuring device. Another example of this can be seen by examining the orbit of the planet Pluto (OK, non-planet Pluto - I hated that demotion for poor Pluto). At any given time, most or all of the other planets are closer to the Sun than Pluto is and can be said to be inside of a sphere defined by the Sun at the center and Pluto as a point on it's surface. If your statement were correct, then the masses of the individual planets and moons and other bodies within that sphere would all act upon Pluto as if they were concentrated at the center (in the Sun) and would not produce permutations in Pluto's orbit. But it is a well known fact that the orbit of Pluto (actually the orbits of all the bodies in the Solar System) are influenced by the other bodies, including the ones that are closer to the Sun. These permutations in the orbits have been well known for well over 100 years. So the idea that the influence of the mass within the Earth can only be assumed to act as if all of it were at the center of the Earth is only valid if the Earth is completely uniform in density, which it is not.

Now, the magnetic flux (field strength) in a closed path surrounding a conductor. I think it is important to differentiate between two types of closed paths that can surround a conductor with a DC current. One such closed path would be the classical path of equal field strength. This is what we envision when we see the classic pattern of iron filings on a piece of paper that the conductor passes through and which are called by the simplistic phrase "magnetic lines of force". This type of path is, by definition a path of uniform field strength. The other type of closed path is an arbitrary one which can have any shape or size. It may be close to the conductor at one point and very distant at others. In a closed path of this type there is absolutely no reason to expect that the field strength will be the same at all points. In fact, it would be a trivial exercise to construct a path where the magnetic field even REVERSES direction as you travel along the path. You can have a large number of such reversals as long as that total number is even. The magnetic field strength can even go to zero as would be required at the point where such a reversal takes place. And Maxwell's laws do take this into account. It is important to realize that this second type of closed path may be perfectly circular. But for any point outside the conductor, you can construct an infinite number of such circular paths all of which have the conductor completely within them.

"So, too, the magnetic field around a wire, at a given distance from center, is independent of the wire diameter, so long as the wire radius is smaller." I fear that you did not express whatever you wished to say very well. The radius of the wire is always smaller than the diameter.

That brings me up to that wonderful idea of "skin effect" or "surface effect". One definition of this seems to hinge around the idea that higher frequency E-M fields do not penetrate as far into a conductor as lower frequency ones or non-varying fields (DC). While this is certainly correct and is one of the reasons why coaxial lines can carry higher frequency signals and wave guide, which has no center conductor, can carry even higher frequency ones, to my mind it is not the only effect that can be labeled "skin" or "surface". Moving charges, which constitute an electrical current, will be subject to electrostatic attraction and repulsion. In the case of a current, these charges are usually all the same polarity and will therefore repel each other. Therefore, even a DC current flowing in a solid conductor will, to some extent, exist as a stronger current near the surface of that conductor and as a weaker one at it's center. Actually, you do not even need to have a current flow to see this effect in action. A Van de Graaff generator accumulates a negative charge on it's hollow metal globe by picking the small negative charges off the moving belt with a conductive brush near the center of that globe. This is only possible because those negative charges (electrons) then repel each other and seek the greatest separation which can then be found on the surface of the globe (sphere). So the outer surface of the globe has a constantly increasing negative charge while the brush, which contacts the belt near the globe's center, actually has a positive charge which attracts even more electrons from that belt. This is one heck of a powerful demonstration of a DC, NO a STATIC "skin effect" or "surface effect".

I therefore argue that NO conductor, even one with a DC current, will have a uniform distribution of the moving charges (electrons) within that conductor. Due to electrostatic repulsion, they will always prefer to congregate near the surface as they travel down the length. A square, a triangular, and best of all a star shaped conductor will have higher densities of current near the outside corners than at other places on the surface and higher current densities at the surface than in the interior or center.

Taking this a bit further I can imagine a conductor shaped like a one pointed star where the cross section would be more or less tear-drop shaped. (Draw a circle and a point outside of it at a distance r from it. Then draw two tangent lines to the circle and through that point.) There would be a much higher current density at that single point than at any other place on or in the conductor. Now if one were to draw a path that was at a constant distance above the surface of this conductor (also tear-drop shaped), I would fully expect, JUST LIKE THE EXAMPLE OF GREATER GRAVITATIONAL FORCE ABOVE DENSER ORE DEPOSITS IN THE EARTH, there would be a greater magnetic field density at a point on that path adjacent to the point of the tear-drop and a lower field density at a point on the opposite side. You can not simply draw a circular or any arbitrary shaped path around this tear-drop shaped conductor and expect the magnetic field to be the same at all points on that path.

And I will leave the precise calculations to someone else. Have fun.



Well, I wouldn't disregard theory out of hand... I certainly have a good grasp of it.  Granted, people call me smart, but I'm no Feynman either.

Regarding wire diameter, the principle is Ampere's law, in same class as Gauss's law and so on, which state that the sum flux through a boundary only depends on what's enclosed within that boundary.  In the Amperian case, it's a closed loop curve, the enclosed area of which carries a current, and the flux is the magnetic field parallel to the path.  Gauss's law applies to electric charge contained within a shell, the flux of which (electric charge) extends through, perpendicular to the shell; but also to gravity for another example, where the acceleration depends on the mass contained within a shell.

As with the Earth: if we could stand at the same altitude (distance from center), and imagine all its mass compacted into an infinitesimal point, we would experience the same acceleration as with all that mass distributed as it is currently.  All that matters is that the mass is contained within a spherical shell beneath our feet.  (The exact shape of the shell, does and doesn't matter, due to more rigorous conditions that I won't go into detail about; suffice it to say, this works best when the shell shares the same symmetry as the problem, i.e. a spherical shell for a spherical mass distribution, and the shell is larger than said distribution.)

So, too, the magnetic field around a wire, at a given distance from center, is independent of the wire diameter, so long as the wire radius is smaller.

And as long as the wire is cylindrically symmetric, the law still applies within it; the magnetic field drops smoothly from its value at the surface of the wire, down to zero at the center.

(The story is different for AC, where self-induction within the wire causes an opposing magnetic field, cancelling out the internal field: skin effect.  Needless to say, this is more complicated, so I won't go into detail; just to note there's interesting things happening when wires or currents are changing over time.)


Which also gives us the tool to understand the other conditions.
Hollow: no field inside.  In effect, all the surrounding currents cancel out here, but we would need a lengthy calculation to show it that way!

Other (non-circular) cross sections: same outside the highest peak, and same inside the lowest valley.  The exact field now depends on angle as well as distance from center, so we can't say what the field is, at every point, so simply.  We might need to use a numerical solution in this case (e.g. Biot-Savart law, replacing the integral with a Riemann sum).

Tim
« Last Edit: August 28, 2021, 11:42:46 am by EPAIII »
Paul A.  -   SE Texas
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Online iMo

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Re: Questions for those who know electromagnetism better than I do
« Reply #27 on: August 28, 2021, 12:01:52 pm »
..I therefore argue that NO conductor, even one with a DC current, will have a uniform distribution of the moving charges (electrons) within that conductor. Due to electrostatic repulsion, they will always prefer to congregate near the surface as they travel down the length. A square, a triangular, and best of all a star shaped conductor will have higher densities of current near the outside corners than at other places on the surface and higher current densities at the surface than in the interior or center.
..
People say the electrons (aka particles) inside a conductor (ie copper) move randomly without DC or AC current. They drift slooowly (like xx millimeters per second) when talking DC current (single digits Amps). With AC they do not even drift, rather oscillate back and forth with zero average drift. Thus it seems that everything we see in our electronics is related to the fields the electrons induce (aka waves).
I still have hard time to imagine myself how you could, with a 10ns DC (ok a pulse has a lot of AC content) pulse, for example, congregate electrons (aka particles carrying a charge) near the surface of a copper rod conductor of, say, 20mm diameter.. :-[
Edit: fixed typos
« Last Edit: August 28, 2021, 02:40:07 pm by imo »
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Offline T3sl4co1l

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Re: Questions for those who know electromagnetism better than I do
« Reply #28 on: August 28, 2021, 02:34:51 pm »
Re: Your paragraph which begins with "As with the Earth:". I think perhaps you confuse the three dimensional integration of gravitational force where a solid body (planet, moon, or star) is divided into layers and those layers into ...

Yes; plain language is horribly imprecise.  I'll also admit it's been a while since I worked problems with the various field equations directly.

The one "out" that I will claim, is to reference the actual things in question: if you want a technical definition of Ampere's or Gauss's law, there are far, FAR better places to find it than from informal forum posts.  Please don't take my word for it, or yours or anyone else's here, for gods' sake... :-DD

Wikipedia is a surprisingly good start on these most well known of laws/theorems, and once one understands the underlying math, the conditions and edge cases will also be apparent (which I glossed over intentionally).

I suppose the one failing is I should've made this "out" more explicit. :)


Quote
Now, the magnetic flux (field strength) in a closed path surrounding a conductor. I think it is important to differentiate between two types of closed paths that can surround a conductor with a DC current. One such closed path would be the classical path of equal field strength. This is what we envision when we see the classic pattern of iron filings on a piece of paper that the conductor passes through and which are called by the simplistic phrase "magnetic lines of force". This type of path is, by definition a path of uniform field strength.

If you'll permit this one indulgence -- strength is "line" density, not the path itself.

I never liked the idea of "magnetic lines of force" anyway.  There's no such thing as a line, you can't count them (no matter how much a certain cgs system wished it were the case :P ).  It's terribly hard to work with (have you ever tried deriving actual field line curves?!), and leads to a lot of easy mistakes -- perhaps another illustrative example has happened here.

The only thing they have going for them, I suppose, is exactly that, the intuition from metal filings.  I suppose it's the more interesting of the effects that are easily observed from such an experiment; but the more meaningful observation really should simply be that more filings are attracted to the poles, where field strength is higher.  Alas, we can't change history, only learn from it.

Another poor example for lines, is spinning a magnet on its axis: this does absolutely nothing, as it happens -- but if you are imagining a bundle of lines trapped in it, surely they must be twisted up by the motion, right?


Quote
"So, too, the magnetic field around a wire, at a given distance from center, is independent of the wire diameter, so long as the wire radius is smaller." I fear that you did not express whatever you wished to say very well. The radius of the wire is always smaller than the diameter.

Too many pronouns, sure.

To be clear, by "independent of wire diameter", I mean, has no dependency on it -- it doesn't matter that it's the diameter or radius, there's no proportion, the factor of 2 is irrelevant.  And the wire radius must be smaller in relation to the radius of the curve, because obviously if some current is outside the loop you're checking, it's not enclosed, so you'll be missing that amount, and now the geometry and relative dimensions matter.


Quote
Moving charges, which constitute an electrical current, will be subject to electrostatic attraction and repulsion. In the case of a current, these charges are usually all the same polarity and will therefore repel each other. Therefore, even a DC current flowing in a solid conductor will, to some extent, exist as a stronger current near the surface of that conductor and as a weaker one at it's center.

Not quite.  There is no imbalance of charge -- the wire is a mass of ions (atoms sans some electrons) bathed in an electron gas (the remaining electrons, such as to make the total charge neutral, of course).

Give or take exactly which electrons you assign as "gassy", i.e., how "ionized" the atoms are.

For which there are theoretical derivations, if it really matters.  Which, it certainly does for the study of semiconductors, but suffice it to say for present purposes, metals are full of electrons, while being extremely neutral overall.

There is a surface layer, associated with ambient electric fields; metals are polarizable.  Since surface charge is supplied from the abundant bulk, this change occurs rapidly (essentially speed of light*).  The DC field due to current flow in a conductor, is extremely small (mV/mm if you're pushing it hard?) so there is essentially negligible change in the surface distribution of charges; and anyway, that change only occurs once, so amounts to exactly zero DC current flow.  Also, the charge density at the surface versus the bulk, is essentially identical; the electron concentration is mind-numbingly vast in metals, and literally a handful of extra electrons at the surface, amounts to nothing.


*If you're applying a fast enough varying electric field that AC effects matter, then you can draw the equivalent circuit and that suffices for analysis: namely, there is some displacement current due to the varying electric field, which in turn draws a current through the metal, which at AC, largely occurs at the surface due to the shielding effect of self-induction (skin effect).

Note that the EM skin effect is a depth corresponding to frequency and material properties, whereas a completely different "skin effect" applies to the distribution of DC surface charges (within a Debye scattering length, I believe?).

*Also, mind that speed of light is a local property.  It changes rather suddenly at and below the surface of a metal (indeed, which is an equivalent way to describe skin effect).


But this does explain when you should expect the current density to vary: if we use a semiconductor instead, its intrinsic (or lightly doped, as the case may be) carrier concentration is very small, and we can indeed saturate its capacity without much trouble (i.e., it doesn't melt or anything, indeed for low enough concentration, it needn't perceptibly warm up at all).  Yet if we apply an electric field to the surface, that will attract even such meager charge carriers as are present (and repel their opposites, electrons/holes), so as to try to maintain charge neutrality -- the electric force is a very strong one -- and this can happen to such an extent that charge density increases markedly.  Congratulations, you've invented a field-effect transistor. :)  Another critical aspect: notice the fields are crossed, E perpendicular to J.

So, this is also to say: metals simply don't transist.  Which seems obvious enough I guess, but it took quite a bit of condensed-matter physics to finally prove what's going on; from a historical perspective say the early 20th century, this wasn't obvious at all.  Heck, the notion of the electron itself was still very new back then.

(Mind, not to imply your state of knowledge is dated or anything; just as an indicator of what level of progress these facts are built upon.)

Cheers!

Tim
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Offline CatalinaWOW

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Re: Questions for those who know electromagnetism better than I do
« Reply #29 on: August 28, 2021, 06:30:43 pm »

Quote
Moving charges, which constitute an electrical current, will be subject to electrostatic attraction and repulsion. In the case of a current, these charges are usually all the same polarity and will therefore repel each other. Therefore, even a DC current flowing in a solid conductor will, to some extent, exist as a stronger current near the surface of that conductor and as a weaker one at it's center.

Not quite.  There is no imbalance of charge -- the wire is a mass of ions (atoms sans some electrons) bathed in an electron gas (the remaining electrons, such as to make the total charge neutral, of course).

Give or take exactly which electrons you assign as "gassy", i.e., how "ionized" the atoms are.

For which there are theoretical derivations, if it really matters.  Which, it certainly does for the study of semiconductors, but suffice it to say for present purposes, metals are full of electrons, while being extremely neutral overall.

There is a surface layer, associated with ambient electric fields; metals are polarizable.  Since surface charge is supplied from the abundant bulk, this change occurs rapidly (essentially speed of light*).  The DC field due to current flow in a conductor, is extremely small (mV/mm if you're pushing it hard?) so there is essentially negligible change in the surface distribution of charges; and anyway, that change only occurs once, so amounts to exactly zero DC current flow.  Also, the charge density at the surface versus the bulk, is essentially identical; the electron concentration is mind-numbingly vast in metals, and literally a handful of extra electrons at the surface, amounts to nothing.


Tim

This is the discussion of why the current density is constant over a surface perpendicular to the wire, which I said earlier is often omitted.  For good reason.  These problems usually show up in first year courses in EM(either freshman year or sophomore year in US colleges), while the discussion of how charge is actually transported in various materials is held until more tools are available to the student.  It has been multiple decades since I had been through this, and I was fumbling through recalling what was going on.  Was headed here but hadn't yet arrived.
 

Offline EPAIII

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Re: Questions for those who know electromagnetism better than I do
« Reply #30 on: August 29, 2021, 01:39:01 am »
Boy, I really wish I had some references on hand about how this "cloud" of electrons inside a conductor behaves.  With no current flow I can easily see how their attraction to the nuclei of the metal atoms will keep them "bound" to those atoms and their charge density will be approximately uniform.

But once a current does start to flow, and I am primarily considering a DC current at this point, then those bonds are effectively broken and, as is obvious from the very fact that a current IS flowing, those electrons are much freer to wander about. And they are still feeling both the attraction to those nuclei and the repulsion from each other. They will move from one atom to the next and while in flight between atoms, they will, while feeling that repulsion, be free to act on it and move away from the center.

I like the reference to a FET. But perhaps a better one may be to the Hall Effect where any conductor can be used and not just semi-conductors. Relatively moderate magnetic fields can be used to induce a Hall Voltage with only moderate DC currents flowing through the Hall sensor. It is not much of a stretch to see where a moderate electrostatic repulsion would also create a potential difference (Voltage) inside a conductor. Those electrons that make up the current flow in most conductors are not just mobile: they are well lubricated.

Paul A.  -   SE Texas
And if you look REAL close at an analog signal,
You will find that it has discrete steps.
 

Offline T3sl4co1l

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Re: Questions for those who know electromagnetism better than I do
« Reply #31 on: August 29, 2021, 12:11:01 pm »
Boy, I really wish I had some references on hand about how this "cloud" of electrons inside a conductor behaves.  With no current flow I can easily see how their attraction to the nuclei of the metal atoms will keep them "bound" to those atoms and their charge density will be approximately uniform.

Not quite sure if this is a sarcastic tone or what, but it's again not hard to find from the keywords --
https://en.wikipedia.org/wiki/Fermi_gas
In fact, not only does the gas remain a gas in the absence of current flow (which is obvious given its statistics), but it remains so even down to absolute zero, where the Pauli exclusion principle creates a pressure forcing electrons into unbound states.  Instead of single atoms with happy filled (well, filled up to neutrality) electron shells, the outer electron shells happen to overlap, indeed so much that they share one or more electrons each, or something like that.

Something is also apparent given Ohm's law: there's no stick-slip motion here, it's smooth conduction all the way from zero to very high current densities (though again, not at all exceeding the charge density, because there's a L-O-T of electrons; pretty sure it vaporizes from ohmic loss before you can get close).

The gas also proposes a mechanism for ohmic loss: the gas is in thermal equilibrium with the crystal lattice -- that is to say, there is an energy coupling between modes (electron-phonon exchange).  Phonons (crystal vibrations) being the predominant mode of heat energy in solids.  Effectively, the electrons collide every once in a while, exchanging momentum with ions, including momentum picked up by the feeble electric field inducing drift (i.e., electric current); thus ballistic motion (unimpeded acceleration) is prevented and the excess energy manifests as heat.

(I'm not sure what a corresponding mechanism is for superconductors; I suppose the same must still apply, it just happens to be insufficient energy on average to disrupt the Cooper pairs, or whatever form the superconductivity takes.)



For point of reference, the thermal velocity of an electron gas is on the order of 10^5 m/s, while the drift velocity of ordinary current density in copper is ~cm/s.  It cools down a bit of course, but evidently retains velocity at 0 K (I don't know what average speed, alas).  An actual real example of "zero point energy". ;)


Quote
But once a current does start to flow, and I am primarily considering a DC current at this point, then those bonds are effectively broken and, as is obvious from the very fact that a current IS flowing, those electrons are much freer to wander about. And they are still feeling both the attraction to those nuclei and the repulsion from each other. They will move from one atom to the next and while in flight between atoms, they will, while feeling that repulsion, be free to act on it and move away from the center.

If it helps, the environment seen by an electron in a crystal, is similar to what a wave sees inside a waveguide: more specifically a periodic potential waveguide, which has a bandstop characteristic (electromagnetic bandgap).  Free propagation is allowed above and below the stopband; as it happens, in crystals, electrons are not free to move below the bandgap (valence band), but holes (absence of electrons) are; and electrons above (in the conduction band) are free.  This is relevant to semiconductors and most insulators (most insulators are just wide-bandgap semiconductors, though I think there's a technical difference in some materials and I forget why), however conductors happen to have electrons naturally sitting in the conduction band, or the bandgap happens to be zero or negative energy width, in either case populating the conduction band.  I also forget the technicals of how this comes to be, but it involves the earlier mentioned effects.


Quote
I like the reference to a FET. But perhaps a better one may be to the Hall Effect where any conductor can be used and not just semi-conductors. Relatively moderate magnetic fields can be used to induce a Hall Voltage with only moderate DC currents flowing through the Hall sensor. It is not much of a stretch to see where a moderate electrostatic repulsion would also create a potential difference (Voltage) inside a conductor. Those electrons that make up the current flow in most conductors are not just mobile: they are well lubricated.

Indeed, the Hall effect also gives a nonuniformity in a conductor.  In this case, it's the effect of crossed magnetic and current fields.  It also depends substantially on carrier density: metals are terrible because they have so many carriers they just swamp out the effect, and sensitive measurements are needed to demonstrate it.  Semiconductors have few carriers that are strongly affected, and therefore semiconductor sensors are predominantly used.

There's even an effect due simply to mechanical acceleration: there is a voltage across a disc (axis to rim) under rotation, even without a magnetic field.

Tim
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Offline bsfeechannel

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Re: Questions for those who know electromagnetism better than I do
« Reply #32 on: August 30, 2021, 03:00:49 am »
You realize that you can use Maxwell's equations to do circuit analysis.  But most folks find the simplified equations of Kirchoff are adequate and far easier to solve.

This is a misconception. KVL and KCL are not simplified Maxwell equations for circuit analysis. They are solutions to Maxwell’s equation for when there are no varying magnetic fields enclosed by—and no varying electric fields along—the path you’ve chosen to analyze your circuit.

When that doesn’t happen, you have the following options.

1. Find analytical solutions to Maxwell’s equations.
2. Find numerical solutions to Maxwell’s equations.
3. Try to do away with the “offending” varying fields by employing techniques like shielding, grounding, impedance matching, decoupling, avoiding loops and excessive long lines, etc.
4. Pray.
 
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Offline CatalinaWOW

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Re: Questions for those who know electromagnetism better than I do
« Reply #33 on: August 30, 2021, 06:05:24 am »
You realize that you can use Maxwell's equations to do circuit analysis.  But most folks find the simplified equations of Kirchoff are adequate and far easier to solve.

This is a misconception. KVL and KCL are not simplified Maxwell equations for circuit analysis. They are solutions to Maxwell’s equation for when there are no varying magnetic fields enclosed by—and no varying electric fields along—the path you’ve chosen to analyze your circuit.

When that doesn’t happen, you have the following options.

1. Find analytical solutions to Maxwell’s equations.
2. Find numerical solutions to Maxwell’s equations.
3. Try to do away with the “offending” varying fields by employing techniques like shielding, grounding, impedance matching, decoupling, avoiding loops and excessive long lines, etc.
4. Pray.

There are two misconceptions (at least) in your response.  I did not say that kcl and kvl are simplified forms of Maxwell's equations.  I was snidely suggesting that while circuits could be analyzed using Maxwell's, they represent special cases where simpler methods are adequate.  And unless you have a different interpretation of enclosed and along, capacitors and inductors violate two of your conditions for use of k l and kcl.

The whole point of the initial comments was that Maxwell's in all their glory are not required for the initial posters problem.  In a nod to another post I will say that only one of the four is required, and not all of the terms in that one are non zero simplifying it further.  This, plus the symmetries of the problem reduce it to algebra.  No differential or integral calculus required.  Engineers and physicists beyond there first couple of years of college should be able to handle the full form for a variety of simple geometries, but as you say, closed form solutions for more general problems range from difficult to currently unsolvable.  Numerical solutions on today's world are tedious but difficult only in geometries that are so complex they tax computational resources.  Much like computational aerodynamics but easier since many em problems are at least linear while the aero guys always have to deal with non linearity.
 

Offline bsfeechannel

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Re: Questions for those who know electromagnetism better than I do
« Reply #34 on: August 30, 2021, 07:36:57 am »
capacitors and inductors violate two of your conditions for use of k l and kcl.
Not if you treat them as lumped components, in which case the circuit path will be between their terminals, away from their internal varying fields.
 
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Offline sandalcandal

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Re: Questions for those who know electromagnetism better than I do
« Reply #35 on: August 30, 2021, 10:47:43 am »
capacitors and inductors violate two of your conditions for use of k l and kcl.
Not if you treat them as lumped components, in which case the circuit path will be between their terminals, away from their internal varying fields.
To add to that. The "inductance" used in KVL and lumped element analysis (the common/typical type of circuit analysis taught in EE) is more precisely described as "partial inductance" which is different to the more generalised mutual-inductance and self-inductance as it is formalised in Physics studies.

This is a good (but quite heavy) paper going over derivation of the lumped element "inductor" model from first principles Maxwell's equations using a partial inductor model: http://eagle.chaosproject.com/sandbox/acstrial/newsletters/summer10/PP_PartialInductance.pdf

Edit: This level of analysis starts to get important for people doing high performance power electronics as well as RF black magic. At the point the lumped element modelling starts needing excess work to keep going, you tend to just go to FEA of the field equations rather than keep attempting analysis using lumped elements.
« Last Edit: August 30, 2021, 10:56:02 am by sandalcandal »
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Offline gcewing

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Re: Questions for those who know electromagnetism better than I do
« Reply #36 on: August 30, 2021, 12:54:28 pm »
My first lecture on "EM Field Theory" started with an intro given by the professor - "..do you really think the electrons transfer the energy from the power station to your household?
An exceedingly silly rant, if it really happened. Of course the marching electrons (a.k.a. current) need the electric field (a.k.a.) voltage, to, as the multiplicative product, deliver the power.
What he was probably getting at is that the power doesn't travel through the wires, it travels through the electromagnetic field surrounding the wires. See https://en.wikipedia.org/wiki/Poynting_vector
 
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Online iMo

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Re: Questions for those who know electromagnetism better than I do
« Reply #37 on: August 30, 2021, 02:05:00 pm »
The fact the "individual electrons" are marching in metals with a speed of a lazy ant, is not so obvious among the folk. Thus the marching electrons (like a water streaming in a pipe) themself cannot transfer power or energy in such a setup. The power is transferred via the e-field, where the field is strongest near the conductor's outer surface and spreads out over the entire universe, with decreasing intensity. The energy transfer inside a conductor is almost zero as the e-field in the conductor is almost zero. The conductor thus only gives the "direction" to the energy transferred by the surrounding field.

In order to calculate the "current" you have to integrate individual field contribution of each electron over the entire space surrounding the conductor. The Poynting's vector shows the direction and the strength of the field.

The energy/power from a battery to a lamp is transferred outside (along) the wires, with highest energy density just above the conductor's surface. When the conductor shows some "resistance" it means Poyinting's vector points towards the conductor's surface and you get energy losses like heat (because the outside field propagates itself with speed of light, but the speed of the field entering the conductor drops down significantly).

For example in a coaxial cable we hams often use the energy is transferred (mainly) in the dielectric between the outer and inner conductor, not in the copper.

Readers discretion is advised..
 

Offline Slartibartfast

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Re: Questions for those who know electromagnetism better than I do
« Reply #38 on: December 15, 2021, 12:33:08 pm »
Now, the magnetic flux (field strength) in a closed path surrounding a conductor. I think it is important to differentiate between two types of closed paths that can surround a conductor with a DC current. One such closed path would be the classical path of equal field strength. This is what we envision when we see the classic pattern of iron filings on a piece of paper that the conductor passes through and which are called by the simplistic phrase "magnetic lines of force". This type of path is, by definition a path of uniform field strength.

If you'll permit this one indulgence -- strength is "line" density, not the path itself.

If you're honest, then you'll admit that his definition - path of equal field strength - is more well-defined than using "line density", a concept which you critisise yourself in the following.

Quote
I never liked the idea of "magnetic lines of force" anyway.  There's no such thing as a line, you can't count them

Sorry I have to say this is not entirely correct. In high-temperature situations you cannot count them, but in a type II superconductor the field condenses into separate flux lines, which in every respect behave like what your imagination expects of field lines. Every line creates a little vortex, which can be imaged and counted. Here is a scanning SQUID microscopy picture of magnetic field lines in a thin superconducting YBCO film.

Cheers Peter
« Last Edit: December 15, 2021, 01:33:55 pm by Slartibartfast »
 

Offline Slartibartfast

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Re: Questions for those who know electromagnetism better than I do
« Reply #39 on: December 15, 2021, 01:05:41 pm »
Boy, I really wish I had some references on hand about how this "cloud" of electrons inside a conductor behaves.  With no current flow I can easily see how their attraction to the nuclei of the metal atoms will keep them "bound" to those atoms and their charge density will be approximately uniform.

But once a current does start to flow, and I am primarily considering a DC current at this point, then those bonds are effectively broken and, as is obvious from the very fact that a current IS flowing, those electrons are much freer to wander about.

This picture is wrong.

The periodical positioning of the nuclei (a.k.a. lattice) forces the available energy levels into bands. In metals, the electrons with the highest energy go into the so-called conduction band, where they have sufficient energy to roam freely. For these electrons, there is no bond, even without an external electric field, without any net current flowing.

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They will move from one atom to the next and while in flight between atoms,

No, they don't. As I said, they don't hop from atom to atom, but roam freely.

You may be confusing this with conduction by holes in doped semiconductors. Since a hole away from an atom does not make sense, here the "hopping picture" is appropriate.

Cheers Peter
« Last Edit: December 15, 2021, 01:11:12 pm by Slartibartfast »
 

Offline Slartibartfast

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Re: Questions for those who know electromagnetism better than I do
« Reply #40 on: December 15, 2021, 01:32:59 pm »
My first lecture on "EM Field Theory" started with an intro given by the professor - "..do you really think the electrons transfer the energy from the power station to your household?
An exceedingly silly rant, if it really happened. Of course the marching electrons (a.k.a. current) need the electric field (a.k.a.) voltage, to, as the multiplicative product, deliver the power.
What he was probably getting at is that the power doesn't travel through the wires, it travels through the electromagnetic field surrounding the wires. See https://en.wikipedia.org/wiki/Poynting_vector

Thanks for poynting out (pun intended) the obvious.  ;D

But there is another dimension to this: Energy (and by derivation, power) is always the product of two properties of the configuration space. Electric current is only one, so there got to be another one (the field). More examples: In radiation it's electric field and magnetic field, in mechanics it's force and velocity. This is the origin of the intensity being the square of the amplitude, and thus the deeper reason of why we use 20 in the definition of dB, when it comes to amplitude but 10 when it comes to power.

Cheers  Peter
 

Offline emece67

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Re: Questions for those who know electromagnetism better than I do
« Reply #41 on: December 15, 2021, 06:00:21 pm »
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« Last Edit: August 19, 2022, 04:52:45 pm by emece67 »
 

Offline m k

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Re: Questions for those who know electromagnetism better than I do
« Reply #42 on: December 15, 2021, 06:30:12 pm »
Just a comment, nothing to do with anything in this thread.

I don't understad the direction of force when a photon interacts with an electron.
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Offline HuronKing

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Re: Questions for those who know electromagnetism better than I do
« Reply #43 on: December 16, 2021, 02:51:43 am »
You wind up dealing with Maxwell's Equations and NOBODY wants to go there voluntarily.  Three people in history have understood Maxwell's Equations:  Maxwell, Feynman and Einstein. 

I get your point in spirit but I'd add 1 name to that list - Oliver Heaviside. It never fails to astound me how he was able to not only re-express Maxwell's cumbersome notation in something much more elegant (and make MANY derivations and solutions of Maxwell's Eqs and coin terms we EEs still use without a second thought), but also reinvented (alongside Gibbs) how all of physics does its math.
 

Offline HuronKing

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Re: Questions for those who know electromagnetism better than I do
« Reply #44 on: December 16, 2021, 04:31:05 am »
My first lecture on "EM Field Theory" started with an intro given by the professor - "..do you really think the electrons transfer the energy from the power station to your household?? Have you ever seen an electron?? Electron is small and lightweight, almost nothing.. Moreover, the electrons travel a few centimeters in a second in a copper wire.. You would never have lit a lamp on your desk if it had worked that way.."

An exceedingly silly rant, if it really happened. Of course the marching electrons (a.k.a. current) need the electric field (a.k.a.) voltage, to, as the multiplicative product, deliver the power. Makes you wonder, if the uphill bicyling professor cannot find out where his efforts go, surely not into the movement (a.k.a. velocity)? Completely analogous, the velocity needs the backwards pulling force, to, as the product, become the power sink that makes him pant.

I'd consider this a way to confuse students, a pedagogical desaster.

What about reactive power and power factor? Complex power is, well, more complex than simply P = IV.
 

Offline emece67

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Re: Questions for those who know electromagnetism better than I do
« Reply #45 on: December 17, 2021, 12:04:55 am »
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« Last Edit: August 19, 2022, 04:52:54 pm by emece67 »
 

Offline Slartibartfast

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Re: Questions for those who know electromagnetism better than I do
« Reply #46 on: February 02, 2022, 05:09:16 pm »
My first lecture on "EM Field Theory" started with an intro given by the professor - "..do you really think the electrons transfer the energy from the power station to your household?? Have you ever seen an electron?? Electron is small and lightweight, almost nothing.. Moreover, the electrons travel a few centimeters in a second in a copper wire.. You would never have lit a lamp on your desk if it had worked that way.."

An exceedingly silly rant, if it really happened. Of course the marching electrons (a.k.a. current) need the electric field (a.k.a.) voltage, to, as the multiplicative product, deliver the power. Makes you wonder, if the uphill bicyling professor cannot find out where his efforts go, surely not into the movement (a.k.a. velocity)? Completely analogous, the velocity needs the backwards pulling force, to, as the product, become the power sink that makes him pant.

I'd consider this a way to confuse students, a pedagogical desaster.

What about reactive power and power factor? Complex power is, well, more complex than simply P = IV.

Sorry, there is nothing complex about P(t) = I(t) * V(t), and yet it is true.

Usage of complex numbers is a simplification for the case of single frequency (sine) AC. If you want to use it in the more general case, you have to deal with the complexities of using a simplification in a case where it is not simple anymore.
 

Offline RJSV

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Re: Questions for those who know electromagnetism better than I do
« Reply #47 on: February 05, 2022, 04:16:52 am »
  eejeffrey has it right: (see comments back a bit), the thickness, of a pipe doesn't matter, for perfect round the fields cancel, in the interior...
  (I was just too timid, but confirmed the wire diameter issue, / non-issue. thanks
 

Offline KE5FX

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Re: Questions for those who know electromagnetism better than I do
« Reply #48 on: February 05, 2022, 06:50:47 am »
 

Offline snarkysparky

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Re: Questions for those who know electromagnetism better than I do
« Reply #49 on: February 05, 2022, 03:51:18 pm »
still waiting for someone to explain to me how a constant E and constant B field at a point in space can imply that any energy is flowing *through* that point.

 


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