Walter Lewin Demo/Paradox: Electromagnetic Induction Lecture 16

[tedward]

Thanks for these! I've read Romer's paper on this, it's great - everyone interested in this problem should read it.

Loved the analysis in the second link - he sums up all the points of confusion we've been discussing here nicely, and we come to the same conclusions.

I also agree with him (and actually DISagree with Lewin) on one point: If you define a path outside an inductor between the terminals and use that in your circuit analysis loop, then Kirchoff's law applies. This is Feynman's approach, and it makes sense to me. Lewin is a purist and would insist on using the actual circuit path - through the coils - in which case you of course need to use Faraday's law. If the books get it wrong, it's only to the extent that they do not discuss the path dependence, or identify explicitly what path they are using, leading to confusion about the true 'voltage drop' across the inductor.

He also shares my discomfort with only using the $E_{charge}$ to define voltage, i.e. the scalar potential.

I didn't watch the video in the 3rd link, but read in the comments he gets into quantum mechanics, which is well out of the scope of what's required here.

 

[tedward]

I don't understand why you mentioned resistor with negligible length so many times, do you think the calculation would be difficult or would be wrong if the resistors had non-negligible lengths? What is the hidden obvious difference, can you explain in detail?

Curious, would like to know the details, look forward to.

Here are the examples I was thinking about. We have flexible resistor wire (12 ohms) in series with conducting wire with zero resistance. In all the cases, total emf, total resistance, and current are the same. The path voltage $V$ is what the voltmeter measures with a flux-free loop (and of course corresponds to the true heat dissipation using $P=IV$. Measured across the resistor path, it reads 12 V in each case. But since the induced voltage $V_i$ depends on the orientation of the resistor relative to induced electric field (i.e. to the solenoid), the scalar potential $V_s$ changes accordingly, as $V_s = V - V_i$

The point that I'm arguing is not that you can't use the scalar potential, you absolutely can if it suits your purposes. But it's a specific convention that you learn in order to define a potential, it doesn't correspond to the true (net) electric field. I don't think most people who are arguing about what the voltage adds up to are really using this convention. I think they just see Mabilde's analysis and say 'well that makes sense to me.' I think if you showed them examples like this most people would take the path voltage as the convention that actually makes sense to them, and that's the one that Lewin uses which is path dependent.

If path dependency is a problem for people, they would likely have convulsions to learn that the 'voltage' across a resistor now depends on how you position it.

 

[hutchphd]

To (hopefully) cap this latest outbreak of recurring Lewin mania, I propose a HAIKU

Lewin's lecture strong
Folks refuse to understand
Sillyness abounds
.
I am unable to think of any other approach. No offense intended..

 

[Averagesupernova]

What you call no flux path voltage is misleading at best. There are field lines varying on the outside of loop also. You seem to be claiming voltmeter leads are unable to be affected on the outside.

You never replied to this. I'm awaiting a response.

 

[tedward]

Doesn't mean you're entitled to one. Re-read the Test Lead post where I address this.

 

[Averagesupernova]

Post # please....

 

[tedward]

#101

 

[Averagesupernova]

Ok so I looked at post 101 at the part you talk about the test leads closely following the DUT. Btw, they aren't drawn that way in your drawings. They are experiencing the same thing the wires in the DUT are. Not really much of an ah-ha moment here. That does NOT mean the voltage where the probes actually land is what the meter reads. Drawing 1 in post 177 actually has 6 volts between the probes. For real volts. Is this what path dependency actually means? Proving a point about voltmeter leads?
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Yes I know I said a few post back that the resistor wire dissipates 12 watts and I still stand by that. You may wonder how that can be since I claim there is only six volts measured across it. I've covered this over and over as well. The resistor wire is in the field also and is part of the loop that is generating the 12 volts. In the exact same space it also dissipates it. This is modeled as a voltage source in series with a resistance. The RSD academy video I linked to earlier in this thread covers that quite well.
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I find this quite ridiculous. I can't decide which is more absurd. If the claim is the voltage at the probe points really is indeterminate or if it boils down to an argument about which way the test leads should run and then knowing what to add or subtract (or not) to the reading.

 

[hutchphd]

I find this quite ridiculous. I can't decide which is more absurd. If the claim is the voltage at the probe points really is indeterminate or if it boils down to an argument about which way the test leads should run and then knowing what to add or subtract (or not) to the reading.

The point tto point voltage is indeterminate and you do need to know where your test leads run.

"If you ask a stupid question, you will get a stupid answer"

 

[Averagesupernova]

And if you are unable to recognize sarcasm you will GIVE a stupid answer.
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In all seriousness, if you recognize the voltage is indeterminate it may be possible to find a way to route the test leads that gives the real answer. Some people have.

 

[hutchphd]

I have lost the trace of rational thought on this thread. I don't even know what the question is......

 

[Charles Link]

I thought I might join this discussion. One related thread is this Insights article by @rude man https://www.physicsforums.com/threads/how-to-recognize-split-electric-fields-comments.984872/
that received very mixed reviews.
There is one criticism in there of how can it be possible to use an $E_s$ in the case of an inductor and have $E_s=-E_m$ in the conductor when $\nabla \times E_s=0$, while $\nabla \times E_m=-\dot{B}$ is non-zero. I just figured out the other day how this is possible : For one loop around the inductor ring we have $\oint \vec{E}_m \cdot ds=-\dot{\Phi}$, while for $E_s$ we are in the adjacent wire in the coil when we go once around, so that $\oint \vec{E_s} \cdot ds=\dot{\Phi}$, rather than the zero that Stokes theorem would insist upon, with $\nabla \times E_s=0$. Meanwhile, with the $E_s$ we get to amplify the $E_m$, so that if we have a coil of $N=1000$, we get a reading of $\int E_s \cdot ds$ that is one thousand times the reading we would get from $\oint E_m \cdot ds$.

This splitting of $E_{total}$ into $E_s$ and $E_m$ has one other difficulty that I recognize: If you have a changing $B$ outside the loop of the circuit, we do have $\nabla \times E_m=0$ for the area inside the circuit, so there is zero EMF. However, if we compute $E_m$ from symmetry by drawing a ring around the $B$ with the ring passing through the circuit, we get a non-zero $E_m$, and thereby must necessarily introduce an opposing $E_s$ or we could have an $E_m$ across a resistor in that circuit, and Ohm's law would be violated. This makes it far easier to work with EMF's in this case, and not introduce the $E_m$ and $E_s$.

My overall opinion is that the $E_s$ and $E_m$ does have a good amount of merit at times, while at other times we are better off just working with EMF's. One other previous related thread on Physics Forums is the following, where in post 192 @cnh1995 has a solution whose methodology I detailed in post 193. His solution uses an electrostatic potential (he called it $V_{ab}$ but it isn't a voltage so it might be better designated as $U_{ab}$), and he gets an answer much easier than I got by solving the Kirchhoff loops. (See post 152). Here is the link: https://www.physicsforums.com/threads/faradays-law-circular-loop-with-a-triangle.926206/page-6

 

[Charles Link]

One other case of interest is if you measure a fractional turn (e.g. one half turn) of the solenoid of many turns. If you use Faraday's law, you don't get any EMF in your (external) voltmeter loop unless you sort of artificially assign one half unit of changing magnetic flux to the voltmeter loop. See also post 104 by @alan123hk that got me thinking about this.

Meanwhile, using $E_s$ and $E_m$, you have the (non-zero) reading of the voltmeter incorporated into $\int E_s \cdot ds$, so that here your $E_s$ and $E_m$, with $E_m=0$ (assume voltmeter lead wires are positioned to have zero $E_m$), actually gets you a correct result without needing to recognize this as a special case, whereas the Faraday loop method needs to be treated as kind of a special case.

When you have one complete turn or any number of complete turns in the voltmeter loop, you immediately recognize that the loop with the meter contains that amount of changing flux, but with the fractional number of turns, it isn't so obvious.

Note: $E_s=E_{electrostatic}$ and $E_m=E_{induced}$.

Edit: I'm currently reviewing this=I think this post with the fractional ring may be in error, and that the Professor Lewin methodology gets it correct. What I was thinking on this is that you could measure just the electrostatic $\int E_s \cdot ds$ if you could attach the leads of the voltmeter so that they have zero $E_m$ throughout their path. I'm starting to conclude that is a topological problem, and this part needs to be retracted.

 

[Charles Link]

One additional comment: In post 166 @tedward refers to a voltage sign convention problem. I have to wonder if he has stumbled upon something that puzzled me a couple years ago=that the electric field $E_m= E_{induced}$ in an inductor (e.g. when there is a voltage drop across the inductor) points opposite the direction of $E_s$ in a capacitor or resistor. It was at this point that I saw the need to introduce $E_s=E_{electrostatic}$ into the calculations to explain the physics of the inductor and what is measured with the voltmeter. (e.g. Place the inductor in parallel with a resistor and have a sinusoidal voltage source driving both of them, and compute the $E's$. When the top part of the parallel combination is at a higher voltage, the $E_s$ in the resistor points downward, but the $E_m$ in the inductor points upward). See https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/ post 7 and post 43, and we even discussed Professor Lewin's paradox in some detail throughout this thread.

 

[tedward]

Btw, they aren't drawn that way in your drawings.

My drawing doesn't match up to what I described in 101, true. It actually doesn't matter, but you would measure the same reading if the leads always ran parallel to the resistor wire (see below).

The resistor wire has an induced voltage of 6V (shown in blue) and a scalar potential (due to the built up charges as shown) of 6V. These effects combine to give 12V (the path voltage corresponding to the NET electric field in the resistor). The voltmeter, being part of the circuit, is subject to BOTH of these effects as well - the induced voltage (from the curved section) and and the electrostatic voltage (charge build up on the VM's internal resistor. The only difference is the voltmeters resistor is lumped as opposed to distributed. The voltmeter is in parallel with the resistor, therefore it acts as a perfect mirror of what it's measuring, just as it does in a DC circuit. There is no way the voltmeter can report anything other than the total path voltage across the resistor, because it experiences the exact same thing. It's the fact that there is no flux in the measurement loop (i.e. between the VM lead and the resistor) that guarantee that they are parallel and have the same drop.

The resistor wire is in the field also and is part of the loop that is generating the 12 volts. In the exact same space it also dissipates it. This is modeled as a voltage source in series with a resistance. The RSD academy video I linked to earlier in this thread covers that quite well.

Yeah you've mentioned the voltage source / tiny batteries in the wires before. It's pure fantasy. A resistor is a resistor, it dissipates energy, that's it. The emf comes from the electric field induced by the solenoid, per Maxwell-Faraday. It's an outside source of energy. Therefore we do not model any voltage source in the circuit. Why RSD Academy describes it like this, I have no idea. He's trying to have his cake and eat it too - as far as I can tell he's using the more familiar path voltage, but invents a voltage source in the wires to makes things come out to zero. If he was actually using the scalar potential (per McDonald's paper which he cites), he would say so instead of making up ridiculous explanations.

 

[tedward]

I have lost the trace of rational thought on this thread. I don't even know what the question is......

All I'm trying to do was explore the implications of the different conventions. But the 'electrical guy' keeps derailing with objections.

 

[Averagesupernova]

@Charles Link the voltage across an inductor thread is a good read. @Dale covers what I've been saying all along very well. Start at post #64 for those who want to get right to the point of what's going on in THIS thread. I have tried to explain this from the circuit theory side all along. I've encountered every type of resistance (ha, a pun), counter arguments, etc. Can't have voltages across wires, a wire is not a voltage source, the resistor I call source resistance isn't real, you name it.
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As someone who has significant experience troubleshooting electronics I can say I've run into currents being induced where they aren't supposed to be. It might be a wire in a harness, a circuit board trace, poor circuit board layout causing cross-talk between inductors placed too close, etc, etc, etc. I get what is going on here. Wires can act like transformer windings. Lewin's experiment did just that. If anyone is unable to accept this I can say they wouldn't have gotten anywhere at several of my former employers. The posts made by @Dale was an eye opener for me as to why I've had such trouble getting my point across.

 

[hutchphd]

This is why somebody needs to ask a single salient question in a well regulated post and I can supply a (hopefully) orrect answer. There is no paradox and one need not define odd mechanical simulacra or invoke angels to explain it.
Not 100 diferent questions in parallel. This is just not a useful process.
For instance

There is disagreement, in this thread as well as others, whether these voltages of the secondary winding(s) each consisting of a half a turn of wire can be measured. This is a completely new rabbit hole. My view is the Mr. @mabilde did it right in his video.

No. There is no "right" or "wrong" way to measure them because they are not defined If you are in a rabbit hole just stop digging. One must specify the exact path along which to consider the EMF (in practice this will specify where to run the "voltmeter" wires).
It is neither right nor wrong it is simply unspecified or not

The posts made by @Dale was an eye opener for me as to why I've had such trouble getting my point across.

I do not find the posts by @Dale . Can you give specific reference? I value his opinion.

 

[Charles Link]

Here is the "link" again to what @Averagesupernova referred to in the previous thread.
https://www.physicsforums.com/threads/why-does-a-voltmeter-measure-a-voltage-across-inductor.880100/

I think there were a couple of logic errors that resulted in disagreements. One example if I read it correctly is post 101 by @tedward where @alan123hk disagrees with him in post 104 when he refers to the multi-tap transformer. In post 177 I think @tedward has it correct, if the voltmeter leads are such that they don't pick up any $E_m$.
In post 190 , (I'm editing this=at first I thought he might have erred, but I believe he has it correct.) If you consider the voltmeter leads as just the two leads leading vertically to point A , the wire on the left sees -6 volts of an electrostatic type, and the wire on the right sees +6 volts also electrostatic. Edit: I still need to study this in more detail=it's tricky...but yes, I think @tedward has it right. additional comment=he's calling it a voltage $V_s$ across the resistor in post 177, when it is properly referred to as a scalar/electrostatic potential difference. $\Delta U_s$ might be a better designation. I still need to study it further though...

Edit: I'm starting to conclude that it isn't fussy where the leads are in post 177 and post 190, =the important thing is that they are on the left side of the page with the voltmeter. If you move the voltmeter and the leads to the other side, I do believe it will read zero volts.

 

[Averagesupernova]

@tedward have a look at the link below. Specifically part way down where it's says: "The three assumptions of CA."
Circuit analysis is all that's needed to make sense of this issue. Naturally we need to know that induction is possible but very little more is needed to understand what's happening here.

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

@tedward you didn't follow this or you didn't pay attention. I'll cut and paste it here for Pete sake.
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The implications are: No electric fields. No magnetic fields. No electric charges. Kirchoff’s Laws apply instantaneously around the entire circuit, so there is no temporal first-next ordering of events.

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

This thread is closed for Moderation

 

[berkeman]

After a number of deletes/edits, the thread will remain closed. It's been a difficult subject to try to address. Thanks for staying level-headed for the most part, up until the end of the thread.