# measuring electromagnetic induction

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?

• 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
• 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
• @Val Part of the questions asked here overlap with this 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 8:07

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, ...).

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 http://i.stack.imgur.com/X5qPb.png.

• 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
• 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