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There is a famous law which says that a potential difference is produced across a conductor when it is exposed to a varying MF. But, how do you measure it to prove? It is quite practical.

Particularly, I had once a problem developing a power supply. When transistor rapidly opened a loop of powerful current, I had very strange effects and discovered that there is a voltage difference across the same wire, without any resistor between the probing points! The longer was the distance between probes, the higher was the voltage. Then I have realized how the real the law is. But, what I could not understand if the voltage were spikes induced in the wire indeed or they were purely fictional, induced right in the oscilloscope probes. I expected that voltage drop must be across the loop-breaking transistor but why was it rather distributed along the resistance-free wire?

Look at dr. Levin's voltmeter. What does it measure? I ask this question partially because he tells about non-conservative measurements, which depends on the path, but does not explain how to set up the paths to measure -0.1 v in one case and +0.9 v in the other, between the same points.

How the typical voltmeter works? There should be some known high resistance and small current through it shows how large the voltage is. But here, in addition to the D-A induced voltage, the EMF may be added because the is also induced current flowing through the voltmeter. How much is the effect?

In school I also had a problem with understanding what if the loop is open? You have a wire. The law will induce potential difference at its ends. But how do you measure the difference? Created such voltage in a short wire, we can be sure that closing the ends with voltmeter probes will not create any current in the voltmeter (just because magnetic field supports the difference. If it just created it then why should it let the polarized charges reunite when a parallel wire is connected?). So, no current and the voltmeter will show 0 voltage despite we know that Faraday law says that must be some. Do you understand what I am talking about? The field creates the potential difference that cannot be measured. It is like gravity stretches a spring but we cannot measure the force created because spring contraction force is balanced by gravity and your dynamometer shows 0. This is my concern that I cannot understand. How do you measure voltage difference when Faraday law precludes the polarized charges from the opposite ends of the wire to come together?

Val
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    Lots of questions here. Could you possibly itemize/list all of the important questions at the end? Otherwise, it's a little hard to follow – Jim Apr 26 '13 at 17:43
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    Also I'd like to say that don't just post large passages. It's hard to read. At-least break into small parts with some quoting and else to make it appealing. – ABC Apr 26 '13 at 17:52
  • Would be fine if down-voters explain their choice... I believe this is because the question is not so clear. This question is so also because the topic is highly confusing. It nevertheless discuss a really interesting topic not so known, historically called the Maxwell-Lodge effect, a kind of classical pendant of the Aharonov-Bohm effect. – FraSchelle Apr 26 '13 at 20:16
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    @Val Part of the questions asked here overlap with this http://physics.stackexchange.com/questions/23835/l. Nevertheless, the interpretation and the understanding of the Levin's lecture is original in SE. Would it be possible for you to edit your question such that it becomes more readable. More people could then help you understanding your point. – FraSchelle Apr 27 '13 at 08:07

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I'm not sure I understood all your points. I suggest you to read this beautifull paper

  • Romer, R. H. (1982). What do “voltmeters” measure?: Faraday’s law in a multiply connected region, American Journal of Physics, 50(12), 1089. http://dx.doi.org/10.1119/1.12923

if you can find it. It's seems it's exactly what Prof. Levin is doing in his lecture. The proof is clear. If you have some difficulties to obtain this paper, I have some notes about this paper that I can share on SE too.

Edit: If you can find this reference, you will see that the only thing which matters is the position of the circuitry relative to the solenoid. More explicitly, this is the topology of the circuit which matters. It will then be clear for you that

  • you would record nothing if you don't enclose the solenoid, subsequently you would measure nothing until you close (i.e. make a turn) the circuit,
  • putting your wires far away from the solenoid does not change the measured voltage, nor putting them close to he solenoid, the only thing which matters is the previous point,
  • doubling the number of turn around the solenoid doubles the corresponding voltage,
  • the wires participate for nothing in the measurement except for encircling the solenoid (as stated in the previous points). This last point is true only for idealised conditions of course (infinite solenoid, no resistive effect in the wire, ...).

Now, some of your questions:

There is a famous law which says that a potential difference is produced across a conductor when it is exposed to a varying MF. But, how do you measure it to prove? It is quite practical.

You have two ways to see induction, as clearly discuss in the Feynman lecture. To be honest, I do not know a better book to start with.

  • you close a loop with a moving bar, whereas a coil was previously inside the loop. Then you record a voltage drop through Faraday's law because the circuit is moving (if you prefer, the path you calculate your integral with).

  • you close a loop and encapsulate a time dependent magnetic field via time dependent current passing through the coil. Then you record a voltage drop because of the time variation of the magnetic field itself.

In both case you have a time varying flux. As Prof. Levin says: "Misterrr Farrraday is happy with that !"

Look at dr. Levin's voltmeter. What does it measure? I ask this question partially because he tells about non-conservative measurements, which depends on the path, but does not explain how to set up the paths to measure -0.1 v in one case and +0.9 v in the other, between the same points.

Yes he does ! It is always from top to bottom (A to D point if I remember correctly), first passing on the left, second time passing on the right. Edit: The previous statement was confusing. When discussing magnetic flux, you need to define convention for following the circuit path, i.e. rotation direction. The path for -0.1 V is the counterclockwise path from A to D, the path giving +0.9 V is the clockwise path from A to D. That explain the signs also.

How the typical voltmeter works? There should be some known high resistance and small current through it shows how large the voltage is. But here, in addition to the D-A induced voltage, the EMF may be added because the is also induced current flowing through the voltmeter. How much is the effect?

I suggest you to read the wikipedia page http://en.wikipedia.org/wiki/Voltmeter. A perfect voltmeter has no EMF. (Edit: see also http://en.wikipedia.org/wiki/Electromotive_force for the definition(s) of an EMF and its different interpretations.)

For me your last paragraph is incomprehensible. Faraday's law relates voltage drop to time varying magnetic flux, so it doesn't apply to open circuit.

Final edit: A way to avoid the flux drop: From http://dx.doi.org/10.1119/1.12923, see https://i.stack.imgur.com/X5qPb.png.

FraSchelle
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  • Why do you post the links to the voltmeter? If they cannot help you answering the question then they don't help the others. What does it mean "passing on this (or that)" side? It is a side on the flat blackboard. Even there we can shape conductros in different ways to change the pathes. In 3D reality we can twist the cables as we like. There is no left and right in reality. Do you say that the cabling must be like on the blackboard? Why? Why do you say that Faraday law is inapplicable to an open wire? Do you mean that voltage drop can occur only in a closed loop? LOL. I feel you are American. – Val Apr 26 '13 at 19:29
  • Why do you say that perfect voltmeter has no EMF if perfect wire has neither. Yet, Faraday law makes things imperfect. Here you should start feeling what I am asking. – Val Apr 26 '13 at 19:30
  • @Val I'm actually French but that's not the debate here I believe :-). The right and left side are indeed the one of the blackboard, and they indeed matter. That even the only thing which matters ! Making one turn more around the solenoid will change the voltage drop for instance. Before any discussion, you should first define a convention for turn and path. I followed the Levin's one. Positive is the clockwise direction if I remember correctly. On the wikipedia page about the Voltmeter, you will see that it records flux. Since you need loop to define magnetic flux, an open wire does not – FraSchelle Apr 26 '13 at 19:42
  • ... verify Faraday's law. It does not mean that there is no voltage drop along an open wire, it means Faraday's law does not apply with the magnetic flux. Does it help you ? – FraSchelle Apr 26 '13 at 19:44
  • @Val I apologize, I wrote a mistake in my last sentence. Hence corrected, it is: "Faraday's law relates voltage drop to time varying magnetic flux (and not field), so it doesn't apply to open circuit." – FraSchelle Apr 26 '13 at 19:46
  • how can voltmeter can be the EMF-proof if it contains only tiny fraction of the loop? The most of the measuring loop are the probe wires. Whether it is left or right it is decision of experementer. – Val Apr 26 '13 at 19:47
  • @Val Sorry, I once more confused you I believe. So, say again: you first need to choose a convention, then you can only discuss differences in a quantity. That's the same as for voltage drop along a wire, except it is now for flux, so you need to define a convention for rotation around the loop. In Levin's lecture, one rotation is clockwise (what I called "on the right"), the other is counterclockwise (what I called "on the left"). Sorry again, my convention was totally misleading. I'll correct the post. – FraSchelle Apr 26 '13 at 19:52
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    I do not see anything about flux in the definition of the law. It says about variable field and conductor, which can exist without the loop. But ok, let's consider that loop must be closed. The voltmeter's loop has the same 1v drop, despite it has megaohm resistors in the circuit. So, presence the voltage is independent of resistors, no matter how large. Open wire is a loop, closed by very great resistor. Thereby, we see that all voltage drops at the meter. But flux must be greater for the voltmeters because their loops cover larger area. Why do we get 1v in sum? – Val Apr 26 '13 at 21:57
  • Why Levin says nothing about the topology? The schematic of such kind is always logical. That is why the word "schematic". It means "abstract view". Real topology may have arbitrary physical layout. If Levin would violate the tradition and presented the physical layout this way, he had to stipulate this very loudly. Because he did not, it is difficult to believe that his schematic is a topology. – Val Apr 26 '13 at 22:00
  • @Val Ah ok my bad, I misunderstood your confusion. Of course the exact value of the voltage will depend on the resistance. Its presence too, since without resistance you would have no voltage drop. That's the famous "non-conservative effect" in Levin's lecture. Won;t you say that the Faraday's law is $\int E dl + d \int BdS /dt = 0$, isn't it $\int BdS$ a magnetic flux ? It also says that the path integral used to evaluate the voltage drop must correspond to the boundary of the surface used to evaluate the magnetic flux, isn't it ? What is the surface associated to an open path ? – FraSchelle Apr 26 '13 at 22:04
  • @Val Sometimes people write the Faraday's law explicitly $$\int_{\partial\mathcal{S}}\mathbf{E}\cdot d\mathbf{l}+\dfrac{d}{dt}\iint_{\mathcal{S}}\mathbf{B}\cdot d\mathbf{S}=0$$ with the notation $\partial\mathcal{S}$ being the boundary of $\mathcal{S}$. Is it more clear ? – FraSchelle Apr 26 '13 at 22:09
  • @Val I just read your last comment about the topology. I do not know any introductory lecture book which discuss electromagnetism using topology considerations, despite it is full of it... The Gauss law and the absence of magnetic monopole are obviously topological in nature. A solenoid is also topological in essence: it relates the $B$ field to the current via the number of turns of the coil, what is called the winding number in math. Perhaps you should not worried too much about that. Topology in the Levin's lecture is just a way to say that $\int Edl$ is not path independent. – FraSchelle Apr 26 '13 at 22:17
  • Yes, I understand the law. I just do not understand why there is a "conductor" rather than "loop" in the definition and how should I attach the voltmeter wires in order to avoid induction. This law also says nothing about resistance. Loop drop is independent of any resistance. – Val Apr 26 '13 at 22:19
  • So he uses topology indeed ! His lecture is all about topology. Mathematicians would say that $E$ is not integrable. Levin clearly shows that the clockwise and the anti-clockwise path from A to D does not give the same result. This is because the circuit winds around the solenoid, see my four points describing the Romer's paper in my answer. – FraSchelle Apr 26 '13 at 22:19
  • @Val I added the link to a picture to avoid the flux drop. – FraSchelle Apr 26 '13 at 22:25
  • Lol, here voltmeters participate in so much loops that I am lost in what is the loop. – Val Apr 26 '13 at 22:38
  • @Val Well, the loop is empty. There is no magnetic flux inside since it does not enclose the solenoid, and the two voltages (clockwise and counter-clockwise) are the same. Easy, isn't it ? – FraSchelle Apr 26 '13 at 22:40
  • I do not see why solinoid must be inside. Furthermore, I read the Wikipedia: Equivalently, it is the voltage that would be measured by cutting the wire to create an open circuit, and attaching a voltmeter to the leads. So, it clearly says that closed loop is unnecessary and you can measure voltage across the ends of an open wire. – Val Apr 27 '13 at 08:41
  • Might be the voltmeter loop topology must match exactly the path we are measuring? This is stronger than just choosing between left and right. – Val Apr 27 '13 at 08:48
  • @Val I don't know where you read this, but it is clear for me that cutting the wire and attaching a voltmeter to the leads is the correct way to make a loop ! Topology has nothing to do with matching a path, it just answers the question: do you circle the solenoid or not ? You may find this answer (and the related question) interesting: http://physics.stackexchange.com/a/62444/16689 – FraSchelle Apr 27 '13 at 08:54
  • Thanks for the reference. But you can attach a voltmeter in infinitude of ways of arbitrary long wires. It is strange that you asking how to close an open wire but all confusion disappears when you hear that there is a voltmeter in the closure. – Val Apr 27 '13 at 11:36