Author Topic: "Veritasium" (YT) - "The Big Misconception About Electricity" ?  (Read 209357 times)

0 Members and 1 Guest are viewing this topic.

Offline EEVblog

  • Administrator
  • *****
  • Posts: 37626
  • Country: au
    • EEVblog
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #350 on: November 26, 2021, 04:19:25 am »
But otherwise, yes, the power is transmitted in the fields. A capacitor is an electic field.
Yes, that explains how power flows through a capacitor. Rings a bell. ;D

I see what you did there  ;D
 

Offline Kalvin

  • Super Contributor
  • ***
  • Posts: 2145
  • Country: fi
  • Embedded SW/HW.
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #351 on: November 26, 2021, 08:30:40 am »
[...]
My counter-argument is this: If Dave's argument is true, we would not have EMI, grounding, ground-loop, signal-integrity etc. issues in our circuits, because if practicing engineers really knew and understood how the EM-fields were actually flowing in the circuits, they would not make that many troublesome designs.

Since in practice we see way too often designs that have these kind of issues, it is clear that a) practicing engineers do not really understand EM-fields, and b) it is not really sufficient to think that the energy flows only along wires.

In a world where EMI etc was the only objective in all designs, then sure, that argument would be absolutely true. Not saying it's false in its reasoning, but maybe not the only conclusion you can draw. We would probably also have to assume that all designs produced from a true understanding of EM would also meet thermal, mechanical, cost, and functional constraints... and still meet all of those constraints when the sales team decide what they actually wanted was a plastic enclosure and have it delivered the week previous. It's always a balance and most engineers are unfortunately humans: mistakes and carelessness happen, things get overlooked with different priorities, it doesn't say anything about their understanding.

In this particular context (Veritasium's video on EM-fields explaining energy transmission, and Dave's response) I just wanted to point out that although practicing electrical engineers know about EM-fields, in reality only small portion of the practicing electrical engineers understand how EM-fields affect their designs.

I do acknowledge that development & production cost and timelines are setting constraints to products. This will not explain why there are so many EMI-problems in the designs that are actually introducing extra development & production costs and make the original timelines fail while trying to fix these EM-related problems, if the practicing electrical engineers really understood how EM-fields are affecting their designs. Most of the EMI-problems would have been easy to prevent with zero cost in the first place if the engineers just understood how EM-fields were behaving.
 
The following users thanked this post: AlienRelics

Offline MIS42N

  • Frequent Contributor
  • **
  • Posts: 510
  • Country: au
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #352 on: November 26, 2021, 11:41:50 am »
I raced off a reply after reading the first three pages of comments, but missed a crucial piece of information. The light bulb can light with ANY current, regardless of how small. So yes, theoretically as soon as electrons move in the wires connected to the battery, a magnetic field is created that will affect the electrons in the wire opposite.

I still take issue with the "energy doesn't flow in wires" title. It should be 'energy doesn't only flow in wires'. Assume the scenario outlined, the switch was closed midday yesterday. Take observations midday today - the light is lit, energy is being converted to work. Measure the magnetic field, it will be unvarying. An unvarying magnetic field does no work. Therefore, all the energy is transported by electron movement. Electrons in the wires.

Change from DC to AC, now the electrons are always changing speed, inducing a varying magnetic field. Simple enough to put a transformer in the path and show that no electrons jump from one winding to the other. So in the transformer, all energy is transferred without wires.
 
The following users thanked this post: AlienRelics

Offline bpiphany

  • Regular Contributor
  • *
  • Posts: 129
  • Country: se
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #353 on: November 26, 2021, 12:13:13 pm »
Fields don't need to be varying for them to have a "flux".
Electrons will be moving through the wires half way to the moon and back. The EM-field "force carrying photons" that makes them move will mainly flow from the battery to the bulb. Along the field lines of the Poynting field, not the long way around. Across the bulb is where most of the work is done (all of it with resistance free wires), and this is where the electrons need to be forced across the load. The flux in the Poynting "energy field" will be focused to where the E and B field are strong and perpendicular. In the steady state this is in the vicinity of the battery and bulb, since this is the only place the E-field has any notable value.
 

Offline Kalvin

  • Super Contributor
  • ***
  • Posts: 2145
  • Country: fi
  • Embedded SW/HW.
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #354 on: November 26, 2021, 02:59:59 pm »
Attached article provides a quantitative analysis which shows at DC condition Poynting  vector has two components, one along the surface of the wire in the direction of power transfer and the other one perpendicular to the wire directed into the wire, representing power loss dissipated inside the wire due to active resistance.

Here is a link to paper "Energy transfer in electrical circuits: A qualitative account" by Igal Galili and Elisabetta Goihbarg, which is referenced in Bud's paper above:

http://sharif.edu/~aborji/25733/files/Energy%20transfer%20in%20electrical%20circuits.pdf

This paper from Galili&Goihbarg contains very interesting reading and simple explanation how the energy flows in a [DC] circuit from a battery into a resistor, and how the Poynting vector shows the direction of the energy flow.

This paper also contains Feynman's own comments about his lecture and Poynting vector: ‘‘This theory is obviously nuts, somehow energy flows from the battery to infinity and then back into the load, is really strange."

The paper from Galili&Goihbarg provides an answer to this Feynman's dilemma:

The surface charge model is essential for students’ understanding of energy transfer in the dc circuit. Past attempts to apply the Poynting vector for this purpose
failed because of the neglect of the surface charge.


Spoiler alert:

Electromagnetic energy does not flow in the wires, as it might be intuitively assumed, but next to them. It enters into the resistors in the circuit at the rate of I**2 R.

Energy flow goes from both terminals of the dc battery to the load and never returns to the battery. Ironically, this understanding might look as if it supports the naı¨ve ‘‘clashing currents model,’’ a well-known misconception regarding the electric current in dc circuits.
 
The following users thanked this post: wilfred, AlienRelics

Offline aneevuser

  • Frequent Contributor
  • **
  • Posts: 252
  • Country: gb
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #355 on: November 26, 2021, 03:37:34 pm »
[Potentially useful information in Kalvin's post immediately prior to this - was posted while I was typing this up though]

This thread has grown a lot since I last looked, so I may have missed some useful contributions in skimming it. However, from what I can see there hasn't been any derivations from first-principles of what the fields around a DC circuit in steady state look like, and why they arise.

For example, suppose that we have a PP3 9 V battery, driving a resistance R via a square circuit as shown in the circuit diagram attached.

1332782-0

It seems to me that there are two key points:

1. A PP3 9 V battery is sufficiently small that its E field will look like a dipole to the rest of the circuit, and so drops in intensity by 1/r^3. Near the resistance, this will be too small to account for the required intensity of the Poynting vector. If this is isn't true for a circuit with 10 m sides, then it certainly is for one with 10 km sides, yet the resistance will clearly still dissipate the same power in the latter.

2. Although we will have E = 0 inside our perfect wires, we must, in fact, non-zero E radially to the wires at the corners, as shown, else the electrons wouldn't be able to turn the corners. These E fields must be due to the charge distributions near the corners.

So both of these points indicate to me that it is not the E field of the battery that is of interest in deriving the correct arrangements of E/B fields to account for the energy flow in this circuit, but is rather the local E fields generated by the local charge distributions on the surface of the wires that we need to know about.

So if we want to explain clearly how the Poynting vector and subsequent energy flow arises in this DC steady state case, then we have to derive the charge distributions on the wires; it's not enough to hand-wave and point at the E field of the battery.

Anyone in wild agreeement/disagreement with those points?
« Last Edit: November 26, 2021, 03:39:44 pm by aneevuser »
 
The following users thanked this post: AlienRelics

Offline bdunham7

  • Super Contributor
  • ***
  • Posts: 7691
  • Country: us
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #356 on: November 26, 2021, 04:27:33 pm »
Anyone in wild agreeement/disagreement with those points?

No, and I think that properly drawn field and Poynting diagrams will look quite different than the 'general explanation' type that we've seen here so far.  The magnetic field around the wire, for example, diminishes as 1/r, so it is going to be quite heavily concentrated near the wire, yet most illustrative drawings just show concentric circles.
A 3.5 digit 4.5 digit 5 digit 5.5 digit 6.5 digit 7.5 digit DMM is good enough for most people.
 
The following users thanked this post: Someone, AlienRelics

Offline Alex Eisenhut

  • Super Contributor
  • ***
  • Posts: 3330
  • Country: ca
  • Place text here.
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #357 on: November 26, 2021, 04:36:51 pm »
But can an electric field travel faster than its tailwind?
Hoarder of 8-bit Commodore relics and 1960s Tektronix 500-series stuff. Unconventional interior decorator.
 
The following users thanked this post: boB, newbrain

Offline bpiphany

  • Regular Contributor
  • *
  • Posts: 129
  • Country: se
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #358 on: November 26, 2021, 04:38:29 pm »
The electric field points from high to low electric potential. From the video, correct for the steady state. The wires are resistance free and hence at the same potential along their stretch.
 

Offline Sredni

  • Frequent Contributor
  • **
  • Posts: 668
  • Country: aq
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #359 on: November 26, 2021, 04:55:58 pm »

So both of these points indicate to me that it is not the E field of the battery that is of interest in deriving the correct arrangements of E/B fields to account for the energy flow in this circuit, but is rather the local E fields generated by the local charge distributions on the surface of the wires that we need to know about.

So if we want to explain clearly how the Poynting vector and subsequent energy flow arises in this DC steady state case, then we have to derive the charge distributions on the wires; it's not enough to hand-wave and point at the E field of the battery.

Anyone in wild agreeement/disagreement with those points?

Absoluteley: surface and interface charges are what make the electric field follow the wire.
See here for a few more references at the end of my answer:

https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant/532550#532550

Incidentally, failure to account for surface and interface charges is what makes people think that there could be a voltage build up in the filament of a solenoid (the famous Lewin experiment). The reason the electric field is zero in the perfectly conducting wire is exactly because the charge that has accumulated at the resistors' ends (and also along the external surface of the conductor) compensates the rotating induced electric field that would be there without the ring.

John D. Jackson, the dreadfully respected author of the bible of Classical EM, wrote a paper about the role of these surface/interface charges, where he noted that this knowledge sorely lacks in most curricula at the time (1996, but I see not much has changed).

John D. Jackson
Surface charges on circuit wires and resistors play three different roles
American Journal of Physics 64 (7), July 1996

A much much easier read might be the note from Chabay and Sherwood:

Bruce A. Sherwood, Ruth W. Chabay
A unified treatment of electrostatics and circuits
American Journal of Physics

This can be find online.
« Last Edit: November 26, 2021, 04:57:50 pm by Sredni »
All instruments lie. Usually on the bench.
 

Offline Electric_Li

  • Newbie
  • Posts: 1
  • Country: us
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #360 on: November 26, 2021, 09:23:50 pm »
I have figured out a proof that shows Derek's conclusion is incorrect. Now I won't dispute the fact that all the physics concepts he explains in the video are correct. I just think he missed two big details. The actual answer to this problem is 1 second.

The first detail he missed is that electromagnetic fields significantly decrease in strength over distance. I did a calculation that estimates the magnitude of the poynting vector going directly from source to load, and finds the energy transferred over 1/c seconds. For the calculation, I assumed a typical bulb resistance and ideal wire. Only the fields from the orthogonal wires contribute to the poynting vector, and they are so far away that I can assume they are effectively a point charge. The value for the magnitude of the poynting vector and for the energy transfer are so small, they are effectively zero. These values increase as the position of the vector moves towards the wires, and it doesn't become significant until you are right next to them. So the path of the energy does follow the wire. It doesn't just jump through thin air wherever it wants to go.

The second detail seems a lot more obvious to me. The diagram for the poynting vector Derek uses is for a steady state circuit. We're talking about transient conditions from when the circuit is closed to when the bulb illuminates. Before the switch is closed, the whole length of wire is neutral, meaning no charge is present, no EM fields are present, and no energy is moving. When the switch is closed, the voltage drop across the circuit pushes charge out of the source into the wires. The charges propagate down the wire at the speed of light (in an ideal wire) until they meet at the load. The poynting vectors would show energy is being transferred during this time, but the energy doesn't reach the bulb until the EM fields get there, which is the same as saying when the charges get there. At this point, calculating the turn on time is simple, Time = distance / velocity. So Time = 1 lightsecond / c = 1 second.

Since EM fields are created and carried by charges, you can't have one without the other. So it's silly to say, "the energy doesn't transfer through the electrons, it travels through the EM fields." You might as well say, "My car didn't move me across town, the gas in the tank moved me." In both statements, the distinction is misleading.


I'd like to hear people's thoughts on my conclusion. Am I right, or did I miss something?

I included some pictures that hopefully make my explanation easier to visualize.


 
 

Offline Bud

  • Super Contributor
  • ***
  • Posts: 6860
  • Country: ca
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #361 on: November 26, 2021, 10:10:36 pm »
Only the fields from the orthogonal wires contribute to the poynting vector

Exuse me?
Facebook-free life and Rigol-free shack.
 

Offline Bud

  • Super Contributor
  • ***
  • Posts: 6860
  • Country: ca
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #362 on: November 26, 2021, 10:23:14 pm »
Anyone in wild agreeement/disagreement with those points?

Thanks for not drawing the battery at the top of the diagram, otherwise all electrons would fall out.
Facebook-free life and Rigol-free shack.
 

Offline bpiphany

  • Regular Contributor
  • *
  • Posts: 129
  • Country: se
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #363 on: November 26, 2021, 10:53:06 pm »
See this post by rfeeces. With his sketch attached below.

https://www.eevblog.com/forum/chat/veritasium-(yt)-the-big-misconception-about-electricity/msg3829871/#msg3829871

The contribution to the field from the perpendicular wire pieces will be absolutely negligible at that distance. The E-field flux is very much localized around the battery and bulb.

 

Offline rs20

  • Super Contributor
  • ***
  • Posts: 2317
  • Country: au
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #364 on: November 26, 2021, 10:54:06 pm »
1. A PP3 9 V battery is sufficiently small that its E field will look like a dipole to the rest of the circuit, and so drops in intensity by 1/r^3.

This would be true (maybe, idk actually) if the battery were just floating in space with no wires attached. However, that's not the real situation here. One of those wires has a voltage 9V higher than the other, and since E = ∇ V, there an electric field between between the two wires.

Near the resistance, this will be too small to account for the required intensity of the Poynting vector. If this is isn't true for a circuit with 10 m sides, then it certainly is for one with 10 km sides, yet the resistance will clearly still dissipate the same power in the latter.

Take a look at the power-through-a-coaxial cable at https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable . It will show that evaluating Poynting vector, even in a 1,000,000km long coaxial cable, gives Poynting vectors that integrate to the expected P=IV. Key part there is noting that there's an E field between the inner and outer conductors, which is not 1/r^3 w.r.t. to distance away from the battery.

I cannot stress enough that if you correctly integrate the Poynting vectors over a plane dividing the battery from the load, you will see the correct value for the power being transferred from battery to load. If you disagree, you made a mistake in the calculation  :)
« Last Edit: November 26, 2021, 10:56:11 pm by rs20 »
 

Offline MIS42N

  • Frequent Contributor
  • **
  • Posts: 510
  • Country: au
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #365 on: November 27, 2021, 01:47:55 am »
The first detail he missed is that electromagnetic fields significantly decrease in strength over distance. I did a calculation that estimates the magnitude of the poynting vector going directly from source to load, and finds the energy transferred over 1/c seconds. For the calculation, I assumed a typical bulb resistance and ideal wire. Only the fields from the orthogonal wires contribute to the poynting vector, and they are so far away that I can assume they are effectively a point charge. The value for the magnitude of the poynting vector and for the energy transfer are so small, they are effectively zero. These values increase as the position of the vector moves towards the wires, and it doesn't become significant until you are right next to them. So the path of the energy does follow the wire. It doesn't just jump through thin air wherever it wants to go.
:
:

Since EM fields are created and carried by charges, you can't have one without the other. So it's silly to say, "the energy doesn't transfer through the electrons, it travels through the EM fields." You might as well say, "My car didn't move me across town, the gas in the tank moved me." In both statements, the distinction is misleading.
"The value for the magnitude of the poynting vector and for the energy transfer are so small, they are effectively zero" was the conclusion I jumped to. But the fine print says the bulb lights if any current flows. It is a totally artificial situation, so "effectively zero" isn't zero. We have to conceded that it is a teeny weeny bit more than zero so the fine print applies. Although the effect is minuscule, it is likely there are real world instruments sensitive enough to detect it.

I sort of agree with "Since EM fields are created and carried by charges, you can't have one without the other". I said earlier the carrier is magnetic only, and certainly there are permanent magnets which create their flux from the alignment of electron motions. That is a notional moving charge. I believe the E field in EM is a mathematical convenience to explain the action of a magnetic field on a charged body. It manifests itself as moving the body as if it were affected by a repulsive or attractive E field. But the real E field is just the repulsive or attractive forces between charged particles. Same but not the same. Sort of like i, the square root of -1. It is mathematically useful but doesn't exist. Closing the switch does change the real E field between the wires, but the effect is orthogonal to the wires so has no observable effect.
 

Offline bsfeechannel

  • Super Contributor
  • ***
  • Posts: 1667
  • Country: 00
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #366 on: November 27, 2021, 04:12:20 am »
I still don't quite get his explaination at 7:35
He shows the Poynting vector S coming out from the battery in a DC circuit. How? There is no EM radiation loss.

You have a fundamental misconception about how power is transferred in DC. In your video you say that the direction of the Poynting vector is away from the source only in AC, and throw in for some reason the skin effect,

I think by now he must have understood why there will be a Poynting vector coming out of the battery in a DC circuit, but he didn't explain in the video why he thinks the skin effect would play a part in the explanation for the energy flow.
 

Offline TerraHertz

  • Super Contributor
  • ***
  • Posts: 3958
  • Country: au
  • Why shouldn't we question everything?
    • It's not really a Blog
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #367 on: November 27, 2021, 07:35:56 am »
The thread got too long before I could read it. Just now watching this fun thing:
   
  The Simplest Math Problem No One Can Solve - Collatz Conjecture
Remined me of Veritassium's video on energy transmission

Which was pretty much right, I thought. It's the fields around the wires... AND the flow of charge in the wires that results from and modifies the fields. Neither one nor the other alone.

One way to consider his experiment, is to cut the wires at both far ends. Then you have two long (ie totally untuned) dipoles adjacent to each other. One with a battery and switch in the middle, the other with a light bulb. You could extend the wire pairs out to infniity, and the only difference would be not getting end reflections back (eventually.)

You could also replace the battery and switch with a balanced AC source, and it would work a lot better. But even with the single step impulse on closing the switch, there's still some energy transfer, beginning with a delay only of lightspeed between the battery and light (parallel wires.) Not _much_ but a little.

However... there is one huge, laughable error in his presentation. Observe his graphic of an EM wave fields in space. Spot the error.



My comment on youtube, buried in 46,785 other comments:
Speaking of 'wrong things', your graphic of the E and M fields for light, radio etc is wrong. One field is phase shifted from the other by 90 degrees. Eg the blue will be maximum while the red is going through zero. It's the rate of change of each one that creates the other.

Did anyone else in this long thread notice that?


« Last Edit: November 27, 2021, 07:40:12 am by TerraHertz »
Collecting old scopes, logic analyzers, and unfinished projects. http://everist.org
 

Offline aneevuser

  • Frequent Contributor
  • **
  • Posts: 252
  • Country: gb
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #368 on: November 27, 2021, 08:18:56 am »
Anyone in wild agreeement/disagreement with those points?

Thanks for not drawing the battery at the top of the diagram, otherwise all electrons would fall out.
I always pay careful attention to all the little technical details like that - to do otherwise is to besmirch the memory of people like Faraday and Maxwell, who figured all this stuff out.
 

Offline aneevuser

  • Frequent Contributor
  • **
  • Posts: 252
  • Country: gb
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #369 on: November 27, 2021, 08:52:07 am »
1. A PP3 9 V battery is sufficiently small that its E field will look like a dipole to the rest of the circuit, and so drops in intensity by 1/r^3.

This would be true (maybe, idk actually) if the battery were just floating in space with no wires attached. However, that's not the real situation here. One of those wires has a voltage 9V higher than the other, and since E = ∇ V, there an electric field between between the two wires.


If a battery is able to maintain charge separation between its external contacts while in a circuit, then its far E field has to look like that of a dipole from the POV of a circuit that is much bigger than the battery. I can't see how it could be otherwise (though I'm always happy to change my mind on seeing a decent argument against my position).

It is true that there are other E fields in the circuit to consider though - these arise due to charge distributions on the wires caused by the battery - that was the whole point of my post. To get the whole field, we add the dipole field due to battery and local fields due to charge distributions. This is simply the principle of superposition (linear Maxwell's eqns blah blah).

I claim (without having actually analysed the setup carefully) that the contributions of the local E fields in the circuit dominate (in most parts of the circuit) over that of the dipole E field of the battery when calculating S at any point, since a 1/r^3 field will die off too quickly for it to be otherwise.

Quote
I cannot stress enough that if you correctly integrate the Poynting vectors over a plane dividing the battery from the load, you will see the correct value for the power being transferred from battery to load. If you disagree, you made a mistake in the calculation  :)
I don't think anyone is disagreeing. I'm certainly not. I suspect that you misunderstood the point that I was making - I have tried to clarify it above.

In short, my rough intuitive analysis tells me that in a circuit as described:

1. there is no E field in the wires as they are ideal conductors so no S vector therein.

2. the E field of the battery is largely inconsequential in calculating S (too small in most places as it goes as 1/r^3)

3. the battery's role is to redistribute charges in the wires to produce a surface charge distribution whose local E field dominates in calculating S (but I'm very hazy on how one would figure out the details of that charge distribution - haven't seen a convincing derivation anyware, and I'm too dim to do it)

4. I'm guessing that the local E field on the wires goes roughly as 1/r from the wires (since it probably looks like a long line of charge on most of the wire)

5. Hence, as B goes as 1/r from the wire, S goes as 1/r^2 from the wire, since S = E x B.

6. So if all that is true, then S is largest near the wires, and dies off inverse-squarely away from them, more-or-less. So the pictures showing lots of lovely S in the middle of square circuits are not incorrect but pretty misleading - most of S will be seen near the wires, and the comment that someone made (here or on another site?) that the wires "guide" S are largely (but not completely) true.

If we could see the magnitude of S as a glowing substance in my circuit, it would be glowing hot near the wires, and much dimmer away from the wires (if my hand-waving is even nearly right). However hot the glow is 1 cm from the wire, it would be 1% of that intensity 10 cm away.

All IMHO, of course.
 

Offline aneevuser

  • Frequent Contributor
  • **
  • Posts: 252
  • Country: gb
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #370 on: November 27, 2021, 09:02:54 am »

However... there is one huge, laughable error in his presentation. Observe his graphic of an EM wave fields in space. Spot the error.

My comment on youtube, buried in 46,785 other comments:
Speaking of 'wrong things', your graphic of the E and M fields for light, radio etc is wrong. One field is phase shifted from the other by 90 degrees. Eg the blue will be maximum while the red is going through zero. It's the rate of change of each one that creates the other.

Did anyone else in this long thread notice that?

I hope they didn't notice that, because it's not true - in the far-field of an antenna, the E and B fields are indeed in phase. It is the time rate of change of one that determines the spatial (twisting) rate of change of the other (in a source free region) i.e. curl(E/B) = k d(B/E)/dt
 
The following users thanked this post: AlienRelics, rs20, newbrain

Offline rs20

  • Super Contributor
  • ***
  • Posts: 2317
  • Country: au
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #371 on: November 27, 2021, 09:14:05 am »
If we could see the magnitude of S as a glowing substance in my circuit, it would be glowing hot near the wires, and much dimmer away from the wires (if my hand-waving is even nearly right). However hot the glow is 1 cm from the wire, it would be 1% of that intensity 10 cm away.

I fully agree with the statement above (if anything, I suspect the falloff might be more extreme than r^2 even), and the reasoning leading up to it.

Sorry I misinterpreted the first time around.  I will just comment that "the local E fields generated by the local charge distributions on the surface of the wires" (from your original post) makes it sound like the E field around a wire is just a function of the charge distributions on that wire alone. I'm not sure if you meant to imply that, but that is a bit wrong -- a wire sitting at ground potential will have zero E field around it if everything nearby is at ground potential. But add nearby wires at non-ground potentials, and you'll see an E field appearing between those wires. It's very much a function of the spacing between the wires and the voltage across those wires; I'd personally not put too much thought into "local charge distributions on the surface of the wires", I don't quite see how that could lead to correct computation of the E field around the wires.
 

Offline aneevuser

  • Frequent Contributor
  • **
  • Posts: 252
  • Country: gb
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #372 on: November 27, 2021, 09:17:48 am »
https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant/532550#532550

This looks like a very interesting post and one that I shall study carefully, in the fullness of time.

Quote
John D. Jackson, the dreadfully respected author of the bible of Classical EM, wrote a paper about the role of these surface/interface charges, where he noted that this knowledge sorely lacks in most curricula at the time (1996, but I see not much has changed).
Oh dear. Yeah, I remember Jackson for EM. Almost as scary as Goldstein for mechanics

Quote
John D. Jackson
Surface charges on circuit wires and resistors play three different roles
American Journal of Physics 64 (7), July 1996
Is a copy of this available on the webs?

Quote
A much much easier read might be the note from Chabay and Sherwood:

Bruce A. Sherwood, Ruth W. Chabay
A unified treatment of electrostatics and circuits
American Journal of Physics

This can be find online.
Hmm. Seems to be fairly well hidden.

Anyway, I'm glad to see that a ferret is really making something of his life -  and in Antarctica too. Must be tough.
« Last Edit: November 27, 2021, 09:37:18 am by aneevuser »
 

Offline aneevuser

  • Frequent Contributor
  • **
  • Posts: 252
  • Country: gb
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #373 on: November 27, 2021, 09:37:01 am »
a wire sitting at ground potential will have zero E field around it if everything nearby is at ground potential. But add nearby wires at non-ground potentials, and you'll see an E field appearing between those wires. It's very much a function of the spacing between the wires and the voltage across those wires; I'd personally not put too much thought into "local charge distributions on the surface of the wires", I don't quite see how that could lead to correct computation of the E field around the wires.
I have to say that I can't follow your point here at all. Sorry. I've only just got up though, so it may all come into focus as the day proceeds.

Regardless, I'm going to stick with local charge distributions on wires being of significance, because, at the very least, we definitely seem to need E fields at the corners to make the electrons change direction (as shown on the original diagram) and they can only come from the local surface charges at the corners, AFAICS. Of course, since I'm not sure what the charge distribution on the outside of a current carrying wire looks like, and don't know how to derive it, maybe I'm way off-base.
 

Offline bpiphany

  • Regular Contributor
  • *
  • Posts: 129
  • Country: se
Re: "Veritasium" (YT) - "The Big Misconception About Electricity" ?
« Reply #374 on: November 27, 2021, 12:45:37 pm »
If we could see the magnitude of S as a glowing substance in my circuit, it would be glowing hot near the wires, and much dimmer away from the wires (if my hand-waving is even nearly right). However hot the glow is 1 cm from the wire, it would be 1% of that intensity 10 cm away.

I fully agree with the statement above (if anything, I suspect the falloff might be more extreme than r^2 even), and the reasoning leading up to it.


This totally depends on the distance between the wires. If the wires run in parallel from the battery to the load the fields are going to look like in the animation in the video. The E-field will be fairly uniform between the two wires (it will spread out a little), and the B-field will decrease rapidly away from the wires. Hence the S-field should follow the two wires fairly tightly.


If the battery and bulb are close together, but the wires go off to the sides, there is going to be very little E-field at all along the wires. The S-field has to, mainly, jump across the gap between the source and sink where both the fields have any strength.

 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf