Why does a voltmeter measure a voltage across inductor?

[OnAHyperbola]

The potential difference across an inductor is supposed to be zero, but a voltmeter measures it to be L*dI/dt.
Also, if the p.d is zero then the electric field in the wires of the coil will be zero and in that case, why should charges flow at all?
What am I missing?

 

[Svein]

In real life any inductor has an internal resistance. It may be low, but it is there. And a current through a resistance creates a voltage drop (or potential difference).
From https://en.wikipedia.org/wiki/Inductor:

"In circuit theory, inductors are idealized as obeying the mathematical relation (2) above precisely. An "ideal inductor" has inductance, but no resistance or capacitance, and does not dissipate or radiate energy. However real inductors have side effects which cause their behavior to depart from this simple model. They have resistance (due to the resistance of the wire and energy losses in core material), and parasitic capacitance (due to the electric field between the turns of wire which are at slightly different potentials). At high frequencies the capacitance begins to affect the inductor's behavior; at some frequency, real inductors behave as resonant circuits, becoming self-resonant. Above the resonant frequency the capacitive reactance becomes the dominant part of the impedance. At higher frequencies, resistive losses in the windings increase due to skin effect and proximity effect."

 

[OnAHyperbola]

@Svein Thanks for your answer. I'm a little more interested in the ideal case, though (where the wires are superconducting)

 

[Dale]

The potential difference across an inductor is supposed to be zero, but a voltmeter measures it to be L*dI/dt.

I am not sure why you think it is supposed to be 0. It is supposed to be L dI/dt. That is, in fact, the defining characteristic of an ideal inductor.

 

[Delta2]

The voltmeter reads the voltage due to the charge densities, hence due to the scalar potential, it cannot read the voltage due to the vector potential. You are right that in an ideal inductor with no ohmic resistance and no capacitance the electric field inside the wires is zero, but because it has no ohmic resistance and no capacitance charges simply do not meet any resistance/impedance and keep flowing.

 

[David Lewis]

Also, if the p.d is zero then the electric field in the wires of the coil will be zero and in that case, why should charges flow at all?

Current flowing through the coil is out of phase with the applied voltage. The coil generates its own voltage when the current changes. If you disconnect the coil from the power source, for example, self induction will try to keep the current going.

 

[Charles Link]

If the inductor is made from a superconductor or is very low resistance, the electric field should be nearly zero inside the inductor or you will get very large currents. Faraday EMF's are the result of an integral of an electric field over the conducting path of an inductor and since the total electric field is nearly zero in the inductor, there must be an (instantaneous) electrostatic E created along the inductor to offset the Faraday E field terms. It doesn't take much electrostatic charge (negligible change in current) to generate an electrostatic voltage. If you attach wires (such as a voltmeter) to the outside of the inductor, they will experience the electrostatic potential of the inductor (from the excess electrostatic charges), but I think they will not be able to sense the Faraday E inside the inductor, since there is ideally no magnetic field outside the inductor. I think @Delta² has it correct in post #5. The voltmeter will simply see this electrostatic potential which is essentially the opposite of the Faraday EMF.

 

[Charles Link]

Just one additional comment though, also for @Delta² . I do think I have seen cases where the voltmeter can measure the E from the Faraday term. If I'm not mistaken, I remember one such case where there was a circuit with a couple of wires to a voltmeter with a localized changing magnetic field into the plane of the diagram. Depending on how the wires were routed around the changing magnetic field, you got two different answers for the voltage that the voltmeter read from two same points of the circuit. Basically the Faraday E would help generate a current through the resistor of the voltmeter.

 

[Delta2]

Just one additional comment though, also for @Delta² . I do think I have seen cases where the voltmeter can measure the E from the Faraday term. If I'm not mistaken, I remember one such case where there was a circuit with a couple of wires to a voltmeter with a localized changing magnetic field into the plane of the diagram. Depending on how the wires were routed around the changing magnetic field, you got two different answers for the voltage that the voltmeter read from two same points of the circuit. Basically the Faraday E would help generate a current through the resistor of the voltmeter.

Ahh well yes seems correct though I haven't experimentally confirmed it, if the voltagemeter and/or the connecting wires are inside a time varying magnetic field then it would be affected by the vector potential.

 

[Charles Link]

Ahh well yes seems correct though I haven't experimentally confirmed it, if the voltagemeter and/or the connecting wires are inside a time varying magnetic field then it would be affected by the vector potential.

In the textbook problem that I saw, (it was a puzzle at first how it worked), the wires simply were routed around a localized changing magnetic field. Depending on whether the loop that included the voltmeter surrounded this region, you would pick up a ## V=-d \Phi/dt ## term along with the voltage of the circuit that was being measured.

 

[Charles Link]

Back to the OP's original question which is really quite interesting=whether it is an inductor that is generating the EMF or a battery that has a chemical interaction as its source=there is an electric field/electromotive force internal to these sources that points from minus to plus. This is not measured by a voltmeter. Instead an electrostatic potential is reached at the positive terminal (relative to the minus terminal) that balances this internal EMF so that equilibrium is reached and no current flows (in the open circuit case). The battery maintains its voltage, instead of spontaneously discharging, because of the internal EMF. A voltmeter connected to the battery terminals measures the electrostatic potential of the battery. If the voltmeter were able to sense the internal EMF that is going on and not see the offsetting electrostatic potential, the voltmeter would read higher at the minus terminal of the battery than at the plus. Instead, the opposite occurs: the voltmeter sees the electrostatic potential, but it doesn't pick up the internal EMF. As a result it measures a voltage drop across the inductor, or alternatively, it sees it as if it was a voltage source with the plus end away from the internal EMF (just like a battery=the plus end is away from the internal EMF, i.e. on the arrowhead side of the internal E). I do think I have this correct, and I think it is in agreement with what @Delta² and others have also stated.

 

[OnAHyperbola]

Okay, so I got thinking after reading all your posts and I thought, let me start at the beginning. So here's the story:
We have an RLC circuit, the wires of which are along an Amperian loop. Now, the changing B-field through the coil produces a non-conservative E-field, directed tangentially to the loop at all points (basically the shape of the circuit--a closed loop E-field). Now, the charges rearrange themselves to produce instantaneous conservative E-field to cancel the induced one.
Now, I'm a positive charge--say a proton-- sitting on the battery. I start my journey through the circuit. First, the battery gives me some potential. During my journey, I encounter an induced electric field that is always in the direction I'm going. (side note: I think E_induced should be (L*dI/dt)/length of the entire loop, irrespective of its geometry). As I move along, this E-field keeps taking away more and more of my potential. There is another, conservative E-field in my way, but its effects cancel entirely by the end of my journey. I encounter a resistance, that reduces my potential a bit, and a capacitor which does the same. By the time I have reached back to where I started, my potential is L*dI/dt less than what it was when I started.
Clearly $$\oint_C E \,dl$$ =L*dI/dt

Three observations:

1. What is so special about the ends of the inductor? It seems to me that wherever you place the ends of a voltmeter (one that measure the scalar potential, as @Charles Link and @Delta² agree), it should measure the same p.d.--L*dI/dt, provided there isn't a battery, capacitor or resistor in between. What kind of conservative field would produce a gradient with same same p.d no matter which two points you choose to measure it between?

2. Let's say the length of the Amperian loop is s Since the induced field, (refer to side note above) is (L*dI/dt)/s at every point. The difference in potential at the ends of the inductor should be $$\int_{}^{} E. dl$$ which comes out to be s'(L*dI/dt)/s, where s' is the length of the inductor's wire.

3. This elusive conservative E-field, must also be in a loop, except for a few chinks (where the capacitor, battery etc are). What the heck kind of charge distribution would produce this crazy field?

So, what say you?

Dazed and confused
OnAHyperbola

 

[Dale]

The voltmeter reads the voltage due to the charge densities, hence due to the scalar potential,

This isn't relevant. The potentials on the terminals of an inductor are standard scalar potentials. All of the vector potentials are internal to the inductor. This is one of the fundamental assumptions of circuit theory.

A voltmeter can easily measure the voltage across an inductor.

http://web.mit.edu/6.013_book/www/book.html

See section 11.3 and especially the second to last sub section

 

[vanhees71]

As soon as induction is involved there's no potential for the electric field anymore, not even locally since according to Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B} \neq 0.$$

 

[OnAHyperbola]

@vanhees71 What you are saying makes sense. Curl of gradient of potential is zero. Ergo, when it's not zero, the potential function ceases to work. It seems to me that this is a potential function that depends on how much length of the loop the charge has traversed, like with friction. I guess what I'm trying to figure out now is: what's the gradient of the conservative field that cancels the induced one?

 

[Paul Colby]

The DC or time invariant component of the voltage is in fact zero across and ideal inductor (##\frac{dI}{dt}=0##) so an ideal digital voltmeter will measure this on it's DC setting. The AC or time varying components of the potential are not zero and an ideal digital volt meter will measure a non-zero value (provided there is a driving current source). All this is evident from the defining equation ##V=L \frac{dI}{dt}##.

 

[vanhees71]

It's easier than that. To understand it, it's best to think about a good oldfashioned voltmeter like a galvanometer, where what's measured is in fact a current running through a coil. To measure a voltage you put it in series with an appropriate resistance, and then you use Ohm's Law to convert the current measured by the galvanometer into a "voltage drop", where here you measure the electromotive force rather than a potential difference.

 

[Charles Link]

Most of the time, I just plug into the differential equation with the term ## L dI /dt ## and give it little extra thought. I think the OP's question is a very good one, and when the inductor is examined in detail, it can be a little puzzling. Consider suddenly applying a DC voltage (e.g. electrostatic) ## V =V_o ## to an inductor. This will be accompanied by an electric field that wants to send a very high current through the conducting wires of the inductor. Without the reverse Faraday electric field, extremely high currents would flow immediately. Instead, one way to look at it is the voltage driving the whole circuit is reduced by an amount ## V=L dI/dt ## . The reverse EMF points in the same direction as the Faraday E field (from minus to plus), but what the voltmeter measures is the applied electrostatic ## V_o ##. The reverse EMF with its Faraday E field explains why the current in the conductive inductor isn't simply a very large value immediately, but the voltmeter does not see the Faraday E. You can get large DC currents in an inductor with no voltage drop and virtually zero electric field in the conducting wires of the inductor. The Faraday E is such that it will be opposite the applied electrostatic E from the voltage ## V_o ## and will offset it to essentially keep the total electric field equal to zero (or nearly zero) in the conducting wires of the inductor (assuming perfect conductor) at all times.

 

[vanhees71]

The equations of circuit theory follow from Maxwell's equations. Let's take your example with an ideal inductance and a resistor in series and close a switch at ##t=0## connecting it with a battery. Then integrating Faraday's Law along the circuit you get (in quasistationary approximation)
$$L \dot{I}+R I=U_0.$$
Nowhere in this derivation you have to apply a wrong concept as interpreting the first term on the left-hand side as some "potential difference", but it originates from the application of Stokes's integral theorem using the definition of the self-inductance ##L##. The physics is that the magnetic field due to the current is changing such as to hinder this change (Lenz's rule), and this is all included in Maxwell's equations.

The general solution of the differential equation is the general solution of the inhomogeneous equation
$$L \dot{I}+R I=0 \; \Rightarrow \; I_{\text{hom}}=A \exp(-R t/L)$$
added to a special solution of the inhomogeneous one, which is obviously given by the stationary limit ##I_{\text{inh}}=U_0/R##. With the initial condition ##I(t=0)=0## you finally get
$$I(t)=\frac{U_0}{R} \left [1-\exp \left (-\frac{R}{L} t \right) \right],$$
i.e., due to the finite self-inductance of the circuit indeed the current is not suddenly switched on but builds up graduately with a typical "relaxation time" ##\tau=L/R## in accordance with Lenz's Law.

 

[Dale]

As soon as induction is involved there's no potential for the electric field anymore, not even locally since according to Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B} \neq 0.$$

That is a bit overstating it. As long as there exists a surface around the component such that the field contribution from induction is negligible on that surface, then it does not matter if there is a local region within that surface where the induced field is non negligible.

 

[Paul Colby]

Working through Maxwell's equations for an inductor is highly recommended and instructive, but, the OPs question seemed like a simpler nemonic might be helpful. All the equations involved are linear ones. This allows solutions to be viewed as sums of harmonic contributions. The DC component even in the case of a "DC" current being applied at t=0 may be so separated into a true DC (and eternal) component plus time varying ones. Clearly, this is true for the more fundamental Maxwell equations as well.

 

[Charles Link]

It seems what goes hand-in-hand with any electromotive force, (e.g. an E field from the Faraday EMF or the chemical potential of a battery) is an electrostatic potential that will offset the EMF. e.g. the EMF generated inside an inductor consists of an E field in the material of the conductor. For a good conductor, this E field by itself would cause enormous currents to flow in the conducting material. This does not happen. Instead, an electrostatic field gets immediately created from a very small amount of electric charge. Perhaps it is an extra complicating factor to introduce this electrostatic E, but it does explain how the current in the conductor(of the inductor) stays very finite (the total E must necessarily be very nearly equal to zero in the conductor). Especially when first learning the concepts, perhaps a very good way to look at it is to simply follow @vanhees71 explanation of post #17 and say that the voltmeter can measure a potential difference and it can also measure an electromotive force.

 

[vanhees71]

No, it's pretty important. There's a nice demonstration in Lewin's physics lectures somewhere on Youtube, where it is demonstrated that there's no potential in this case. It makes a difference for the how you shape the circuit in such cases. Of course, if you have a fixed coil, the geometry lumped into ##L##, and you can neglect the inductance of the rest of the circuit,so that the geometry of the wiring is neglible for the measured EMF. Still we are in a physics discussion and one should keep the distinction between an EMF and an electrostatic potential very clear!

 

[Dale]

Still we are in a physics discussion and one should keep the distinction between an EMF and an electrostatic potential very clear!

Circuit theory is physics too, and the distinction does not exist. As long as the lumped element assumptions hold there is no distinction between an EMF and an electrostatic potential.

I think there is a tendency to over complicate things. Just because a more complicated theory exists doesn't mean that we have to use it when analyzing a simple system.

 

[Charles Link]

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.

 

[vanhees71]

Circuit theory is physics too, and the distinction does not exist. As long as the lumped element assumptions hold there is no distinction between an EMF and an electrostatic potential.

I think there is a tendency to over complicate things. Just because a more complicated theory exists doesn't mean that we have to use it when analyzing a simple system.

We agree to disagree. To derive circuit theory from the underlying fundamental theory, i.e., Maxwell's Equations, you'll see that there is no electrostatic potential anywhere, and it must not be anywhere, because in this case it simply doesn't exist. That's a mathematical fact.

The importance is nicely demonstrated by a lecture by Lewin on Youtube, where he demonstrates the path dependence of the EMF nicely with an experiment. It was discussed some time ago in this forum, but I'm unable to find it. Perhaps some other reader remembers this very illuminating lecture and discussion on physicsforums!

 

[Charles Link]

We agree to disagree. To derive circuit theory from the underlying fundamental theory, i.e., Maxwell's Equations, you'll see that there is no electrostatic potential anywhere, and it must not be anywhere, because in this case it simply doesn't exist. That's a mathematical fact.

The importance is nicely demonstrated by a lecture by Lewin on Youtube, where he demonstrates the path dependence of the EMF nicely with an experiment. It was discussed some time ago in this forum, but I'm unable to find it. Perhaps some other reader remembers this very illuminating lecture and discussion on physicsforums!

Please also read my post #25. The circuit path makes a difference in computing the EMF in the circuit, but I think this is the result of where the wires are strung. An example of this is the textbook problem of a changing magnetic field around which is a continuous resistor ring. Depending on how you attach the leads to two points on the resistor, you can read different potentials.

 

[vanhees71]

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.

 

[Dale]

To derive circuit theory from the underlying fundamental theory, i.e., Maxwell's Equations, you'll see that there is no electrostatic potential anywhere, and it must not be anywhere, because in this case it simply doesn't exist. That's a mathematical fact.

http://web.mit.edu/6.013_book/www/book.html

See section 11.3 and especially the second to last sub section for a slightly different take on the mathematical facts.

 

[vanhees71]

What has Sect. 11.3 to do with our debate and where is it different from the usual math? There is Poincare's Lemma, which cannot be changed to something else! Which equations are you talking about?

 

[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.