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#562 – Electroboom!

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Sredni:

--- Quote from: Jesse Gordon on November 26, 2021, 06:24:43 am ---Sadly, my understanding hasn't changed.
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Dude, the reason your script looks like it stinks to you is because it stinks!

--- End quote ---

Oh, goodness me. What a wondrous, copious and melliflously superfluous word salad. All to hide the fact that you cannot lump Lewin's ring. Here, let me repeat the question you are so eloquently avoiding to answer.

This is Lewin's ring: two resistors in a single loop that goes around a circular region (let's consider it of the same size as the loop, so you can see there is no 'room to twist' the wires) of variable magnetic field. The resistors are required to be on the opposite sides of the variable magnetic field region.

https://i.postimg.cc/pLmfyHxZ/Lewin-ring-is-unlumpable.jpg

Please, show everybody you can draw a circuit path (make it green, meaning it's 'flux-free') that joins the resistors' terminals to the "lumped transformer secondary" terminals and DOES NOT INCLUDE the variable magnetic field region in its interior. Like I did for the lumpABLE circuit I decided to see as lumpED (in my post "Lumpable (lumped and not lumped) and not lumpable circuits for dummies").

In addition, you can also show everybody you can draw the path inside your "lumped transformer secondary" that DOES INCLUDE the variable magnetic field region (make it orange) but IS NOT part of the green circuit path.
I will show you that if you can do that you will run into contradiction.

Too bad my armchair is at the quantum mechanic's shop for repairs and I cannot fly away to another galaxy.

thinkfat:
Oh boy, where to even begin...

Okay, first of all: I'm not in denial of observable reality. In fact, I took the reality YOU observed, turned it into a nice, colorful diagram and explained each measurement YOU did using only Ohm, Maxwell and Faraday. I showed in which loops voltage is induced and why not in others, explained every number you obtained.

Here it is again:

Second: I did not "dream up" any "magic electric field". It's existence was found and explained by James Maxwell. It's his second field equation.

--- Quote from: Wikipedia ---The Maxwell–Faraday version of Faraday's law of induction describes how a time varying magnetic field creates ("induces") an electric field
--- End quote ---

Third: It is very obvious how to measure it: create a wire loop around any area with changing magnetic flux, create some discontinuity (make a cut, put a resistor in the loop), measure the electric field across it. The probe is just some wire coiled up. It senses non-conservative electric fields. Just the same, each turn around a transformer core "measures" this electric field.

I hope this is not too much to stomach.

You mentioned repeatedly how the magnetic flux is fully contained inside a transformers core (which is not exact but, eh, good enough). Then explain how a conductor (charge particles) outside of the core and thus outside of the magnetic flux, can interact with it. "Spooky action at a distance"? The answer "Dude, because Faraday" is not enough. What is the mechanism behind it?

Btw how's that lumped circuit coming along @Sredni asked you to come up with?

jesuscf:

--- Quote from: Sredni on November 26, 2021, 09:29:46 pm ---Oh, goodness me. What a wondrous, copious and melliflously superfluous word salad. All to hide the fact that you cannot lump Lewin's ring. Here, let me repeat the question you are so eloquently avoiding to answer.

--- End quote ---

Says the guy that when solving Lewin's ring, the first thing he does is to lump the EMF, and then he follows up by using KVL to find the voltages in the resistors!

Jesse Gordon:

--- Quote from: Sredni on November 26, 2021, 09:29:46 pm ---
--- Quote from: Jesse Gordon on November 26, 2021, 06:24:43 am ---Sadly, my understanding hasn't changed.
...
...
...
Dude, the reason your script looks like it stinks to you is because it stinks!

--- End quote ---

Oh, goodness me. What a wondrous, copious and melliflously superfluous word salad. All to hide the fact that you cannot lump Lewin's ring. Here, let me repeat the question you are so eloquently avoiding to answer.

This is Lewin's ring: two resistors in a single loop that goes around a circular region (let's consider it of the same size as the loop, so you can see there is no 'room to twist' the wires) of variable magnetic field. The resistors are required to be on the opposite sides of the variable magnetic field region.

https://i.postimg.cc/pLmfyHxZ/Lewin-ring-is-unlumpable.jpg

Please, show everybody you can draw a circuit path (make it green, meaning it's 'flux-free') that joins the resistors' terminals to the "lumped transformer secondary" terminals and DOES NOT INCLUDE the variable magnetic field region in its interior. Like I did for the lumpABLE circuit I decided to see as lumpED (in my post "Lumpable (lumped and not lumped) and not lumpable circuits for dummies").

In addition, you can also show everybody you can draw the path inside your "lumped transformer secondary" that DOES INCLUDE the variable magnetic field region (make it orange) but IS NOT part of the green circuit path.
I will show you that if you can do that you will run into contradiction.

Too bad my armchair is at the quantum mechanic's shop for repairs and I cannot fly away to another galaxy.

--- End quote ---

That's not exactly Lewin's ring, the resistors are too big, but I think I understand what you're asking - and I'll draw a better representation of the Lewin Ring and respond to that after you answer some questions.

Here's some questions I've been asking for days and you're refusing to answer:

You have admitted that in the V1 and V2 voltages in the diagram directly below will both read the same voltage, and you have also admitted that V2 will be suitable as an element in a KVL loop:

Question One:

How can V1 not also be suitable since they both measure the same voltage? Your own trusted source says that if an unambiguous physical measurement of the voltage across the two terminals can be obtained, then KVL holds! How can V1 not work for KVL?

https://i.postimg.cc/fTgyDNp0/20211119-030105.jpg

Question Two:
Considering your own trusted source, if a ONE TURN TRANSFORMER SECONDARY on a toroidal transformer has a voltage that is unambiguously physically measurable, then it should qualify as an element for a KVL loop, Correct?

For example, in the following transformer diagram with a safety wall in the center, why would the output winding not be suitable as an element in a KVL loop?

https://i.postimg.cc/qqG0vgRV/20211124-213839.jpg

If you think the above secondary winding is not suitable in an element in a KVL loop, then why? It is possible to unambiguously measure the voltage, right? So then according to the following textbook, which you yourself cite, KVL should hold, right?

https://i.postimg.cc/sf4j3HbF/Desoer-Kuh.jpg

If you honestly answer the above questions and I'll do my best to draw up Lewin's loop and show how I would measure the voltage across the half turns.

Sredni:

--- Quote from: jesuscf on November 27, 2021, 03:36:52 am ---
--- Quote from: Sredni on November 26, 2021, 09:29:46 pm ---Oh, goodness me. What a wondrous, copious and melliflously superfluous word salad. All to hide the fact that you cannot lump Lewin's ring. Here, let me repeat the question you are so eloquently avoiding to answer.

--- End quote ---
Says the guy that when solving Lewin's ring, the first thing he does is to lump the EMF, and then he follows up by using KVL to find the voltages in the resistors!

--- End quote ---

No, says the guy who uses Faraday's law and not KVL, but you can't tell the difference.

Ok, this had to be just an appendix to my post "What about the other forms of EMF? Uh, what about 'em(f)?", but since you are impatient, I will anticipate it.

Lewin's ring with a localized EMF: adding a battery
The following example shows that the localized EMFs are part of the path integral on the left of Faraday's equation (in its integral form), while the delocalized inductive EMF is the surface integral on the right. By not being part of the PATH integral, the delocalized inductive EMF is at all effects invisible in the closed circuit represented by the ring.

So, let's solve for the voltages and current in a Lewin ring with the addition of the localized EMF from a battery (shown in the following drawing as a 9V battery)

page 1 https://i.postimg.cc/D0mW70jF/screenshot-4.png

Now, pay attention at how the full closed path is partitioned, because it's that exhaustive partitioning that show established that "the inductive EMF is NOT on the path". Or, as Kirchhoff would have said it: "die induktive elektromotorische Krafte ist NICHT auf dem Wege".

page 2 https://i.postimg.cc/zBGvk31W/screenshot-5.png

Oh, look. I have started with Faraday's law, in its local, differential form. Used Stokes theorem to obtain the integral form of Faraday's law, then applied Faraday's law to obtain an equation that involves, the nonlocalized EMF that I call EMF(t), the localized EMF Vbatt(t) and the current I(t).
Then I solve this Faraday equation to find the current I(t).

No sign of KVL, at all. Because KVL is DEAD.
But the most interesting part is that the source of EMF that is not delocalised, the very localized and lumped EMF battery, shows up as a piece of the path integral on the left hand side. That is the nest where the localised EMF Kirchhoff was arguing about, are born. You know, the "electromotorischen Krafte" he writes about in his formulation.
(This will be more clear when I post the next message)