# I Why does a voltmeter measure a voltage across inductor?

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1. Jul 26, 2016

### 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?

2. Jul 26, 2016

### 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."

3. Jul 26, 2016

### OnAHyperbola

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

4. Jul 26, 2016

### Staff: Mentor

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.

5. Jul 26, 2016

### 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.

6. Jul 26, 2016

### David Lewis

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.

7. Jul 26, 2016

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.

8. Jul 26, 2016

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.

9. Jul 26, 2016

### Delta²

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.

10. Jul 26, 2016

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.

11. Jul 27, 2016

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.

12. Jul 27, 2016

### 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 Einduced 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

Last edited: Jul 27, 2016
13. Jul 27, 2016

### Staff: Mentor

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

14. Jul 27, 2016

### 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
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B} \neq 0.$$

15. Jul 27, 2016

### 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?

16. Jul 27, 2016

### 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 $V=L \frac{dI}{dt}$.

17. Jul 27, 2016

### 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.

18. Jul 27, 2016

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.

19. Jul 27, 2016

### 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.

20. Jul 27, 2016

### Staff: Mentor

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.