A New Interpretation of Dr. Walter Lewin's Paradox - Comments

[rude man]

I think we are very fortunate in this case to have a long solenoid, with current $I(t)=\alpha \, t$, that is able to generate exactly what we need in terms of a uniform B with a $\frac{dB}{dt}=$ constant, into the paper, so that we have a practical apparatus to make such a magnetic field. Otherwise it becomes a case where the EMF can be computed from Faraday's law, but not the electric field $E$. $\\$ I do think it is likely the solenoidal geometry proved very important for Faraday and others in coming up with the understanding of magnetism that we presently have. $\\$ It is not immediately obvious from Biot-Savart or Ampere's law, but detailed calculations do show that $B$ is completely uniform inside a long solenoid with $B=\mu_o \, n \, I \, \hat{z}$.

If the contour and B field are circular and concentric, you can compute the E field everywhere along the contour, whether the contour is inside or outside the B field (solenoid).

If the path is irregular and the B field is circular you can use your variable-radius technique to determine E everywhere along the contour.

Agreed?

 

[vanhees71]

@Charles, that is what I responding to. I think you're right, you can't assume symmetry of the E field around the off-axis solenoid; all you can assume is the circulation of E = -d(phi)/dt. And as I said it's like trying to apply Ampere's law to a finite wire: the circulation of H = I always, but that H is not uniform around any circular path. I remember getting faked out (temorarily only of course ha ha) just this way.

As far as vanHees' post is concerned, there is no E_s field anywhere around your path so the gradient term -∇Φ is zero so that does not seem to offer any further enlightenment. I can only think that you have to solve ∇xE = ∇xE_m at every point along your chosen path which is probably prohibitively difficult.

In your case, going with polar coordinates, ∇xE = [∂E_φ/∂r+ E_φ/r - (1/r) ∂E_r/∂φ] k = -∂B/∂t. With the third term on the LHS non-zero plus the boundary values that would be more than the feeble math knowledge of a dumb EE like myself could handle!

EDIT :even if your path doesn't include any B so that ∂B/∂t = 0 the problem is probably not much easier!

Why is $-\vec{\nabla} \phi$ necessarily 0? In my opinion the problem description is to incomplete to make a statement about the source part of $\vec{E}$ simply because the sources of the em. field are not specified.

 

[vanhees71]

@Charles, that is what I responding to. I think you're right, you can't assume symmetry of the E field around the off-axis solenoid; all you can assume is the circulation of E = -d(phi)/dt. And as I said it's like trying to apply Ampere's law to a finite wire: the circulation of H = I always, but that H is not uniform around any circular path. I remember getting faked out (temorarily only of course ha ha) just this way.

As far as vanHees' post is concerned, there is no E_s field anywhere around your path so the gradient term -∇Φ is zero so that does not seem to offer any further enlightenment. I can only think that you have to solve ∇xE = ∇xE_m at every point along your chosen path which is probably prohibitively difficult.

In your case, going with polar coordinates, ∇xE = [∂E_φ/∂r+ E_φ/r - (1/r) ∂E_r/∂φ] k = -∂B/∂t. With the third term on the LHS non-zero plus the boundary values that would be more than the feeble math knowledge of a dumb EE like myself could handle!

EDIT :even if your path doesn't include any B so that ∂B/∂t = 0 the problem is probably not much easier!

Why is $-\vec{\nabla} \phi$ necessarily 0? In my opinion the problem description is to incomplete to make a statement about the source part of $\vec{E}$ simply because the sources of the em. field are not specified.

Yes, thank you, I have completely solved this one. I do believe I have seen textbooks that present the problem of a uniform $\frac{dB}{dt}=\beta$ into the plane of the paper and ask you to compute $E$ around an arbitrary circle. $\\$ The long solenoid with current $I(t)=\alpha t$ does have uniform $\frac{dB}{dt}=\beta$, (in the $\hat{z}$ direction), throughout its interior. The surprising thing is it is incorrect to pick an arbitrary circle to compute $E$ and assume uniformity of $E$. The circle must be centered on the axis of the solenoid, or $E$ is not constant, (in the $\hat{a}_{\phi}$ direction), around the circle. The computation of the EMF $\mathcal{E}=\oint E \cdot dl$ works, but a uniform $\frac{dB}{dt}$, surprisingly, doesn't have sufficient symmetry to compute $E$ from the uniform $\frac{dB}{dt}$ inside the circle. $\\$ The $E$ gets computed everywhere by drawing circles of varying radii, that are all centered on the axis of the solenoid. For these circles, $\mathcal{E}=\oint E \cdot dl=2 \pi \, r \, E(r) =-\pi r^2 \frac{dB}{dt}=-\beta \pi r^2$.

Let's see again. I still don't get this obvious contraction of fundamental math, in this case Stokes's integral theorem.

Let's do it once more: As usual, the local treatment is the most simple approach. You have given the (approximate) magnetic field
$$\vec{B}=\beta t \vec{e}_3.$$
Then we have
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}=-\beta \vec{e}_3.$$
The solution of this is
$$\vec{E}=-\beta \vec{r} \times \vec{e}_3-\vec{\nabla} \Phi,$$
where $\Phi$ is undetermined since the sources are not given.

For the EMF this is of course unimportant since any closed-loop integral over a gradient in a simply connected region (which is the case for the interior of the solenoid you are discussing) gives zero. Thus we have
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=\int_{A} \mathrm{d}^2 \vec{f} \cdot (\vec{\nabla} \times \vec{E}) = -\int_{A} \mathrm{d}^2 \vec{f} \cdot \beta \vec{e}_3.$$
For any (!) circle of radius $R$ parallel to the $x_1 x_2$ plane completely in the interior of your solenoid gives thus $-\pi \beta R^2$. What's, in your opinion, wrong with this simple argument?

 

[Charles Link]

For any (!) circle of radius RRR parallel to the x1x2x1x2x_1 x_2 plane completely in the interior of your solenoid gives thus −πβR2−πβR2-\pi \beta R^2. What's, in your opinion, wrong with this simple argument?

I completely agree with it in the calculation of the EMF. What I was also expecting is that a computation of $E$ would necessarily also follow from symmetry=(i.e. if the $B$ field and $\frac{dB}{dt}$ is uniform over the region of interest). As you pointed out, if the sources aren't provided, there is no guarantee that the problem has the necessary symmetry to be able to calculate the induced electric field $E$. $\\$ (I found this kind of surprising, but the example of a solenoid with the uniform magnetic field inside of it shows that the sources (and any symmetry they have and/or don't have) needs to be taken into account to compute the $E$).

 

[vanhees71]

I see what you mean now. You want to calculate $\vec{E}$ in the hand-waving way how sometimes the electric field of a statically charged sphere or infinitely long cylinder is calculated by making use of the integral laws. Well, this is only possible for very very symmetric situations and a solid portion of physics intuition not to make mistakes with hand-waving arguments. In general, this is something for magician genisuses rather than us usual mortals, who better use math. Of course, for sufficiently symmetrical problems the math becomes usually very simple, as this example shows.

Another observation is that almost everything which is dubbed "a paradox" in physics is simply due to the fact that some people rather obscure the didactics by trying to avoid math, making the understanding of the problem at least difficult if not impossible and then leading to apparent contradictions. The prime example is the so-called twin paradox, usually presented in the first few lectures about special relativity, instead of simply stating that an ideal clock always shows its proper time. It's anyway wrong to present special relativity as a collection of apparent paradoxes rather than stressing that it resolves the paradoxes of Newtonian physics, but that's another topic.

 

[vanhees71]

BTW: This also applies to Walter Lewin's paradox either, and this is a kind of paradox which can be even easier avoided than the twin paradox, because it's due to sloppy language, i.e., because some people call an EMF a voltage, although a voltage is a potential difference. As Faraday's Law, $\vec{\nabla} \times \vec{E}=-\partial_t \vec{B}$ explicitly states (again math is your friend) this is precisely a situation where the fundamental concept is that $\vec{E}$ in this case has no potential and thus that line integrals between two points are path dependent, and that's also very nicely explained by Lewin in his famous lectures (I hope they are still available on Youtube although he stupidly ruined his reputation by stupid emails to students).

 

[Charles Link]

@vanhees71 Post 52 of this thread contains Professor Lewin's video: https://www.physicsforums.com/threa...asure-a-voltage-across-inductor.880100/page-3

 

[vanhees71]

That's not precisely the one I meant, but it seems to be good too (I'll watch it completely later this weekend). What I had in mind was a collection of an entire introductory experimental physics course with a lot of experiments done by Lewin in the lectures. Originally it was at MIT, but it's also still on Youtube, as your link shows. I think, this one is the movie I had in mind:

 

[Charles Link]

@vanhees71 The one you just posted, (which I didn't have time to watch yet), is apparently the one that is the subject of this discussion: https://www.physicsforums.com/threads/walter-lewin-paradox.948122/#post-6003552 $\\$ @mabilde also has a very interesting video in this thread.
In the video that is in post 52 of the "link" I provided, I did not spot any errors of any significance. I will need to watch the one you just posted.

 

[rude man]

Why is $-\vec{\nabla} \phi$ necessarily 0? In my opinion the problem description is to incomplete to make a statement about the source part of $\vec{E}$ simply because the sources of the em. field are not specified.

Referring back to post 22, I was referring to the circular E fields within the solenoid, which are in air, for which obviously there can be no static component, ergo no potential gradient.

If you're thinking of the axial E fields along the solenoid - there are two E fields, one is emf-generated and one is static. They cancel each other so there is no net axial E field in the solenoid.

 

[Charles Link]

If you're thinking of the axial E fields along the solenoid - there are two E fields, one is emf-generated and one is static. They cancel each other so there is no net axial E field in the solenoid.

@rude man This statement is unclear as written, but I believe you might be referring to the electric fields inside the conductor windings of the solenoid. In that case, I believe what you said is correct, but you weren't very clear on what you were referring to. Yes, for the current to be finite in an ideal conductor, the total electric field must be nearly zero. $\\$ Also, the fields point in the $\hat{a}_{\phi}$ direction. If by "axial" you mean around the axis, then, yes, I agree.

 

[rude man]

If you're thinking of the axial E fields along the solenoid - there are two E fields, one is emf-generated and one is static. They cancel each other so there is no net axial E field in the solenoid.

@rude man This statement is unclear as written, but I believe you might be referring to the electric fields inside the conductor windings of the solenoid. In that case, I believe what you said is correct, but you weren't very clear on what you were referring to. Yes, for the current to be finite in an ideal conductor, the total electric field must be nearly zero. $\$ Also, the fields point in the $hat{a}_{phi}$ direction. If by "axial" you mean around the axis, then, yes, I agree.

By "axial" I meant along the axis, no in the theta direction. Don't see what was unclear about my statement. So I assume you disagree that way?

 

[Charles Link]

By "axial" I meant along the axis, no in the theta direction. Don't see what was unclear about my statement. So I assume you disagree that way?

You are referring to the electric field "inside the conductor". The "inside the conductor" part was not clearly stated. (Outside the conductor, inside the solenoid, the electric fields do not cancel).

 

[rude man]

You are referring to the electric field "inside the conductor". The "inside the conductor" part was not clearly stated. (Outside the conductor, the electric fields do not cancel).

No I'm not. I'm referring to the net axial E field, comprising equal and opposite $E_m$ and $E_s$ fields. Shouldn't be much argument as to what "axial" means IMO.

The E field in the wire is zero if the wire has zero resistance and is entirely $E_m$ if the resistance is finite.
$E_m$ is in any case related to current i by $E_m$ = $Ri/2 \pi a$, R=resistance per turn, a = coil radius.

 

[Charles Link]

No I'm not. I'm referring to the net axial E field, comprising equal and opposite $E_m$ and $E_s$ fields. Shouldn't be much argument as to what "axial" means IMO.

The E field in the wire is zero if the wire has zero resistance and is entirely $E_m$ if the resistance is finite.

Yes=inside the wire=inside the (ideal) conductor. We are going in circles here, but you confirmed my question.

 

[arydberg]

I did this experiment and it shows what Dr. Lewin stated. I took a 500 foot roll of wire used for dog fencing. ( it measured 5 ohms). Unspooled it, Cut it in half and rewound 250 feet back on the spool. I brought out a pigtail and connected it to a 1meg resistor. The other end of the resistor went to the 2nd 250 foot wire. I then wound the 2nd piece on the spool in the same direction. and connected a 100 K resistor to the beginning and end of the coil.

I put 2 strong magnets in the center of the coil and snatched them away and got a pulse on my Oscilloscope. The pulse on the 1 meg resistor was 10 times that on the 100 K resistor. I have 2 pictures and hopefully included them.

 

[rude man]

I did this experiment and it shows what Dr. Lewin stated. I took a 500 foot roll of wire used for dog fencing. ( it measured 5 ohms). Unspooled it, Cut it in half and rewound 250 feet back on the spool. I brought out a pigtail and connected it to a 1meg resistor. The other end of the resistor went to the 2nd 250 foot wire. I then wound the 2nd piece on the spool in the same direction. and connected a 100 K resistor to the beginning and end of the coil.

I put 2 strong magnets in the center of the coil and snatched them away and got a pulse on my Oscilloscope. The pulse on the 1 meg resistor was 10 times that on the 100 K resistor. I have 2 pictures and hopefully included them.

Dr. Lewin's data was never in contention. That was not the issue. It's his explanations that were wrong, in particular the statement that "Kirchhoff was wrong". Kirchhoff's laws hold in all cases. They refer to voltage drops, which is not necessarily the same as measurements using voltmeters.

A voltmeters always correctly measures the voltages it "sees" but this voltage can be artificially induced by the voltmeter and its leads and is thus not the voltage in the absence of the voltmeter and its leads. For example, in Lewin's setup there is a voltage between any two points along a wire not including a resistor, yet the voltmeter reads zero.

The sum of voltages along any closed path is always zero irrespective of the nature of the emf generating them.

 

[anorlunda]

Kirchhoff's laws hold in all cases.

That's an overly broad statement. There are important assumptions. If those are violated, then Kirchoff's Laws can not be used. See the Insights Article.

https://www.physicsforums.com/insights/circuit-analysis-assumptions/

 

[rude man]

I think I disagree with your #1. The Lewin setup includes time rate of change of magnetic flux outside a conductor being non-zero if I'm interpreting your statement per your intention, yet there Kirchhoff's laws certainly hold.

I agree with the rest. Quasi-stationariness must be assumed, ortherwise lumped-circuit anaylis laws have to be superseded by Maxwell's equations. A radiating circuit is one example, as is a distributed circuit.

But that is not the discussion here.

Reference https://www.physicsforums.com/insights/circuit-analysis-assumptions/

 

[anorlunda]

The Lewin setup includes time rate of change of magnetic flux outside a conductor being non-zero if I'm interpreting your statement per your intention, yet there Kirchhoff's laws certainly hold.

At minutes 30-35 in the video, he says that Kirchoff's Laws only apply when the external magnetix flux is zero. And since the flux is nonzero in his experiment (assumption #1) you can't use Kirchoff's Laws or circuit analysis to describe that experiment. Well duh.

Bottom line, you can't say that KVL and KCL apply always.

By the way, be careful when you say you don't agree with those assumptions. They are repeated in many standard textbooks. Peer reviewed journals and standard textbooks are the bible here on PF.

 

[rude man]

At minutes 30-35 in the video, he says that Kirchoff's Laws only apply when the external magnetix flux is zero. And since the flux is nonzero in his experiment (assumption #1) you can't use Kirchoff's Laws or circuit analysis to describe that experiment. Well duh.

Bottom line, you can't say that KVL and KCL apply always.

By the way, be careful when you say you don't agree with those assumptions. They are repeated in many standard textbooks. Peer reviewed journals and standard textbooks are the bible here on PF.

Voltage is the line integral of the electrostatic field. And the circulation of that field is zero.
I suggest perusal of the two papers by Princeton's K. McDonald I cited in my Insight article on this subject.
And if I may counter with my own "standard textbook": Fundamentals of Electric Waves by Stanford's H H Skilling. Let me know if you need chapter & pages.

 

[Charles Link]

@anorlunda and @rude man I think you both may be arguing the very same thing, and it is open to debate whether the EMF generated in a loop by a changing magnetic field is part of Kirchhoff's voltage laws (KVL), or if it happens to be an exception that Professor Walter Lewin has highlighted. Others have previously argued this fine detail: See https://www.physicsforums.com/threa...ge-across-inductor.880100/page-5#post-5533643 . $\\$ Right around post 83 @Dale and @vanhees71 went back and forth on this a couple of times, but I think everyone is in agreement on how this problem gets solved, and it is very useful that Professor Walter Lewin has pointed out this special case, even if he says a couple of things that perhaps also aren't 100% accurate.

 

[anorlunda]

I remember that thread. It got really heated among several people who really know their stuff. You may be right that it's semantics.

I really don't care enough about Professor Lewin to go down that rabbit hole, so I'm going to exit this conversation.

 

[rude man]

Pace nobiscum.

 

[vanhees71]

Well, there's nothing to fight about. I think there's no paradox at all (I don't like the word "paradox"; it just indicates a lack of careful analysis based on "common knowledge", which is contrary to the very basic principles of basic science). Just use Maxwell's equations, and everything is fine. Also avoid to talk about "voltage" as soon as emf's from time-varying magnetic fields are involved. BTW Contrary to the statement in the above cited Insight article, of course there are both electric and magnetic fields in stationary circuits (in fact there's only one electromagnetic field in nature, but that's another story). The only thing is that one can eliminate them from the considerations using the stated assumptions and lump everything in currents, voltages and emf's. That's because Kirchhoff's laws are nothing else than the integrated version of Maxwell's equations under the simplifying assumptions made.

 

[Charles Link]

@vanhees71 makes a very good point, you can no longer use the term "voltage" when you include a circuit loop that has a changing magnetic field inside of it. $\\$ The "voltmeter" does not measure a "voltage" difference in this case, between the two points on the circuit that it probes.$\\$ Instead, the voltmeter needs to be considered for what it actually is=a couple of wires with a large resistor through which a small current flows. In this case, the voltmeter really doesn't "measure". Instead, it gives a reading which is the (multiplicative) product of the small current times the large resistor. The placement of the wires that form the leads of the voltmeter can yield different results depending on whether the circuit loop that they form encloses the changing magnetic field, in which case there is an EMF around that circuit loop.

 

[vanhees71]

Another equivalent view is that the EMF drives the electrons in the wires making up the volt meter (think of an old-fashioned galvanometer for simplicity), leading to the current @Charles Link mentioned in the previous posting.

This becomes clear if one uses the complete (!) integral form of Faraday's Law of induction. Its fundamental form is, as anything in electromagnetism, the local form in terms of derivatives (SI units):
$$-\partial_t \vec{B}=\mathrm{\nabla} \times \vec{E}.$$
Now if you integrate this over an arbitrarily moving area $A$ with boundary $\partial A$ you can first use Stokes's theorem. The only correct version of this simple treatment is
$$-\int_{A} \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B} = \int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}.$$
Now one likes to express in terms of the magnetic flux through the area
$$\Phi_{\vec{B}}=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
Now, if the area and its boundary are moving, you cannot take the partial time derivative out of the integral in the previous equation but you get an additional line integral along the boundary curve of the surface, which you can lump to the integral on the right-hand side. Taking Gauß's Law for the magnetic field, $\vec{\nabla} \cdot \vec{B}=0$ into account the resulting equation gets
$$-\dot \Phi_{\vec{B}} = \int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B})=:\text{EMF},$$
where $\vec{v}$ is the velocity field along the moving boundary of the area we've integrated over.

Now if you choose the area such that its boundary is along the wires connecting the volt meter with the rest of the circuit, what it measures is in fact the electromotive force, i.e., the line integral along the closed (!) boundary. It's obviously the line integral over the force on a unit charge $\vec{E}+\vec{v} \times \vec{B}$, and this shows that indeed that's the physical picture on what's measured given above: The force on the charges inside the wires connecting the volt meter with the rest of the circuit (including the wires making up the coil in the volt meter, if you take the model of a old-fashioned galvanometer setup).

This considerations also explain why the reading of the volt meter is beyond the simple Kirchhoff circuit theory: It's reading cannot be understood without taking into account the correct geometry of the connection of the volt meter with the rest of the circuit since this you need to calculate the line integral defining the EMF, which is what the volt meter measures. The Kirchhoff theory becomes applicable only if you make the wires connecting the volt meter very short, so that the magnetic flux through the corresponding current loop becomes negligibly small. Then the reading is what you expect according to the Kirchhoff circuit theory, i.e., the EMF through the element of the circuit you want to measure (which may be an Ohmic resistor, a capacitor, or coil).

Note: Another source of confusion is the very name "EMF" for the line integral: Here force is obviously not the modern notion of "force" (which is represented by the Lorentz force per unit charge, $\vec{E}+\vec{v} \times \vec{B}$) but its meaning is more in the sense of "energy". Indeed the EMF is a line integral of the force along a closed loop. The very fact that the quantity is a line integral along a closed loop shows that it is NOT a "voltage". If there'd be a potential for the force integrated over, the integral over any closed loop is 0 (modulo the caveat that the region under consideration is simply connected!).

 

[Henryk]

There is no paradox whatsoever!. I admit, it took me a while to understand what is going on.
First thing first, the loop with two resistor is a red hearing. So, let's remove it and we get a circuit like that:

Now, we have a loop containing two voltmeters encircling flux change of 1 Wb/s. Obviously, the induced EMF is 1 Volt and the direction is indicated by the circular arrow. With the way the voltmeters are connected, the one on the right would show a positive voltage, the other negative voltage, just like in the video.
How much will each of them show?. That would depend on the internal resistance of the voltmeters. Portable meters have resistance of 10 Mohm, if both have this value, one will show 0.5 V, the other -0.5 V. Change the internal resistance of the left voltmeter to 1 Mohm and the other one to 9 Mohm and you will get -0.1 v and 0.9 V. No paradox, just a red herring.
However, Dr Lewin makes a statement in his video that I would disagree. He says that the Kirchhoff (second) law is not valid. The way I was thought physics, it is still valid. I understand that the Kirchhoff law says that for a loop $\sum I_k R_k = \sum EMF$ and that actually agrees with the Faraday law.
Now, I would also like to point out that the supposed tutorial contains some false statements. One of the false statement is

"which is non-conservative in the sense that its circulation is non-zero. E_m can be created by a chemical battery, magnetic induction, the Seebeck effect, and others."

This statement is not correct. The non-conservative electric field can only be created by changing magnetic flux. The field inside a battery is conservative. How is it created.

The key to understand operation of a battery is thermodynamics and equilibrium condition for particle exchange. Thermodynamics tells us that a system is at equilibrium with a reservoir with respect to particle exchange if the chemical potentials are equal. Let's take, for example, an alkaline battery. It consist of a zinc cathode, MnO anode and KOH electrolyte. KOH in solution dissociates into K⁺ and OH^- ions. At the cathode, the following reaction takes place ( see https://en.wikipedia.org/wiki/Alkaline_battery )

Zn(s) + 2OH⁻_(aq) → ZnO(s) + H₂O + 2e⁻

The reaction of solid Zn with OH^- ions produces ZnO, water and free electrons. Where do the free electrons go? they go to the Zn metal charging it up negatively, i.e. increasing the chemical potential of electrons in the metal. The reaction stops when the chemical potential of electrons in the Zn metal become equal to the chemical potential of the electrons attached to OH^- ions. The net result is formation of a potential difference at the electrode/electrolyte interface. This is not unlike creation of the depletion layer in a p-n junction of a semiconductor.
Similarly, there is a potential step created at the anode. The total voltage of an (open circuit) battery is algebraic sum of the two voltage steps.
Seebeck effect, photovoltaic cell EMF can also be understood considering the thermodynamics, that is, EMF is created by a gradient of chemical potential of electrons and the field is conservative.

 

[rude man]

@vanhees71 makes a very good point, you can no longer use the term "voltage" when you include a circuit loop that has a changing magnetic field inside of it. $\\$ The "voltmeter" does not measure a "voltage" difference in this case, between the two points on the circuit that it probes.$\\$ Instead, the voltmeter needs to be considered for what it actually is=a couple of wires with a large resistor through which a small current flows. In this case, the voltmeter really doesn't "measure". Instead, it gives a reading which is the (multiplicative) product of the small current times the large resistor. The placement of the wires that form the leads of the voltmeter can yield different results depending on whether the circuit loop that they form encloses the changing magnetic field, in which case there is an EMF around that circuit loop.

Yes. Your resistive voltmeter shows the field equivalent of Ohm's law which is ir = d(E_s + E_m) with d the length of r. This reduces to ir = dE_s if d << voltmeter wire lead lengths. In my various posts I had made this assumption.

 

[rude man]

BTW Contrary to the statement in the above cited Insight article, of course there are both electric and magnetic fields in stationary circuits (in fact there's only one electromagnetic field in nature, but that's another story)..

E_m is also an electric, not a magnetic field. Two E fields: E_s and E_m. One begins and ends on charges; the other does not.