Why does a voltmeter measure a voltage across inductor?

In summary, the potential difference across an inductor is expected to be zero, according to circuit theory. However, in reality, real inductors have side effects such as resistance and parasitic capacitance which can affect their behavior. Additionally, in an ideal case where the wires are superconducting, the electric field inside the coil would be zero and charges would continue to flow due to self-induction. In this ideal case, a voltmeter may not measure the voltage due to the vector potential, but can still measure the electrostatic potential generated by excess electrostatic charges on the inductor. However, in certain scenarios, a voltmeter can measure the voltage due to the Faraday EMF if it is affected by the vector potential
  • #106
A minor "typo" in post #103. In M.K.S. ## \ ## ## \mathcal{E}=-d \Phi/dt ##. The "c" in the denominator is only in c.g.s. units.
 
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  • #107
@OnAHyperbola Please read my post #68 if you haven't already. I think you might find it interesting. With so much other discussion going on, it is likely you missed it. Until just a moment ago I hadn't seen your post #101 amidst all of the other discussion. Note: The ac power lines use f=60 Hz in the U.S. ## \ ## In Great Britain and some other places they might use f=50 Hz, but hopefully you can follow the idea that Dr. Lewin's example shows precisely how a stray EMF can enter into your voltmeter and/or the circuit you are working with.
 
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  • #108
vanhees71 said:
I don't care, whether it's circuit theory or not.
Then it is odd that you would repeatedly claim that it is circuit theory in the face of well reasoned objections. When I am in such a position (making a mistake on a topic that I don't care about) I just say "oops", learn, and move on. I have to say that this interaction with you has bothered me greatly.

vanhees71 said:
You only need Maxwell's equations in a very simple way
Yes, I agree.

vanhees71 said:
the quasitationary approximation that the current is observed on every branching (this is what distinguishes for me circuit theory from the full Maxwell equations
Again, you don't get to push personal definitions here. The fundamental assumptions of circuit theory are more than that and this has been explained and referenced in detail.

vanhees71 said:
Perhaps Dale could point to the place, where this analysis is beyond usual circuit theory
The place pointed to by the line labeled "changing magnetic flux".

vanhees71 said:
so Lewin's example obviously only misses item 3, i.e., that magnetic couplings are only within a single element of the circuit. I.e., for calling the problem one of "circuit theory" it's not allowed to have a "magnetic flux" within a loop of the circuit as here.
Yes, exactly!

vanhees71 said:
Well, as I said, it's only semantics. Lewin's analysis, as expected, is of course correct although it goes beyond the strict definition of "circuit theory"
Yes, as I have agreed several times. His analysis is correct, the problem is exactly his semantics. The "for the birds" bit is just semantics, but it is such awful semantics that it needs to be objected to.

KVL is not for the birds, it is for circuit theory and it is valid whenever the assumptions of circuit theory are valid.
 
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  • #109
OnAHyperbola said:
I think one can go about pretending there really is a potential drop across the inductor as long as one knows what's really happening behind the scenes and the assumptions that one is going in with.
Yes, I think that is the take home message. I will try to post a little about assumptions tomorrow.
 
  • #110
Charles Link said:
One item that puzzled me here, [I think I have it figured out to my satisfaction, but it is somewhat complex and everyone (even the most astute ones) are likely to come up with interpretations that differ somewhat], is that the electric field E from the EMF points in the direction in the inductor (just like the EMF inside a battery) from the minus end to the plus end. For an electrostatic component, such as a capacitor, the electric field E points from plus to minus. A voltmeter (or oscilloscope) has no trouble reading the voltage of either of these components. For the inductor, ## V=\int E \cdot dl ## and is very well defined, but the sign is actually opposite what you would get if the E as a function of position were an electrostatic field. Thereby the EMF E is a somewhat different entity than an electrostatic field. The voltage V is however very well defined from an EMF. The sign actually gets reversed when computing the voltage (please check this carefully=I do think I have it correct), but in any case it can be written as a potential (voltage) even if the curl E is non-zero. When sensed with a voltmeter or oscilloscope, these devices are unable to distinguish whether they are reading an EMF or reading an electrostatic voltage.
Repeating my post #25. Also @OnAHyperbola my post #68 I think you might find of interest. If you have figured out from all of this how changing magnetic fluxes can affect a circuit and can accurately predict what the voltmeter will read when there is a changing magnetic flux present, (and it will depend on how you string the wires from the voltmeter to the circuit), I think you have learned quite a lot.
 
  • #111
Charles Link said:
A minor "typo" in post #103. In M.K.S. ## \ ## ## \mathcal{E}=-d \Phi/dt ##. The "c" in the denominator is only in c.g.s. units.
I never ever use SI units in electromagnetism, if I can avoid it. I use exclusively rationalized Gaussian units (aka Heaviside-Lorentz units) as in HEP and high-energy nuclear physics.
 
  • #112
Dale said:
Then it is odd that you would repeatedly claim that it is circuit theory in the face of well reasoned objections. When I am in such a position (making a mistake on a topic that I don't care about) I just say "oops", learn, and move on. I have to say that this interaction with you has bothered me greatly.
Me too, because it's just a misunderstanding between a physics and engineering definition. I'll try to avoid the label "circuit theory" from now on, because it seems to have another meaning than what I learned in the E&M lecture. Sorry for any confusion this might have caused in this discussion.

Just a note concerning this book by Nielsson. From a physicist's point of view, it's pretty inaccurate and even dangerous in its language concerning percisely our debate. I don't know, what happens if you try to solve a problem with moving parts like in a generator or motor, but I guess that's also not "circuit theory" in this sense. As far as I can see, that's not even treated in the book...

My point of view is also found in at least one textbook for electrical engineers, which I like very much

K. Simonyi, Foundations of Electrical Engineering, Pergamon Press (1963)

although I know in better detail the German Edition ("Theoretische Elektrotechnik") of 1993. There's a large chapter on circuit analysis, and there everywhere the correct Kirchhoff rules, including the time derivative of the magnetic flux is used.
 
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  • #113
@Charles Link Yes, I did read your post #68 and it is a very good example illustrating the effects of a changing magnetic flux on the voltmeter reading; one should be careful what the wires are going over/around and how it could leave you scratching your head if you haven't noticed :)
 
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  • #114
vanhees71 said:
I never ever use SI units in electromagnetism, if I can avoid it. I use exclusively rationalized Gaussian units (aka Heaviside-Lorentz units) as in HEP and high-energy nuclear physics.
Then it isn't a typo. My mistake. 95% of the time or more, the voltage equations of the electrical engineer will be in SI units. The conversion from gauss to Webers/m^2 is an easy one, but the voltage, current/charge, and resistance can be rather cumbersome in converting from c.g.s. to SI. Maybe this might partly explain the different viewpoints between @Dale and @vanhees71 =a EE vs. someone doing more theoretical work. In E&M calculations, I also favor the c.g.s. units, since we used the book by J.D. Jackson Classical Electrodynamics in graduate school. However, it always takes me a few minutes, (basically by comparing the two Coulombs law formulas ## v=q/r ## vs. ## V=Q/(4 \pi \epsilon_o R) ## and using e=q= 4.801 E-10 e.s.u.'s and e=Q=1.602 E-19 Coulombs., and letting R=1 meter so that r=100 cm) , to write out the conversion factors when I am looking for the voltage that I'm going to measure with my SI voltmeter. The difference in voltages is a factor of 300, but I can never remember if you multiply or divide. :-) :-) :-)
 
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  • #115
vanhees71 said:
Just a note concerning this book by Nielsson. From a physicist's point of view, it's pretty inaccurate and even dangerous in its language concerning percisely our debate.
But it isn't written for a physicist's point of view, it is written for electrical engineering students taking their first course on circuits with no background knowledge of Maxwell's equations. The assumptions are standard, although the wording does vary from text to text depending on the previous knowledge of the target audience.
 
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  • #116
OnAHyperbola said:
as long as one knows what's really happening behind the scenes and the assumptions that one is going in with
I wanted to go back and highlight this comment of yours. I think it is fantastic that you have this level of scientific clarity in high school.

In any scientific endeavor you will use some sort of equation and some assumptions. For instance, in high school you may have been exposed to the "SUVAT" equations, such as ##v=u+at##. This is a perfectly valid physics equation, but you need to understand its assumptions. There are some basic assumptions from Newtonian mechanics, such as that it is only good for velocities much slower than the speed of light, and (perhaps more importantly) there are some additional assumptions such as that the acceleration is constant. If you use the equation when the assumptions are valid, then you will get results that are physically meaningful. If you blindly "plug and chug" without consideration for the assumptions then you will get nonsense, not because the formula is wrong but rather because you applied it in the wrong situation.

Often, the assumptions will help you select the right formula to use, and the better assumptions you can make the simpler your formula can be. If you understand that, then you will have an academic and scientific advantage over your peers who focus exclusively on the variables.
 
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