The selector is the presence or better the absence of magnetic flux inside the measurement loop.
The beauty of it is that [correct voltage of far branch] + [emf with correct sign] = [voltage read by multimeter]
That also explains why you cannot measure partial turn voltage, but only integer multiples.
PS: after some more headscratching, I think I understood why you cannot see any voltage across the loop wire: every path that involves the piece of wire between the resistors and that does not go around the center of the electric field will sum up to 0. Therefore, if you just naively connect you voltmeter probes to the points "A1" and "A2", that path will only observe the voltage across the wire according to Mr. Ohm. No need to argue with electric charge being counteracted by another electric field.
And Lewin's supporters went the obfuscation way again! Paragraphs and paragraphs of gibberish.
And Lewin's supporters went the obfuscation way again! Paragraphs and paragraphs of gibberish.
People who studied it, prefer to call it physics.
But I am sure it is gibberish, to you.
Alright, IIRC R1 = left, R2 = right, A1 is top of R1.......D2 is bottom of R2. So if we are measuring R2 with a voltmeter on the right, we are also measuring the equal voltage comprised of the contribution of the loop and R1. Assuming clockwise current flow, the voltage calculated across R2 is I*R2 and the other matching voltage would be the EMF (A2-A1) - (I*R1) - EMF (D1-D2). Or, if I've botched the signs, something like that. Is that right?
So how does this work when you have only a partial turn or a wire that goes straight through a core like a current transformer?
My understanding of the difficulty of measuring partial turns was a bit different, I think, but perhaps the two can be reconciled--or I'm wrong. So how does this work when you have only a partial turn or a wire that goes straight through a core like a current transformer?
I have no idea what the points A2, A1, D2, D1 refer to - probably they are present in the original Lewin drawing but I do not have it at had now. But I am a bit troubled by your use of "EMF (A2-A1)" and I believe therein lies the rub. You are still trying to apply Kirchhoff and you are implicitly assuming the wires are like 'batteries'. No, you need to let go of that because there is no longer the Eind field in the wires. It has been obliterated by the Ecoul field. This is what KVLers have trouble accepting.
A partial turn of wire is not a closed circuit, so it cannot have a current flowing.
But this falls out of the boundaries of magneto-quasistatics.
Your follow-up question - because I've developed this ability to read minds - is "what about partial inductances?
Here is the best demonstration I have found that Lewin is wrong and Electroboon/Mehdi is right. The video is by 'fromjesse'
The problem with this demonstration is that KVL only "holds" in a very particular placement of the probes. If you try to measure from the outside of the ring the voltages will be different. Confirming what Lewin said: you can measure two different voltages at the same time when you have a varying magnetic field.
Here is the best demonstration I have found that Lewin is wrong and Electroboon/Mehdi is right. The video is by 'fromjesse'
This is no different from Mabilde's probing.
Now, don't be alarmed if what follows will sound gibberish (and probably pompous and faggotish) to you.
What the Mabilde probing does is to run the probes at right angle with the induced electric field inside the loop. This will eliminate the contribute of the Eind field along the probes, leaving it untouched in the arc that is probed. BUT they are not considering the contribute of the coloumbian electric field Ecoul that is present in the arc AND in the probes. And it is not perpendicular to them, so they are not cancelling the effects on the probes themselves.
In the end, what they measure in absence of resistors is not the actual voltage along (note that I use alond, because path matters) the probed arc, but only the contribute of the coloumbian field along the probes (because in the arc Ecoul = Eind but they are opposite and they cancel so there is no contribute there). And being Ecoul conservative, this value is equal to Ecoul in the arc.
They are measuring a PARTIAL contribute to the actual voltage, namely the part that is ascribed to the coloumbian field Ecoul alone.
I'm not a 'KVLer'
QuoteA partial turn of wire is not a closed circuit, so it cannot have a current flowing.Why is it relevant whether or not current is flowing?
QuoteYour follow-up question - because I've developed this ability to read minds - is "what about partial inductances?Your Kreskin abilities may be failing you, as I was not thinking that.
And this is what a KVLer would do. Why do you compute partial contribute to the emf at all? And only on the horizontal branches? Write the equations by listing the voltage drops on one side, and the linked emf on the other (zero if absent)
What is the definition of inductance?
And this is what a KVLer would do. Why do you compute partial contribute to the emf at all? And only on the horizontal branches? Write the equations by listing the voltage drops on one side, and the linked emf on the other (zero if absent)
It was not my idea to draw the diagram this way, it is how Lewin drew it, AFAIK the 'horizontal branches' aren't really intended to represent the actual physical layout. As for the way I've written the EMF part, I was simply trying to interpret what you wrote to make sure I understood it correctly. Are you saying that geometrical accuracy issues aside, you simply cannot break down the EMF into two parts?
But, anyway... What do you think would change in the Lewin ring if, instead of a perfect conductor, you had a highly conductive copper conductor?
I tell you what: almost nothing. And certainly nothing of relevance. The only difference is that, instead of zero electric field and zero voltage drop in the conductors, you will see an almost negligible electric field E = j /sigma_copper and an almost negligible voltage drop of a handful of microvolts. Against the hundreds of millivolts of drops at the resistors.
It was not my idea to draw the diagram this way, it is how Lewin drew it, AFAIK the 'horizontal branches' aren't really intended to represent the actual physical layout.
As for the way I've written the EMF part, I was simply trying to interpret what you wrote to make sure I understood it correctly. Are you saying that geometrical accuracy issues aside, you simply cannot break down the EMF into two parts?
QuoteWhat is the definition of inductance?Hmmm... L = V/(dI/dt) or something like that?
So where are you going with that? Are you going to claim that unless current is flowing, you can't have a definite voltage?
There's also the EMF in the resistors that you need to account for, since they're part of the loop. At this point it doesn't really make sense to break EMF down into parts any more, at least not for the sake of the computation: the total EMF around the loop is given as 1V.
But, anyway... What do you think would change in the Lewin ring if, instead of a perfect conductor, you had a highly conductive copper conductor?
I tell you what: almost nothing. And certainly nothing of relevance. The only difference is that, instead of zero electric field and zero voltage drop in the conductors, you will see an almost negligible electric field E = j /sigma_copper and an almost negligible voltage drop of a handful of microvolts. Against the hundreds of millivolts of drops at the resistors.
Really, almost nothing? The attached image is from 'fromjess' experiment which shows a significant voltage drop between two points in the ring. Hard to see, but in the attached image I think it is 184mV. You can clearly see from the video that you can measure a sizeable voltage between any two arbitrary points around the copper ring. As he says it in the video "so we can measure positive and negative voltages all around this dial". How do explain that now?
Picture yourself in a boat on a river.
KVL holds for any placement of the probes, if you account for the probes as being part of the circuit.
Check this video by 'RDS Academy' which explains it in more depth. The conclusion is clear: Lewin messed up the measurement:
But, anyway... What do you think would change in the Lewin ring if, instead of a perfect conductor, you had a highly conductive copper conductor?
I tell you what: almost nothing. And certainly nothing of relevance. The only difference is that, instead of zero electric field and zero voltage drop in the conductors, you will see an almost negligible electric field E = j /sigma_copper and an almost negligible voltage drop of a handful of microvolts. Against the hundreds of millivolts of drops at the resistors.
Really, almost nothing? The attached image is from 'fromjess' experiment which shows a significant voltage drop between two points in the ring. Hard to see, but in the attached image I think it is 184mV. You can clearly see from the video that you can measure a sizeable voltage between any two arbitrary points around the copper ring. As he says it in the video "so we can measure positive and negative voltages all around this dial". How do explain that now?
That measurement has nothing to do with the fact that the ring is made of a finite conductivity conductor. You would have basically the same measure even if they were superconducting traces.
And I have already explained why he gets those reading in my answer to you above.
It is not my fault if you cannot make sense of it.
Both fromjesse and RSD Academy resort to deleting comments that tell them they are wrong, which goes to show how insecure they are about what they dish out to their viewers.
KVL holds for any placement of the probes, if you account for the probes as being part of the circuit.
No. It doesn't. He even shows a video from our fellow EEVBlogger Joeqsmith where he clearly demonstrates that the positioning of probes affects the measurement.
This is what we call cognitive dissonance: KVL holds for any placement of the probes, as long as you place them in a very specific and unique position.
Give me a break.QuoteCheck this video by 'RDS Academy' which explains it in more depth. The conclusion is clear: Lewin messed up the measurement:
He concludes that Lewin doesn't know Ohms law. I wonder how could he have fooled MIT, which I thought was one of the most, if not the most, prestigious technology institutes in the world, for 43 years, as he doesn't know Ohms law, doesn't know how to model a circuit, doesn't know how Kirchhoff law works, doesn't know how to probe a circuit.
How could they have given him these awards:
1978 – NASA Award for Exceptional Scientific Achievement
1984 – Alexander von Humboldt Award
1984 – Guggenheim Fellowship
1984 – MIT Science Council Prize for Excellence in Undergraduate Teaching
1988 – MIT Department of Physics W. Buechner Teaching Prize
1991 – Alexander von Humboldt Award (again)
1997 – NASA Group Achievement Award for the Discovery of the Bursting Pulsar
2003 – MIT Everett Moore Baker Memorial Award for Excellence in Undergraduate Teaching
2011 – first recipient of the Educator Award for OpenCourseWare Excellence (ACE)
Lewin must be a master con man, no doubt. Or the MIT is a joint. Or both.
The RDS Academy dude also accuses Lewin of defying the scientific and engineering establishment and textbooks. I've never seen a reputable textbook both of engineering and physics which claims KVL holds under a varying magnetic field. In fact I've seen exactly the opposite. They call our attention to the fact that this cannot happen.
You forgot this 'achievement': Victim In Walter Lewin Online Course Sexual Harassment Case Comes Forward
https://www.huffpost.com/entry/walter-lewin-sexual-harassment-mit_n_6532698
Really, almost nothing? The attached image is from 'fromjess' experiment which shows a significant voltage drop between two points in the ring. Hard to see, but in the attached image I think it is 184mV. You can clearly see from the video that you can measure a sizeable voltage between any two arbitrary points around the copper ring. As he says it in the video "so we can measure positive and negative voltages all around this dial". How do explain that now?