Why does a voltmeter measure a voltage across inductor?

[Charles Link]

Back to the resistor ring I mentioned in post #27=You get an EMF and a current around the ring. You can read a voltage drop as 1/4 of the EMF for the ring or 3/4 for the same two points depending on how you run the wires to your voltmeter. In any case, in both cases, the voltmeter reading is unambiguous in that in can be accurately predicted with theoretical calculations and verified by experiment. The two points on the ring can not be assigned any precise potential difference, but it is a well defined voltage reading that is observed for any given configuration of the wires connecting to the voltmeter.

 

[vanhees71]

Do you have an accurate definition of this problem? If I understand it right, it's exactly Lewin's example. The following writeup of his is even more clear than the lecture:

http://videolectures.net/site/norma...vative_fields-do_not_trust_your_intuition.pdf

 

[Charles Link]

Do you have an accurate definition of this problem? If I understand it right, it's exactly Lewin's example. The following writeup of his is even more clear than the lecture:

http://videolectures.net/site/norma...vative_fields-do_not_trust_your_intuition.pdf

A quick look at it=yes, that is the type of scenario I am referring to. I have previously worked a couple of variations of this problem, (I've seen it on two or 3 occasions), but the same concepts are employed in the solution. You pick up the EMF from the changing magnetic field in your circuit equation for a given loop (including a loop containing the voltmeter) if the loop encloses the region of changing magnetic field. Perhaps @vanhees71 that this is the answer for @Dale that the circuit equations are always well defined (e.g. for the purpose of computing the current that flows), but you can't necessarily assign a potential to every point in the circuit.

 

[OnAHyperbola]

I am going to have to agree with @vanhees71 here. EMF and electrostatic potential are completely different fish. Clearly, we are going to have to deal with EMF in a separate manner. I read somewhere that EMF is to a date as electric potential is to the time between two dates. That sounds right.

I think the EMF is this ever-present thing that is there at all points on the loop and that the voltmeter picks up on. The electrostatic potential in between any two points on the superconducting wire is zero, but the voltmeter picks up on the EMF. But if you follow this model, what the voltmeter should pick up if it is connected to either end of the battery, for example, should be the sum of the potential and EMF=V_Battery-Ldi/dt

I had watched the Lewin lecture and the lecture supplement a while ago. There are two resistors with different resistances on either side of a closed circuit and a changing B-field through it. The voltmeter connected to R₁ reads a different value from the one connected to R₂, and the sum of their absolute values gives the EMF.

 

[Dale]

What has Sect. 11.3 to do with our debate

It describes one of the conditions under which circuit theory can be applied. The induced E field does not need to be negligible everywhere, just on some surface surrounding the inductor.

 

[Dale]

Perhaps @vanhees71 that this is the answer for @Dale that the circuit equations are always well defined (e.g. for the purpose of computing the current that flows), but you can't necessarily assign a potential to every point in the circuit.

Well, I would have a slightly different take. The circuit equations only hold under certain circumstances/assumptions. When those assumptions are violated then you run into trouble if you try to apply them. However, when those assumptions hold then you can use circuit theory just fine.

I think that it is important to understand the underlying assumptions. Then you can avoid using circuit theory when it doesn't apply and avoid blaming circuit theory for not being able to solve problems that violate it's assumptions. It is a useful theory and by keeping its assumptions in mind then you can avoid over complicating things when it applies and you can avoid over simplifying things when it doesn't apply.

I feel that this thread is diving into the over complicating side.

 

[Charles Link]

I think it was quite useful to explore the circuit theory in a little more depth how we did through this discussion and observe in more detail the origin of voltage sources in electrical components. It was interesting to compare the inductor with its EMF to the voltage of a capacitor. In particular, as I mentioned in post #25, the Faraday electric field $E$ in the inductor points toward the plus voltage end of the inductor [please verify this for me=I checked the calculation a couple of times, but it helps to have it validated-e.g. when you apply a voltage to the inductor, the Faraday $E$ (where $\nabla \times E =-dB/dt$ )will oppose any currents that the driving voltage attempts to send through the inductor] while for a capacitor , the electrostatic type electric field points from the plus voltage end to the minus. The voltage of both of these components can be readily observed with a voltmeter or oscilloscope, and follow precisely what the circuit equations tell us, but it was informative to look more closely at the origins of these voltages from an electric field standpoint to be able to establish a consistency in going from circuit equations to calculations done using Maxwell's equations.

 

[Charles Link]

And one question for @OnAHyperbola : Assuming my post #37 is correct, when you put a voltmeter across the inductor (of post #37), are you measuring what is going on inside the inductor or are you simply measuring the driving voltage?

 

[OnAHyperbola]

@Dale So there isn't a conservative field in the wires, just the induced one? That actually makes sense.
@Charles Link I'm measuring what's going on inside the inductor.

 

[Dale]

I'm measuring what's going on inside the inductor

Then I really misunderstood your opening post.

If you are looking inside the inductor then you cannot use circuit theory, you have to use Maxwell's equations. You have to specify the geometry and the boundary conditions and so forth and there is no such thing as L. The behavior is determined by the geometry. You also have to use a similar level of detail to model the voltmeter. Which is the point of Dr Lewins paper above.

If you do all of that then you can calculate what the voltmeter will read.

 

[Charles Link]

@OnAHyperbola you "probe" the inductor with a voltage source, you basically can determine its impedance. If you probe the combination of the voltage source plus inductor with a voltmeter, you really don't get much info about what electric fields exist and where. You simply get info on the voltage at the point between the voltage source and the inductor. The circuit theory equations for the inductor tell you what to expect for your voltage reading. The Maxwell equations are a little more complex, but tell you what the B and E fields are doing. Using for a long solenoid $B=\mu_o n I$ where $n$ is number of turns per unit length and $EMF=-d \Phi/dt$ where $\Phi =BAN$ where N is number of turns, you can compute $L=\Phi/I$ and $EMF=-LdI/dt$. This is all relatively straightforward. The part that puzzles me is a more careful look at the inductor shows the $E$ from the Faraday EMF points opposite the direction than the $E$ does for a resistive or capacitive component when the voltmeter measures a positive $V$. The driving voltage (which is the same as the measured voltage) should generate an electric field in the inductor that precisely cancels the electric field from the inductor EMF which would make sense for the case where your inductor is a very good conductor or a superconductor. Meanwhile, you've asked some very good questions and I think you're kind of on the right track of answering at least some of them. This last simple scenario of a voltage driving an ideal inductor should hopefully yield an answer that is entirely consistent in both the circuit theory and Maxwell's equations.

 

[OnAHyperbola]

@Charles Link Yes, that is puzzling. If you look at Lenz's law, the induced E-field is the same direction as those in the resistor and capacitor. All of these act to lower the potential. A voltmeter should measure a negative value across the inductor. Once the charges have gone around the loop (assuming the starting potential was zero) the potential should be -L×di/dt. Twice around should be -2L×di/dt. 100 times should be -100L×di/dt.

 

[Charles Link]

@OnAHyperbola It is my observation (I haven't yet verified it with any textbook or other source) that the voltage from an EMF type source is $V=+\int E \cdot dl$ while the voltage from an electrostatic type $E$ field is $V=-\int E \cdot dl$. It seems you might also concur with this observation, but it would be good to get an item like this confirmed with a more authoritative source than a couple of us doing a couple of calculations. I do think we have it right, but I'd like to see it addressed in a textbook.

 

[Delta2]

You can't read off a potential, because there's none! You can read off an EMF though! Otherwise that's the kind of example I have in mind. I'll try to google this very nice example from Lewin's lecture.

You make it sound like the scalar potential dies when the vector potential arrives. It is $E=-\nabla V-\frac{\partial A}{\partial t}$ at all times. The scalar potential and the charge densities that produce it are still there in the case of the inductor and the voltmeter reads the scalar potential (unless we do winding of the voltmeter connecting wires into the helix of the coil).

 

[Charles Link]

@Delta² Could you perhaps take a look at my question of post #43. It is sort of key to having any kind of consistency with how I'm thinking this stuff works...

 

[Delta2]

Not sure what you trying to say there but in the case of an inductor the E-field from the vector potential is in opposite direction to the E-field from the scalar potential.

 

[Charles Link]

Not sure what you trying to say there but in the case of an inductor the E-field from the vector potential is in opposite direction to the E-field from the scalar potential.

@Delta² That's a good start to what I was looking for. The other item is that, as I mentioned in a couple of the previous posts on this thread, the total E field in the conductor (or superconducting) inductor (solenoid) will be zero because the E field from any driving voltage (which could be electrostatic but it doesn't need to be) will pretty much cancel the Faraday E field in the inductor. If you agree with this part, I think I have a reasonably consistent description of what the measured voltage consists of. I'm not completely sure whether it isn't an EMF that gets measured at times, but that would be computed with the "+" sign in the integral $V=\int E \cdot dl$.

 

[Delta2]

@Delta² That's a good start to what I was looking for. The other item is that, as I mentioned in a couple of the previous posts on this thread, the total E field in the conductor (or superconducting) inductor (solenoid) will be zero because the E field from any driving voltage (which could be electrostatic but it doesn't need to be) will pretty much cancel the Faraday E field in the inductor. If you agree with this part,

yes I totally agree here.

I think I have a reasonably consistent description of what the measured voltage consists of. I'm not completely sure whether it isn't an EMF that gets measured at times, but that would be computed with the "+" sign in the integral $V=\int E \cdot dl$.

Don't get this part, how can you directly measure the EMF produced by an inductor? if you put a voltmeter at its endpoints, wouldn't that measure the scalar potential difference in the endpoints?

 

[Charles Link]

yes I totally agree here.

Don't get this part, how can you directly measure the EMF produced by an inductor? if you put a voltmeter at its endpoints, wouldn't that measure the scalar potential difference in the endpoints?

If you take an inductor that already has a DC current running through it (setting up the initial conditions) and connect it to resistor, isn't the driving voltage in the R-L circuit from the inductor's $EMF=-LdI/dt$ which is the Faraday EMF=essentially an E-field from the faraday term $\nabla \times E=-dB/dt$? $V=\int E \cdot dl=-d \Phi/dt$, etc... That EMF (which comes from the $dB/dt$) can be readily measured with a voltmeter. Similarly for a transformer EMF and any other driving voltage of this form... I think it is impossible for a voltmeter or oscilloscope to distinguish between an EMF type driving voltage and an electrostatic one...Another earlier part of this discussion was that $\nabla \times E$ is non-zero, so is it mathematically valid to use this, and also is it perhaps incorrect to call this EMF voltage a potential? The whole thread has really been quite interesting...

 

[Delta2]

In my opinion the EMF that originates from any sort of structure with $-d\Phi/dt$ drives the buildup of charge densities at the end points , so it creates a scalar potential and then this scalar potential is responsible for what is happening in the rest of the circuit.

 

[Charles Link]

In my opinion the EMF that originates from any sort of structure with $-d\Phi/dt$ drives the buildup of charge densities at the end points , so it creates a scalar potential and then this scalar potential is responsible for what is happening in the rest of the circuit.

I think this could be a possibility. Although I've had plenty of E&M courses and also been through the circuit theory in college, (1974-1981), this is the first time I've looked this closely at the inductor and observed the $V=\int E \cdot dl$ with a plus sign, i.e. that the Faraday E behaved differently from the electrostatic in computing a voltage. It has proved for interesting discussion in any case. The other item that surfaced was back around post #25, where @vanhees71 was able to find a "link" to the item I mentioned a day or two ago about a changing magnetic field inside a continuous resistor ring and connecting a voltmeter. The way the wires are wrapped from the voltmeter to the ring can put the voltmeter circuit in a loop that has an EMF from the changing magnetic field and thereby it gets a different reading, for the same two points, etc.

 

[OnAHyperbola]

@Charles Link You are absolutely right in that the voltmeter won't distinguish between an EMF type driving voltage and an electrostatic one. I think I may have found an answer to our conundrum. And, yes, it is incorrect to call this EMF a potential difference; it is an electromotive force, but much, much more importantly, it is a change in magnetic flux THROUGH A SPECIFIC LOOP wrt time. I think this conversation is very enlightening:

(look at the newest comment on the YouTube page)My question was, how is there a drop in potential if there is no field inside the inductor? There isn't. The voltmeter doesn't care that there isn't. It can't think. The wires of the voltmeter and inductor make an Amperian loop, through which the magnetic flux changes in time. The changing flux through this loop creates an emf around the loop and a feeble current in it. EMF/R=I , which the voltmeter interprets as a p.d.
That seeming drop in potential/voltage/EMF/whatever didn't even exist until the voltmeter came along to measure it and make its own loop. It is our measuring it that forced it to exist in a measurable way. (This is just my take)
Also, I was wrong when I said the voltmeter should read a p.d when attached to any wire in the circuit. No changing flux there==> no emf around the loop==> 0 reading

 

[vanhees71]

It describes one of the conditions under which circuit theory can be applied. The induced E field does not need to be negligible everywhere, just on some surface surrounding the inductor.

I'm really puzzled about what you mean. Of course, the induced E field is not neglible. What's neglected in quasistationary circuit theory (i.e., for the case that the typical wavelength of the fields involved is large compared to the geometrical extensions of your circuit) is the displacement current. Faraday's Law is fully implemented, i.e.,
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}$$
is integrated along an arbitrary surface. For circuit theory the best choice is to use a surface with the wires, resistances, coils, capacitors, and sources as its boundary. Then you use Stokes's Law. If the circuit is at rest this results in Kirchhoff's Law including $-\dot{\Phi}/c$, where $\Phi$ is the magnetic flux along the surface.

Take the above example of a coil $L$ and a resistor $R$ in series connected to some voltage source/battery $U$. Integrating in direction of the current density, you get
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=R i-U=-L \dot{i}.$$
As you very clearly see the closed-path integral of $\vec{E}$ is not 0 if the current is not stationary, and thus the electric field is not conservative in this case. See also Levin's example where the source is given by an induced EMF and the practical importance for the path dependence when measuring "voltages" in such cases.

For Kirchhoff's other law you use the Ampere-Maxwell law, reduced to the Ampere Law by neglecting the "displacement current", leading to the quasistationary condition
$$\vec{\nabla} \cdot \vec{j}=0,$$
leading to "current conservation" at branchings within the circuit. It's nothing else than the conservation of electric charge.

 

[Delta2]

Take the above example of a coil $L$ and a resistor $R$ in series connected to some voltage source/battery $U$. Integrating in direction of the current density, you get
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=R i-U=-L \dot{i}.$$

Essentially you using faraday's law in integral form here right? If so why do you evaluate the integral of left hand side as only $Ri-U$?? Shouldn't you include the term $\int E\cdot dl$ for the path inside the coil wire?Do you let the E-field to be zero inside the wire of the coil? Or the path of integration does not include the spirals of the coil and it goes as straight line through them?

 

[vanhees71]

Essentially you using faraday's law in integral form here right? If so why do you evaluate the integral of left hand side as only $Ri-U$?? Shouldn't you include the term $\int E\cdot dl$ for the path inside the coil wire?Do you let the E-field to be zero inside the wire of the coil? Or the path of integration does not include the spirals of the coil and it goes as straight line through them?

That's the very point! Of course, my path on the left-hand side goes through the wire of the coil, but (assuming $R_{\text{coil}}=0$, i.e., lumping its resistance into the overall resistance $R$) there's no "voltage drop" across it. The inductance enters the equation from the right-hand side, i.e., the flux of the magnetic field inside the coil (I neglected the inductance of the rest of the circuit as much smaller than that of the coil).

I insist that here you DO NOT measure a potential difference. To the contrary one has a closed loop here! It's an EMF, which is more general than electrostatic potentials.

Also it's true that there is of course the four-vector potential, but in my experience for physics discussions it is wise not to refer to it. In classical electrodynamics you can very often avoid it and discuss everything in terms of the observable (and thus physical) fields $\vec{E}$ and $\vec{B}$ (and in macroscopic electrodynamics also $\vec{D}$ and $\vec{H}$). Introducing the four-potential usually involves debates about their physical interpretation, which is pretty tricky, because they are gauge-dependent quantities.

 

[Dale]

I'm really puzzled about what you mean. Of course, the induced E field is not neglible.

It doesn't matter, it is not relevant to the thread.

The surface described is not a surface enclosing the whole circuit, it is a surface enclosing a single component, such as an inductor. The induced E field on such a surface certainly can be negligible, even for a strong inductor.

 

[vanhees71]

Stokes's Law is also very clear about the surface and the boundary involved, including the mutual orientation of both. In my example the path was along the circuit. Now you can choose any surface, for which this path is the boundary.

 

[Charles Link]

Essentially you using faraday's law in integral form here right? If so why do you evaluate the integral of left hand side as only $Ri-U$?? Shouldn't you include the term $\int E\cdot dl$ for the path inside the coil wire?Do you let the E-field to be zero inside the wire of the coil? Or the path of integration does not include the spirals of the coil and it goes as straight line through them?

Perhaps @vanhees71 needs to answer this, but I believe the integral that he is doing is the Faraday $E$ over the path of spiral path of the inductor. This is the EMF of the inductor and it acts as a voltage in the circuit. Also notice the sign on his integral: it gives a $V=\int E \cdot dl$ just as my claim in post #43. Editing... And in fact, I think he answered your question in post #57.

 

[vanhees71]

I do the integral along the entire "wire", including the spiral path of the inductor. As I said, I used the usual approximations or circuit theory (ideal coil, negligible self-inductance of all parts of the circuit except the coil as well as the magnetic field outside the coil), and again THERE IS NO $V$! It's $-\dot{\Phi}/c=-L \dot{i}$!

For more details, see p. 102 ff of

http://th.physik.uni-frankfurt.de/~hees/physics208/phys208-notes-III.pdf

For more basic E&M (including worked-out examples of elementary circuit theory)

http://th.physik.uni-frankfurt.de/~hees/physics208.html

 

[Charles Link]

I do the integral along the entire "wire", including the spiral path of the inductor. As I said, I used the usual approximations or circuit theory (ideal coil, negligible self-inductance of all parts of the circuit except the coil as well as the magnetic field outside the coil), and again THERE IS NO $V$! It's $-\dot{\Phi}/c=-L \dot{i}$!

I'm using the letter V to indicate the EMF. Perhaps I should use $\mathcal{E}$.