Відео

Thank you for pointing out the new video by Trevor. It seems he, SiliconSoup and I have 'specialized' in covering different angles of the Lewin ring: Trevor is more focused on areas and geometry, SiliconSoup on charge density, while I prefer to look at fields and measurement loops (this last part will be clear when I'll upload the "What a voltmeter measures" video.)

Thank you for your kind words, 'armchair fellow' . (For everybody else: 'armchair fellow' is kind of an insider joke. SiliconSoup and a few others will understand it). 😊

Yeah, I get that a lot. 😭 Out of curiosity, do you think the audio level in my previous video "Dispelling the Lewin KVL paradox" to be acceptable? I was almost shouting, there.

@@copernicofelinis I have some ideas for you if you're looking to improve your audio! I'm guessing there's some stuff with your microphone, recording and finalizing process, etc. that can be changed to help the volume out a fair bit.

​@@no1unorightnowI'm open to suggestion. I have already tried 'fixing my audio' using win10's diagnostic tool but to no avail. I am now thinking that my mic got damaged when I left the headset in a room where I used ozone.

The answer to the question in the title is in the video linked at the end of the description. ua-cam.com/video/qTzqIYrc_Hs/v-deo.html

I have just finished updating a corrected version of the subtitles. The problem with the voiceover is twofold: first, my abysmal pronunciation (compounded by the fact that the recording was initially meant only to estimate the length of the video); second, the low sensitivity of the mic. I have a Logitech headset that behaved spectacularly years ago in Skype - loud and clear - but as of now with Seven and 10 only produces low levels (something like 1-2 notched over a max of 10) and a lot of noise. I had to record this voiceover with my smartphone, keeping the phone in one hand and a pillow between me and the laptop to muffle the sound of the fans. I also did it in the middle of the night, and my voice was naturally on the whisper side, sorry. I have an updated version of this video with a couple more figures (not important) and corrections on the name Helmholtz and the formula overlay frame, but I don't think I will re-upload it anytime soon. I will be more careful with the audio levels (and try to improve my pronounciation) in the next videos. Try using headphones, they can solve the whispering problem. As for the mumbling, I need to work on that 😢. If the headphones won't do the trick, try the subtitles or better yet the transcript. I still have a couple of typos to catch, but they should help. Oh, and thanks for the feedback.

Thanks for the detailed reply, and sorry if I was abrupt but it was frustrating to find such interesting content which I had to abort! It made sense when you explained that the audio was recorded at night. This was the only problem for me. If you are able to deliver speech at a daytime volume, any old microphone or smartphone will be adequate. The only fancy addition might be some post-production compression, but a smartphone probably has that built-in. Speech doesn't require much bandwidth, dynamic range or s/n ratio. I was unable to follow the content because of the late night whispering :-)

Is it possible to do this from within UA-cam? Do you know? I have the whole voiceover in the form of text in a web page but I do not have an ad-free website where to upload it to. But a few months ago I had to open a WordPress account to talk to Simplenote's assistance. If I find where I put the credentials, I might upload it there. PS Thank you for pretending the problem is in your bad hearing and not in my pronunciation ;-) .

​@@copernicofelinisno no my friend, you're good! I just came home and tried with headphones, it helps a lot, especially with some of the weird mistakes in UA-cam automatic captions. I'm afraid I don't know anything about how to work with UA-cam's subtitles but I'm sure that someone will read this and can provide some information. Again thanks for the video and as my wife will confirm by how I can get startled when she is "suddenly" behind me, giving me a heart attack, my hearing really is that bad 😅

Try now: I edited the subtitles. It should be better, if the changes have been applied.

​@@copernicofelinisWhoa, that's a lot better indeed. Thank you so much, I appreciate it dearly.

I wish I could say it was a one-off typo, but I just realized I have been spelling "Helmholtz" wrong for YEARS. Possibly decades! I just had one of those moments that in the movies are shown with a progressively closer close-up of the actor's face, while the background expands around it. Thank you so very much for pointing it out. PS I commented (positively) to your video years ago. 😊

That is on the back burner. Right now I am planning to upload a video on the IEC definition of voltage, another one on what a voltmeter actually measures, and one on when probing does matter and why. Then I would like to do a video on the role of surface and interface charge in establishing the E field inside a circuit. And only after that consider the energy transfer question (which has a lot to do with surface charge and also requires making some complicated figures).

Great and yeah we need to understand fundamentals correctly. most people do just read the definitions and move on i really like this video

And also I never really understood the term "flux". The magnetic flux and mmf stuff if you can try to put it in your bucket list .

@@lokeshvirat5915 well, there are two important 'dual' quantities that can be defined on a vector field: a line integral that quantifies how much the vectors of the field are going along a given line (when the line is closed this is the circulation), and a surface integral that quantifies how much the vectors of the field go through a given surface (this is flux). The circulation is associated with the projection of the field along the (local) direction of the path, while the flux is associated with the projection of the field along the direction orthogonal to the (local) surface. There are very powerful theorems that can relate the circulation of a given field (how much goes around a given closed curve) to the flux of a related field (how much goes through the surface delimited by that closed curve). And if you look at Maxwells equations in their integral form, you will see that they relate fluxes and circulations of E and B. But this is too long a story to be told in a comment.

@@copernicofelinis I'm gonna subscribe and turn on notifications , so I don't miss on this ones

I agree 😊. This voiceover was a recording I did to find out what the length of the video would have been. I thought I could do that in 15 minutes, and it ended up being 33 minutes. Once I cut out all the pauses I had already invested too much time in it to throw it away. I'll try to improve my pronunciation and timing in the next video.

​@@copernicofelinis Apart from that, I find your explanations really clear and well backuped with figures and equations. Keep it going!

​@@Antonio-lp8hxthanks.

You should see a doctor

Thank you. I appreciate corrections. Apologies for mauling your language. 😁

@@copernicofelinis No worries! 🙂

I am well aware of these "probing done right" videos, and what they try to recover is the path integral of the coulombian electric field alone (the middle picture in the diagram with three rings). This is the conservative **part** of the electric field that admits a scalar electric potential. The problem with this approach is threefold: 1. It describes only "half" of the physical system: one should also give the vector potential A to give the complete picture. 2. It forces you to consider a different physics for loops that link a variable flux and loops that do not, making the physics schizophrenic. In particular one has to invent reasons for the failure of Ohm's law inside every coil. 3. Voltmeters measure the difference on voltage, not the difference in electric potential; it is only in the conservative case that these two quantities coincide. I will show this in the next video. So, no, that is not good probing because they put their probes in the middle of the changing magnetic flux region, and because the quantity they recover tells only half of the story.

@@copernicofelinis "It describes only "half" of the physical system:" No, they simply are measuring correctly. "It forces you to consider a different physics for loops that link a variable flux and loops that do not" That is not the case either - they simply show how to measure correctly. "Voltmeters measure the difference on voltage, not the difference in electric potential" Aaaaaannnddd you might want to refresh your memory on that one - the definition of voltage for example - cause that IS the difference in electrical potential. You have not addressed anything but instead are making up ridiculous claims. By now it seems more like you subsrcibed the a false notion of physics and are now becoming more and more reluctant to accept that you were wrong, to the point of throwing a tantrum.

@@ABaumstumpf let's see: what is the potential difference between point A and the top terminal of the smallest resistor, according to Mabilde? How many volts? And what is the resistance of the copper conductor for that same path? Show me how Ohm's law work in this case.

@@copernicofelinis "Show me how Ohm's law work in this case." You are the one making the false claims. You are claiming that the most well known theories of electrondynamics are wrong - the burden of proof is on you. What Mabilde has down was simply showing that Lewin had an error in his measuring-setup and if corrected all his spurious claims fall apart. And now you are getting emotional cause you made the same mistake.

@@ABaumstumpf what setup? Look at the diagrams of the two half circuits where I show the computation of voltage as E x length. There are NO PROBES. The reason the voltmeters in the video show the correct values as computed from first principles is because the probing there is correct. But the values of those voltages are computed without the need of any probes. Oh, BTW, Ohm's law works perfectly fine with my values of voltage. No need to invent a different physics. Mine is the physics taught at Berkeley and MIT. ( Invoking the authority principle is a good thing in a UA-cam video because it allows viewers to know that what they see is not some 'original research' but is at least based on real science).

Each voltmeter shows the line integral of the TOTAL electric field along the path that goes through its probes and its internal resistor. This voltage is the same as the voltage along the tested branch of circuit only if there is no changing magnetic field linked. Each voltmeter is connected to both branches of the circuit: one branch will form a loop with the probes that does not link the core, the other will form a loop that does link the core. The voltage shown by the voltmeter will be equal to the voltage along the branch that does not form a loop around the core. All of this will be shown in the next video. What is important to notice here is that the field is a composition of two components and that this resultant is almost zero inside the conductor portions of the ring (in accordance with Ohm's law).

@@copernicofelinis The low resistance only reduces the Coulomb component along the conductor, it does not reduce the Faraday electric field along the conductor.

​@@woodcoast5026 no. The coulombian component is almost equal to the induced component . They are both around 1.1V/m, their difference is the negligible 0.0002 V/m that is compatible with the high conductivity of copper. The coulombian electric field (almost completely) obliterates the induced electric field in the conductor.

@@copernicofelinis That does not "obliterate" the Faraday field.The Faraday field is still there. The two components are still there forming the two components of the composition .

@@woodcoast5026 yes, that is correct, but the operation of superposition is something that resides in our heads only. What atoms and electrons 'see' and 'feel' is just the resultant field which is nearly zero. And it is nearly zero because the coulombian field has nearly almost entirely canceled the induced field in the copper. The same happens in the electrostatic case: the electrostatic field in a conductor is exactly zero because the external field has been obliterated by the field component generated by the displaced surface charge. Even more interesting: the interior of the conductor *never* gets a chance to 'see' or 'feel' the external field, because the surface charge kills it before it can penetrate the conductor.

It was an April fools joke lol

It was a joke lmao 😂

Just a joke bro :P

The conclusion is that Copernico went back in time so far that he disappeared. He built a time machine.

Toroidal transformers are everywhere, like parsley. Here it is used as a means to generate a sinusoidally changing magnetic flux. If I had had a powerful enough magnet I could have attached that to my grandma's sewing machine and used that instead.

@@copernicofelinis so could you theoretically use a sinusoidal alternating magnetic field to replace the moving magnets of a generator? Like make a solid state generator that use 2 alternating sinusoidal magnetic fields at 90* to eachother.

​@@freemanrader75 I believe the point of a generator with moving magnets is that to convert mechanical energy (used to move the magnets) into electrical energy. If we remove the magnets and replace them with an electrically generated variable field, we end up with a transformer.

Just showing that when a variable magnetic field is present inside the circuit path (the ring with the two inductors) voltage is no longer uniquely defined, and depends on the path. In this case, it depends on which side of the ring the voltmeter is located. Lookup "Kirchhoff is for the Birds Lewin" to see a full fledged lecture on what is going on.

That is correct. If you allow me some nitpicking, what is important is that there be no appreciable variable magnetic field *linked* by the measuring loop. There is a changing magnetic field 'in the vicinity' because the core is in the vicinity' of the probes, but the loop formed by voltmeter, probes and 'nearest capacitor' does not link any of it. That is where one CAN APPLY KVL. Notice that each voltmeter can be seen as part of two measurement loops: one is that described above, the other is the loop formed with the 'farthest' capacitor. This second loop, though, does link the variable magnetic field in the core and KVL does not apply. One needs to use Faraday's law to recover the correct voltage across the farthest branch of the ring.

With an infinitely long solenoid all the flux is contained as well. With a long enough solenoid the flux in any finite region around the core is negligible (because it spreads out over all space). As a matter of fact, Lewin measured the field around his solenoid and confirmed it is negligible where his ring sits. Anyway, how do you explain these results? There is no B field where the ring is, and the self inductance of the loop is negligible at 50 Hz.

@@copernicofelinis Invoking infinity on a tabletop seems a bit "theoretical". After all we are doing experiments. I might add the explanation is due from the theorist / experimenter. Critics are allowed to criticise.. I should not cite self inductance but rather mutual inductance if there were any flux that is . I did not see lewin actuall measure the flux around his solenoid which was not infinite . Even in gigh school 50 yars ago I did this experiment , I did not know EM theory at 15 hut I studied coils about 10 cm and 10 or so turns . One energised weakly with a signal generator the other on a scope . I picked up plenty of signal at more than 10 .

@@georgejo7905 if the ring does not go around the core you won't get any appreciable signal at 50 Hz. I can also devise an experiment where there is no primary coil and the change in magnetic field is caused by a magnet falling through the ring with the resistors. in that case there is not even mutual inductance. Did you read the stack exchange pages I linked in the description of the first video?

@@copernicofelinis Sorry not on stack exchange . No other way to link ? just write it? not sure which video

@@georgejo7905 stack exchange is not a video network. You do not need to be part of it to read the pages. The pages I mean are linked in the description of my first video. Just click there and you will read the html page, no subscription or shady s**it required (maybe they try to put cookies like 99% of the websites)

Input resistance of the scope is 1 meg; probes and copper interconnects in the ring have negligible resistance. Scope measurements confirm what I have measured earlier (not shown in the video) with my multimeter (10meg minimum input impedance). Even the ESR of these cheap caps (which is usually a few ohms) is not playing a part in this circuit. (The ring with inductors is another thing...)

@@copernicofelinis I have made this circuit with resistors. Are the grounds for the channels you are using on your scope for this demonstration electrically connected together by the manufacturer at the scope .

@@woodcoast5026 yes, of course. It does not affect the measurements, as can be inferred from the fact that I get the same values when I measure the voltage with a voltmeter (which is floating).

You can use a voltmeter to measure the AC voltages, like I showed in my other video.

Ideally all the magnetic field is inside the toroidal core. The physical circuit (resistors and copper wire) can occupy a region of space where there is no magnetic field at all. (In practice if a current is flowing due to the induced electric field, this current will generate a magnetic field around it, but in this case it is so small that we can neglect it - and in doing so we also neglect the self inductance of the ring)

Hello C F, you stated incorrectly that Kirchhoff loop rule can be applied to the meter and adjacent resistor. It cannot , because two conductors in parallel do not form a loop. The rule for them is not Kirchhoff it is the Ohms law for conductors in parallel .

@@woodcoast5026 two resistors in parallel form a loop. The rule for parallel resistors comes from applying KVL to such a loop. The voltage across either resistor is the same exactly because of KVL.

@@copernicofelinis You just stated "two resistors in parallel form a loop" that is an absurd statement. When two resistors are in parallel the current runs from one end of the split path to the other end of the split path. KVL applies to conductors in SERIES. The meter and resistor are not in series they are in PARALLEL.

@@woodcoast5026 a loop is any closed path in a circuit. See here: www.electrical4u.com/nodes-branches-and-loops-of-a-circuit/

@@copernicofelinis It is not a loop, it is not a closed path. It goes through the meter and resistor in parallel , and then continues around the ring.

I've been going through this without any induced EMF and it doesn't seem to work with just the pure interpretation of Faraday no matter what I try. Because looking at the resistor voltage measurements, which are supposed to be the only voltages in the circuit and hence are truly representative of the actual electric field in the loop. By working out the net electric field in the loop by superposition, adding the electric field contributions of the voltages across the two resistors around the loop. This means the loop electric field actually runs in the OPPOSITE direction to the current and the electric field in the resistors is in the SAME direction as the current. Have I missed something and been a dumb dumb?

@@biskwit2416 Current going clockwise. R Right, the current goes into the top of R Right and out of the bottom of R Right. E field high at top of R Right. R Left , the current goes into the bottom of R Left and out of the top of R Left. E field High at bottom of R Left.

@@woodcoast5026 R Right current in top Voltage high, bottom Right low. E-field clockwise. From R right top, Voltage high, along loop, field ant-clockwise.

@@biskwit2416 Yes, current flows along an E field from High to Low. There are two fields , one for R Left, and one for R Right.

@@woodcoast5026 Take off R left and short. Just have the right resistor with current clockwise has say PD of 1V running from top to bottom so clockwise. So E-Field is clockwise. Go from top terminal of R at 1V going left around the copper and around to 0V R terminal. Field from 1V to 0V must be anticlockwise. Clockwise through the resistor ant-clockwise in the copper. Is this possible?

That's my broken English. I tried to say: "...and the internal resistance of the meter is the load". It came out as unintelligible maybe, but iirc the automatic translation got it right, despite my pronunciation. (Not so much "portion", which apparently I pronounced "Persian")

@@copernicofelinis yes I hear it . Load

Yes, in fact the multimeter shows practically zero. The 0.0004 volt is nearly zero considering the error of measurement and the fact that a couple of inches of copper wire with 20 mA in it will drop a handful of microvolts. I will add some numbers in the description. I wanted to add titles and drawings, but I was not able to install OpenShot (cx_freeze error) so I just stitched the videos together.

M.I.T. lecture goes into charged sphere in great depth ua-cam.com/video/JhV-GOS4y8g/v-deo.html

Great cheers 👍

It is not a probing problem. You do not even need any probes. Remove the voltmeters entirely. Ohms law V=I*R. The current through each resistor is the same and the voltages are different, and so no doubt the voltages are different. Use an infrared non contact thermometer to display the temperature of each resistor due to power dissipation. The power formula W=V*I rearranged V=W/I , for the different temperatures present, that confirms the presence of the different voltages.

If I had a dime for every time I've read this objection, I'd be running my own Space Agency. :-) Yes, when we break the ring and attach a voltmeter in the gap we just created, we measure (practically almost exactly) the whole EMF. There is no inconsistency at all with the fact that the voltage along the conducting part of the ring - and even the resistors, now - is (almost exactly, because there is a tiny current flowing through the high impedance voltmeter) zero. What the voltmeter on the side of the gap is measuring correctly is the voltage *across* the gap, along any path that does not run around the variable flux region. The crucial point is that in order to be sure the voltmeter is not affected by the changing flux, you must be sure your measuring loop (only that!) does not run around the changing flux region. If you open the ring and you want to correctly measure the voltage *along* the ring with the resistors (or along any path jumping from one end of the cut to the other but running around the magnetic flux region) you must create a measuring loop that comprises this particular path but that does not enclose the magnetic core. In fact, if your measuring loop sits in a region of space free of variable magnetic fields, all the voltages between the same pair of points will be all equal, irregardless of the path of integration. In this setting, the voltage along the probes and the internal resistance of your voltmeter will also be the voltage across the voltmeter and probes, which will also be the voltage along the ring with the resistors. Therefore, by putting the voltmeter on the other side of the gap and the core you can measure the voltage along the opened ring. You get a different voltage from the one across the gap but that's alright: voltage can depend on path. If we remove the resistor an leave a single open circuited coil, the conclusion is that the voltage across the coil is the emf, while the voltage along the coil is (ideally) zero. Again, voltage depends on path, and this dependence is essential in deriving the well known L di/dt formula for the voltage across the terminals of a lumped self-inductance. If you make a multi-turn transformer you will be able to measure different voltages by tapping different turns - but your voltmeter outside the transformer will correctly measure only the voltage *across* the jump between turns, and not the voltage *along* the portion of coil between its tips. If you want to compute the voltage along the coil you need to apply Faraday's law and add (or subtract) to that reading the emf multiplied by the number of times you go around the core. You will get zero in the ideal case, or just a tiny ohmic drop in copper in a real setup. This is why you can't measure any voltage build up in the turns - basically, there is none!

@@copernicofelinisGood day Copernico!! Thank you again and I'm sorry for your frustration in dealing with the same old questions! In saying that, I'm not sure you did what I asked.!!!! The whole crux of the argument seems to be that the scope probes, thus positioned in your experiment, are part of the loop where the field is free of changing flux, hence do not interfere with the results. I asked you to remove the ring totally and just leave the probes in situ, with gnd to gnd where they were originally connected to the ring and the same the other end where you will have to join tip to tip. You will get an EMF similar to what you measured in the ring. That's what I don't get. Does this "pickup" miraculously go away when the probe is attached to the reconnected ring? As far as there being a voltage difference between windings, but no voltage along it, that is something I find unfathomable. I think I'll leave it at that and thank you for your civility and patience. I'm a bit knackered so will read through it again soon! Take care👍

@@biskwit2416 and if I had another dime for this, as well, I'd be headed for Mars, not just a mere 100 km above. Don't worry, questions are fine and this is one question one needs to go through to understand the ring. Of course you will read a voltage on the two voltmeters: the emf will be split in the same ratio as the internal resistances of your voltmeters. So, for example with an emf of 1V, if you have a nice voltmeter with 90 meg input impedance and a thrift store voltmeter with just 10 meg input resistance, you will read +0.9V on the first voltmeter and -0.1V on the second one. Again, voltage depends on path. If the voltmeters were identical you would read +0.5V and -0.5V. But here is the catch: with only two voltmeters and no other component, each voltmeter only sees one measurement loop, and that loop goes around the variable flux region. You can't apply KVL to that, and in fact voltages in that loop depend on the path. Incidentally this single loop shows that the multivalued voltage is a properties of the loop alone, and has nothing to do with probing (the probes here are the circuit itself!) When you consider the ring with the resistors and you attach the voltmeters, each voltmeter sees three different measurement loops: one with the nearest resistor (which is magnetic free and for which you can apply KVL); a second one with the distant resistor that goes around the core, and finally a third one with the other voltmeter (this one as well goes around the core). The only correct 'direct' reading is that associated with the first measurement loop. For the other two you need to apply Faraday's law and subtract (or add, depending on the signs) the emf to find the correct voltage along the other branches. The beauty of it is that correct voltage of the near resistor is equal to the 'correct' voltage of the second resistor minus the emf. There are no inconsistencies. The other loops that go around the core are, so to speak, shunted by the nearest resistor and what you measure is the result of the interplay of all branches which is basically identical to the effects of the emf on the ring alone. You can solve the circuit for the different possibilities: circuit without voltmeter, circuit with finite input impedance voltmeters, and even measurements loops alone (without all other components) to see that the complete solution is the result of the interplay of all branches but the only branches that count are basically those with the small resistors. (Do not expect dramatic changes in the currents due to the aforementioned shunting, the currents are all in the nanoamps range) The presence of the voltmeter branches adds a minor loading effect of a few nanovolts (and a change in current of a few nanoamps). Yes, I did solve the circuit in all flavours.

@@biskwit2416 heres a little exercise that has no variable magnetic field in it. Consider a 1000V generator with an internal resistance of 50 ohm. Attach a 10 meg voltmeter to its terminals. What do you read? And what is the current through the voltmeter? Now shunt the voltmeter with a 0.5 ohm resistance. Do you still read 1000V? What is the current through the voltmeter? And through the 0.5 ohm resistor? Finally consider only the generator with its resistance and the 0.5 ohm load. Has the voltage and the current in the 0.5 ohm resistor changed much without the voltmeter?

Yeah, I could write a book by putting all those comments together. :-] What I know comes from electromagnetism books. Three books in particular have this problem clearly explained. Here are the relevant quotes (from one of my comments in another channel's video, slightly edited to remove non-relevant parts): The first book is *Purcell and Morin* , "Berkeley Physics vol 2: Electricity and Magnetism" Section 7.5 "A consequence of Faraday's law of induction is that Kirchhoff's loop rule (...) is no longer valid in situations where there is a changing magnetic field. Faraday's law has taken us beyond the comfortable realm of conservative electric fields. The voltage difference between two points now depends on the path between them." And then it adds: "problem 7.4 provides an instructive example of this fact." And what is problem 7.4? Lewin's ring, around a toroidal core like the one depicted in this video. The problem is solved in the end of the book. And concurs with Lewin, of course. Let's move to MIT with *Haus and Melcher* , "Electromagnetic Field and Energy". This book is entirely online at web DOT mit DOT edu SLASH 6.013_book SLASH www SLASH (I hate you, youtube!) The chapter on magnetoquasistatics opens with Lewin's ring around a toroidal core. There is even a video demonstration (Demo 10.0.1: Nonuniqueness of Voltage in an MQS System). Again, in line with what Lewin says. The third book has a more applicative slant: *Ramo, Whinnery and vanDuzer* , "Fields and Waves in Communication Electronics". This book explain clearly why voltage has to be multivalued and when and how we can treat it as single valued to simplify our treatment of circuits. It lliterally takes you by hand expanding the integrals and showing you precisely where the L di/dt formula come from. In particular at page 174 of the third edition it says: "let us take a closed integral of electric field along the conductor of the coil, returning by the path across the terminals. Since the contribution along the part of the path which follows the conductor is zero, all the voltage appears ACROSS the terminals." Moreover, on page 179, it explains how in lumped circuit theory we pretend to treat voltage as if it were a potential difference: "In the above we seem to be treating voltage as potential difference when we take voltage of a node with respect to the chosen reference, but note that this is only after the circuit [path] is defined and we are only breaking up the integral of E.dl into its contributions over the various branches. As illustrated in the preceding section, we do have to define the path carefully whenever there are inductances or other elements with contribution to voltage from Faraday's law." Finally, to better understand the difference between voltage and potential difference and how the two concepts are linked in the presence of a variable magnetic field, I would suggest *Popovic and Popovic* , "Introductory Electromagnetics", sec. 14.4 "Potential difference and voltage in a time-varying electric and magnetic field". You can find an extended quote searching "Assigning a notion of voltage even when there is a changing magnetic field" on Electrical Engineering Stack Exchange.

I would say that each voltmeter and its probes are part of multiple loops. The current in the voltmeter is the result of the composition of the effects (not the currents) of all loops. For example, in Lewin's ring, voltmeter A and its probes form three loops: one with R1, one with R2, and one with the other voltmeter. If you consider these loops alone, *deleting the other components* , you get three different currents (namely 0, emf/(R2+Rvm1), emf/(Rvm1+Rvm2) ) which are different from the actual current flowing in the voltmeter when all elements are present (it's a rather complicated function, but we can approximate it with emf/Rvm1 R1/(R1+R2) ). Of course we can look at the measurement loops *with the other elements present* . What we can say in that case is that in the loop with R1 - which does not contain variable flux - we can apply KVL, so the voltmeter will measure the voltage along the branch with R1 correctly (voltage is the same for the path going along the branch and for all paths in the space across the terminals, as long as I stay in the magnetic-free region), while the other loops will produce measurements of the other branches that are altered by the linked emf. The nice part is that these measurements will be altered so as to produce the same voltage associated with the first loop. I've read many times the objection "but the other loops go around the changing flux, so you will induce a current in the probes!!!" The point is that the voltmeter (as well as its probes) will see almost none of that current because it is shunted by the nearest resistor.

@@copernicofelinis Hi Copernico. I analysed the circuit again. I worked it out and posted my new comment on the subject in the comment section for this video.