You realize of course both sides feel the same way about the other side, right?
You realize of course both sides feel the same way about the other side, right?
There are no sides here.
That KVL doesn't hold under varying magnetic fields is an established fact, experimentally confirmed and predicted by the theory. We are not siding with Lewin or anyone or anything.
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.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!No, says the guy who uses Faraday's law and not KVL, but you can't tell the difference.
A rose by any other name is still a rose! You are the one that can not tell they are the same...
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.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!No, says the guy who uses Faraday's law and not KVL, but you can't tell the difference.
A rose by any other name is still a rose! You are the one that can not tell they are the same...
You see Sredni? They understand that their claim that KVL, the real KVL, always holds is false. Haven't I told you? But to save face they're now calling Faraday's law KVL.
-- 2+2 = 5.
-- No, two plus two equals four.
-- Yeah, that's what I wrote.
-- No you wrote two plus two equals five.
-- No, I didn't.
-- Yes, you did. How to you call this number: 5?
-- Four.
I have a simple challenge for team 'Lewin'.
I have a simple challenge for team 'Lewin'.
It's not team 'Lewin'. It's team 'Classical Electrodynamics'.
But what's the point of the exercise?
Both 'teams' are doomed to find the same values for what is read by the voltmeters.
Anyway, if I copied it correctly - and there is no guarantee I did with all those turns making me dizzy - I would say
Vr1 = -24mV, Vr2 = +216mV
Vm1 = -136 mV, Vm2 = -56 mV
Provided I counted the 'rings' right and didn't change some sign here and there.
And if I got one sign wrong somewhere, that would proof that there can be say 200 mV in two inches of copper wire that carry a current of 100 nA?
Excellent! You got the correct results. Now, what is the voltage between nodes A and D, VAD?
Excellent! You got the correct results. Now, what is the voltage between nodes A and D, VAD?
I didn't find a way to make it glow, but let's see...
IT DEPENDS ON THE PATH.
The only established fact so far is that bsfeechannel has no idea of what he is talking about!!!
The only established fact so far is that bsfeechannel has no idea of what he is talking about!!!
C'mon, man! There's no shame in being wrong. You can bet your bottom dollar that all of us held at some time in the history of our lives the same misconceptions you and Jesse Gordon are now expressing. We normally get aware of voltages when we poke circuits with the probes of our meters and we have this intuitive, but misleading, perception that voltages are generated by the components. What components do is to shape the conditions in which the electromagnetic field exists in that region of the space. Voltages are the consequence of the existence of the fields. Once you switch your "paradigm" to think in terms of fields--not just circuits--you immediately expand the capabilities of your analyses.
For some of us, the shift is very difficult. But it is worth it.
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.
Oh, no. Now you answer MY question, not a question of your choice.
And my question is:
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/kX0TSBw6/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.
You realize of course both sides feel the same way about the other side, right?
There are no sides here.
That KVL doesn't hold under varying magnetic fields is an established fact, experimentally confirmed and predicted by the theory. We are not siding with Lewin or anyone or anything.
Oh boy, where to even begin...
Okay, first of all: I'm not in denial of observable reality.
Excellent! You got the correct results. Now, what is the voltage between nodes A and D, VAD?
I didn't find a way to make it glow, but let's see...
IT DEPENDS ON THE PATH.No it doesn't depend on the path, because we are calculating it.
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.
Oh, no. Now you answer MY question, not a question of your choice.
And my question is:
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.OK here's how anybody can unambiguously physically measure the induced voltage in half of an air-core transformer secondary turn
Sure, now use your vast knowledge of electromagnetic fields and help Sredni calculate the voltage VAD from the problem above.
It's like trying to explain color to someone who is blind since birth.
I have already calculated "VAD" twice: its value depends on the path.
On the path set by the first voltmeter I get -136mV; on the path set by the second voltmeter I get -56mV.
I can compute on any path I want, if I know how the dB/dt region is partitioned by the path itself. Do you have any idea how easy it is for me to compute it? The only difficulty is making sure I am following that spiral labyrinth you set up hoping to get us confused.
Now, I solved your challenge.
Will you solve a few very elementary problems I pose? Or will you evade my questions as Jesse Gordon is doing?
Sure, now use your vast knowledge of electromagnetic fields and help Sredni calculate the voltage VAD from the problem above.
My "vast knowledge" won't help you. I asked the same questions long ago, got the right answers, but ended up like you: confused.
I can show you the door to your enlightenment, but I can't walk you through it. You'll have to do it yourself.
It's like trying to explain color to someone who is blind since birth.
I have already calculated "VAD" twice: its value depends on the path.
On the path set by the first voltmeter I get -136mV; on the path set by the second voltmeter I get -56mV.
I can compute on any path I want, if I know how the dB/dt region is partitioned by the path itself. Do you have any idea how easy it is for me to compute it? The only difficulty is making sure I am following that spiral labyrinth you set up hoping to get us confused.
Now, I solved your challenge.
Will you solve a few very elementary problems I pose? Or will you evade my questions as Jesse Gordon is doing?
You have solve nothing!
---snip---
This is the challenge: Calculate V1, V2, Vx, and Vy.
Excellent! You got the correct results.
It's like trying to explain color to someone who is blind since birth.
I have already calculated "VAD" twice: its value depends on the path.
On the path set by the first voltmeter I get -136mV; on the path set by the second voltmeter I get -56mV.
I can compute on any path I want, if I know how the dB/dt region is partitioned by the path itself. Do you have any idea how easy it is for me to compute it? The only difficulty is making sure I am following that spiral labyrinth you set up hoping to get us confused.
Now, I solved your challenge.
Will you solve a few very elementary problems I pose? Or will you evade my questions as Jesse Gordon is doing?
You have solve nothing!
---snip---
It's a no, then?
You confirm will evade questions just like Jesse has done till now?
By the way: your challenge wasQuoteThis is the challenge: Calculate V1, V2, Vx, and Vy.
I calculated them (it took me some six-seven minutes, mostly to decide to get off the couch and find paper and pencil, and to make sure I had copied the spirals right), and you said it yourself:QuoteExcellent! You got the correct results.
And now you are on your victory lap as if you were the only one to get the correct results. (And all this after I wrote, in the same message I posted the solution in: "Both 'teams' are doomed to find the same values for what is read by the voltmeters."
Go figure.
I wonder if 'everyone else reading' should be made aware that your solution forecasts that a copper wire a few inches long will drop 20 mV when a current of 0.24 mA flows through it. But hey, you have tiny little batteries in your wires, right? Like they explained coils in high school.
So, to be clear, you won't try to solve the simple circuit quizzes I pose?
You must be the kind of guy than never pays up when he loses a bet.
Once again, what is the value of VAD?
Once again, what is the value of VAD?
Once again:
It depends on the path.
How did place the probes to measure it? Midway through the disk, because you chose an highly symmetrical setup?
How would you place the probes in the solenoid was made in triangular shape and the ring with the resistor shaped as Mickey Mouse's head silhouette, tilted and off-center?
You must be the kind of guy than never pays up when he loses a bet.
Once again, what is the value of VAD? Go argue with results:
https://www.eevblog.com/forum/amphour/562-electroboom!/msg3842183/#msg3842183