I Walter Lewin Demo/Paradox: Electromagnetic Induction Lecture 16

[TSny]

@TSny I could have been more clear. The lines bloom out on the outside of the coil. But they bloom IN towards the center of the bore on the inside of the coil. I'm not sure if the phrase bloom in is the wisest choice of words. Lol. The fact is that all of the flux lines originate AT the wires. Current increases and they move away from the wires.

I see. So my arrows should have pointed radially inward for the case where the current is increasing. If the coil is a long solenoid, then the field at any instant of time should be approximately uniform over the cross-section of the solenoid. Suppose the current increases linearly with time so that the field lines move inward as new lines are created at the source current. It seems to me that the lines would quickly get very crowded together near the axis of the solenoid and I have a hard time visualizing how the field could remain uniform across the cross-section as the current increases. Anyway, I don't want to get into the middle of this. I was just trying to visualize your description of what's going on and understand why you are saying that voltage is only generated in the arc portion of the pie-shaped loop. Thanks.

Edit: Add picture with field lines moving inward

 

[Averagesupernova]

I'll clarify a bit. The lines move towards the center on the inside. They don't all arrive at the center. It gets crowded for sure.

 

[TSny]

I'll clarify a bit. The lines move towards the center on the inside. They don't all arrive at the center. It gets crowded for sure.

Ok. So the speed of a field line decreases as it moves inward in such a way that the density of lines is always uniform within the solenoid. The lines asymptotically approach the center.

 

[tedward]

Ok I'll post it again and try to explain it.

https://www.pengky.cn/zz-generator-...lternator-principle/alternator-principle.html
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The wire loop of the rotor only generates when it is CUTTING through the flux lines. Notice when when the sine wave goes through zero the wire that forms the rotor is not cutting the flux. The voltage generated is based on the sine of the angle of the rotor at a particular instant.
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It doesn't matter what is causing the cutting action. Growing and shrinking flux is just as effective as actual mechanical motion between the wire and the flux .
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This has all been said to death here, there's really no more I can say.

Wrapping my head around this a bit. You're using the word 'flux' to simply mean the magnetic field itself - depicted your diagram. That is not what that word means. To use the concept of flux, you have to define an area, i.e. the circuit loop. Magnetic flux is the total AMOUNT of magnetic field that passes THROUGH the loop. If we measure magnetic field in Teslas (T), then magnetic flux is measured in Tesla-meters-squared (Tm^2). If that amount changes with time, you get an induced ELECTRIC field, that points around the circle (along the wire, either clockwise or counterclockwise). The magnetic field lines themselves DO NOT HAVE TO TOUCH THE WIRE. You could have a small solenoid with a large circuit loop, and no magnetic field in between. What pushes current around the loop is the INDUCED ELECTRIC FIELD created by the solenoid. There is no 'cutting' or 'bowing'.

The video you linked to demonstrates what is called MOTIONAL EMF. It is a separate physical phenomenon, but is often lumped in induction because they give rise to the same law. Here you have a conductor MOVING THROUGH the magnetic field. The magnetic field exerts a force DIRECTLY on the free charges (no induced electric field here), causing an emf and therefore current. I guess in this case you could call it cutting, the way a (horizontal) blade cuts across (vertical) grass. Both of these effects are often called 'induction', but only one actually induces an electric field. They're both described by Faraday's law but they are different effects.

You've learned a bad model, that's all. It's hard to unlearn something that you've sworn by before, but if you don't you're never going to get this stuff.

 

[tedward]

Suppose the magnetic field of the coil is coming out of the page inside the coil:
View attachment 322304
Consider a pie-shaped loop inside the field region:
View attachment 322309Is this at least somewhat along your line of thinking?

This is a much better illustration of what's going on - flux is the amount of field that passes through any defined area. Whether the field lines 'bloom' in or out a bit towards the ring is irrelevant. The integral that defines flux (below) only counts the part of the magnetic field $\vec B$ that is perpendicular to the area - that's what the 'dot product' in between the vectors means. The part that points radially doesn't contribute anything.
$$\Phi_B = \iint \vec B \cdot d \vec A$$

But to really see what's going on, you need to see the induced field. I'll put up a diagram if I an find a good one

 

[tedward]

Here are a couple diagrams, which include both changing magnetic field $\vec B$ and the induced electric field $\vec E$.

This first one shows the changing magnetic field pointing into the page, and the induced electric field circles counter-clockwise WITHIN the flux area.

This one shows a circuit with the magnetic field inside the loop, and the resulting induced electric field that drives current. Not that the magnetic field lines DO NOT TOUCH the wire itself, there's nothing in the definition that requires that.

Finally here's one that shows the direction of the induced electric field (in free space) as it relates to the increase or decrease of the magnetic flux inside.

 

[hutchphd]

always uniform within the solenoid.

Yes approximately uniform but changing density according to any changing current through the solenoid windings (just to be clear.....)

The notion of cutting ines of flux is a topological argument: useful because both they and the wire path are, in fact, closed curves. Any change in flux will require a flux loop to be "cut" and then reattached on the other side of the wire path. The notion of lines of flux is useful but only when correctly used. Cut is a verb requiring either motion of the wire path or of the flux lines
The use of lines of flux as a visualizationn tool is very useful IMHO but requires some discipline and practice.
Another annoyance in this argument is "voltmeter". It is not that smart. What a voltmeter measures fundamentally is current in a conductive loop and it converts that into a "voltage" value (typically using a local high value internal resistor). No magic fields required.

 

[TSny]

Yes approximately uniform but changing density according to any changing current through the solenoid windings (just to be clear.....)

Yes

The notion of cutting ines of flux is a topological argument: useful because both they and the wire path are, in fact, closed curves. Any change in flux will require a flux loop to be "cut" and then reattached on the other side of the wire path. The notion of lines of flux is useful but only when correctly used. Cut is a verb requiring either motion of the wire path or of the flux lines
The use of lines of flux as a visualizationn tool is very useful IMHO but requires some discipline and practice.

The visualization of moving B-field lines "cutting" conductors is interesting. I have occasionally seen illustrations such as this video between times 18:15 and 18:36. But I'm not used to thinking of an induced emf in a stationary conductor as "caused" by moving B-field lines cutting across the conductor. Of course, we are all familiar with the related idea of "motional emf" that occurs when a conductor moves such that it cuts across static B-field lines. Here, I think of the underlying "cause" of the emf as due to the Lorentz force law $\mathbf{F} = q \mathbf{v} \times \mathbf{B}$. For the case of a stationary conductor cut by moving field lines, I think of the emf as "caused" by the presence of an induced electric field $\mathbf{E}$ (as described by the Maxwell equation $\mathbf{\nabla} \times \mathbf E = - \frac {\partial \mathbf{B}}{\partial t}$) rather than as "caused" by the field lines cutting the conductor. [Edit: see more video here].

Another annoyance in this argument is "voltmeter". It is not that smart. What a voltmeter measures fundamentally is current in a conductive loop and it converts that into a "voltage" value (typically using a local high value internal resistor). No magic fields required.

There is a very interesting paper in The American Journal of Physics [50, 1089 (1982)] by Robert Romer with the title "What do voltmeters measure?: Faraday's law in a multiply connected region". It is very relevant to the "Lewin Paradox". Lewin himself has referred to this paper. Here is a quote from the paper that gives a nice way of describing what a voltmeter measures :

A voltmeter (whether a conventional indicating meter or an oscilloscope) is most often an ohmic device, usually of high resistance, which gives an indication (a deflection of a meter needle or an electron beam) proportional to the (small) current that passes through it. A little thought convinces one that the volt-meter reading (call it V) is equal to the line integral of E, $\int \mathbf{E} \cdot \mathbf{dr}$, where the path of integration passes through the meter, beginning at the red (or “+””) lead and ending at the black (or “-“) lead. (I have been unable to think of any device that is normally considered to be a voltmeter, whether a real device or an imaginary one, one with high resistance of low, or even nonohmic in character, for which it is not the case that the reading is proportional to the line integral just described. Even the imaginary miniature worker who implements the common thought-experiment definition of “potential” by carrying a tiny test charge from one point in space to another, measuring the work required, is just measuring the line integral of E along whatever path is chosen.)

So, when applying Faraday's law $\oint \mathbf{E} \cdot \mathbf{dr} = -\dot \Phi$ to a closed loop, where there is a voltmeter in the loop, the part of $\oint \mathbf{E} \cdot \mathbf{dr}$ that passes through the meter will represent the reading of the voltmeter.

 

[tedward]

The visualization of moving B-field lines "cutting" conductors is interesting.

Interesting, I've never seen that visualization before. I always picture the magnetic field vectors as fixed in place, with the magnitude fluctuating. But as you said, it's the induced E-field that actually pushes charge. These are after all just visualizations, and any visualization that leaves it out is missing the point. I don't think that part is emphasized enough when Faraday's law is taught even in undergrad classes. It took me a long time to figure out that 'emf' is just the sum of this induced field.

There is a very interesting paper in The American Journal of Physics [50, 1089 (1982)] by Robert Romer with the title "What do voltmeters measure?: Faraday's law in a multiply connected region".

I've read the Romer paper, it's very clear. That's why voltmeters on the outside the circuit (i.e. outside the region of flux) measure exactly what they're supposed to - the path voltage between the two points they are connected to. That's also why this argument of the voltmeter wires 'canceling' the intended voltage reading doesn't make sense, at least as far as this voltage convention is concerned. If there's flux in your loop though (as in Mabilde's setup), that influences your measurements.

 

[tedward]

Now that we're past basic physics, I thought it might be interesting to discuss the two voltage conventions in a bit more detail. One of the things the diagrams that Mabilde or RSD Academy always do is assume the resistors are of negligible length. I think that hides the true nature of what the 'scalar potential' convention of voltage really is. What happens when the resistor's length cannot be ignored?

Consider a wire loop around a solenoid with a 12V emf. But one-quarter of the wire is a resistor of uniform resistivity, with a total resistance of 12 ohms. The rest of the wire has negligible resistance. First we'll use the path voltage convention $$V = \int \vec E \cdot \ d \vec l$$ Here V is the voltage drop between two points measured in a clockwise path. We get the following diagram:

Note that the 12V across the resistor is the sum of two effects: the induced voltage $V_i$ corresponding to 1/4 the total emf, and the scalar potential $V_s$ due to the static charges built up at either end (the charges shown in the diagram). Of course here, the electric field, and hence the voltage, in the conducing wire is zero. The sum of voltage around the loop is 12V, which equals the emf of the solenoid.

Now let's see how the scalar potential convention handles this. I'll use $V'$ to denote this voltage. We can define $V'$ as follows: $$V' = \int \vec E_s \cdot d \vec l $$ Here $E_s$ is the the electrostatic field - i.e. the electric field due to the static charge only. We can also write it in terms of the path voltage and induced voltage: $$V' = V - V_i$$ Here $V_i$ is the induced voltage corresponding to that section of the path. This is what is sometimes ambiguously referred to as the 'emf in the wire', but it basically corresponds to the integral of induced electric field as it would appear in free space: $$V_i = \int \vec E_i \cdot d \vec l$$ Using the scalar potential, we get this diagram:

Here we can see that the voltage in the resistor is only due to the static field, and the voltage in the conducting wire gives the opposite value, so the loop sum is indeed zero. But what should be clear here is that the scalar potential, instead of including the emf in the wire as people seem to regard it, actually subtracts the induced emf $V_i$ from the overall analysis. The definition of 'scalar potential' drops the induced voltage completely precisely because it is non-conservative.

One drawback (in my opinion) that we see immediately is that the static potential across the resistor is not the $V$ in $V=IR$, Ohm's law - that voltage is the path voltage. You can still apply Ohm's law but you now have to redefine it to include the induced voltage $Vi$, in other words: $$V' = IR-V_i$$ If you only use resistors with negligible length, this fact is obscured, as the scalar potential and path voltage would essentially be equal across the resistors. Somehow I doubt if the 'voltage' shown here was shown in a video attacking Lewin's position, that very many people would be convinced, as it wouldn't agree with their intuitive notions of what voltage is. There are some other major inconveniences to scalar potential but I'll address them in a separate post.

The same relationships exist for the electric field. As the net field is the superposition of the static and induced fields, the static field is the net field minus the induced field: $$\vec E_s = \vec E - \vec E_i$$
Below is a diagram of the various electric fields and their relationships.

 

[Averagesupernova]

If you're wondering if the resistive wire is dissipating 12 watts, 6.75 watts, or .75 watts the answer is 12. No matter how you diddle the voltmeter leads, that's what the resistor will do. You have to model the resistor wire as a 3 volt source in series with the 9 volt source in series with its own 12 ohms. This is what tells me that path dependency is a bunch of BS. Unless you want to unravel ohms law now too. If Lewin's setup is any indication of what your circuit will do, flipping the voltmeter leads will cause the meter to read either 3 or 12. Measuring with the @mabilde setup will always read 9. If the circuit is opened you will read 3 across the resistive wire and 9 across the regular wire with the @mabilde setup. So, we measure voltages and nothing adds up to what the resistor wire will heat to. Yet another paradox. (But not really).

 

[tedward]

If you're wondering if the resistive wire is dissipating 12 watts, 6.75 watts, or .75 watts the answer is 12. No matter how you diddle the voltmeter leads, that's what the resistor will do.

Uh... yes(?). Use $P = I^2 R$ or $P=IV$, which underscores that the true $V_r$ = 12V. What's your point?

You have to model the resistor wire as a 3 volt source in series with the 9 volt source in series with its own 12 ohms. This is what tells me that path dependency is a bunch of BS. Unless you want to unravel ohms law now too.

Can't make heads or tails out of this. There are 12 V (path voltage) across the resistor, 12 ohms of resistance, and current is 1A.

If Lewin's setup is any indication of what your circuit will do, flipping the voltmeter leads will cause the meter to read either 3 or 12.

?? Flipping voltmeter leads changes the sign of the measurement, period.

Measuring with the @mabilde setup will always read 9.

No disagreement - he's "measuring" scalar potential. Measure on the outside and you will measure the actual path voltage (which corresponds to correct power dissipation) of 12V (see below)

If the circuit is opened you will read 3 across the resistive wire and 9 across the regular wire with the @mabilde setup. So, we measure voltages and nothing adds up to what the resistor wire will heat to. Yet another paradox. (But not really).

No idea what your talking about here. I drew the solenoid bigger to clearly see the magnetic flux in Mabilde's measurement loop.

 

[hutchphd]

Uh... yes(?). Use P=I2R or P=IV, which underscores that the true Vr = 12V. What's your point?

There is a small matter of the signs

 

[Averagesupernova]

Lewin's setup is any indication of what your circuit will do, flipping the voltmeter leads will cause the meter to read either 3 or 12.

Actually I made a mistake here. Voltmeter leads will read either 12 or zero.
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@tedward
You just don't seem to grasp a few things. It's obvious you are not an 'electrical guy'. The voltmeter leads form the same conductors as if the loop went 360 degrees. That's the reason you measure 12 volts. The reason that there is 12 watts in the resistor I already explained due to the way the circuit has to be modeled. In any circuit whether part of it is in a changing magnetic field or not, the voltage source has to have it's internal resistance accounted for. A 12 ohms section of wire has a source resistance of 12 ohms when exposed to a changing magnetic field. If the whole loop were a 12 ohm section of wire with a
12 ohm load resistor well out of the field you would only see 6 volts at the load.

 

[Haborix]

Is this horse dead enough, yet?

 

[tedward]

There is a small matter of the signs

I used the voltage 'drop' convention, so $V_r = \int \vec E \cdot d \vec L$, instead of the potential difference convention of $V_{ab} = -\int \vec E \cdot d \vec L$. Is that what you're referring to?

 

[tedward]

Is this horse dead enough, yet?

All I'm trying to do now is see if people, especially EE's, actually use this scalar potential convention. It seems very counter intuitive to me. The arguments back-and-forth have gotten tiresome though, I agree.

 

[Haborix]

Fair enough. It seemed to me you all had reduced this to semantics, or a mathematical equivalent, and the added value for continuing was very low for all parties involved. I think there are some good nuggets in the thread for others who come along; I just hope they don't get stuck in a vortex of never-ending confusion.

 

[alan123hk]

All I'm trying to do now is see if people, especially EE's, actually use this scalar potential convention. It seems very counter intuitive to me. The arguments back-and-forth have gotten tiresome though, I agree.

I don't understand what you are trying to discuss or prove. Isn't the definition of the conservative field potential that does not vary with the path and the voltage that varies with the path already clear?

Anyone can use them according to their needs. Do you think that electrical engineers only use the definition of electric potential and never the definition of voltage? Kirk T. McDonald is a physicist. In his article, he also uses the concept of potential to analyze electromagnetic circuits to express something. Are people going to keep pestering and saying what's the inconvenience and disadvantage of this because it doesn't directly correspond to the power loss in the resistor, etc.?

I suggest that if you think that there is a violation of the laws of physics that leads to wrong results, please point it out directly, it will help focus the discussion.

 

[tedward]

I don't understand what you are trying to discuss or prove. Isn't the definition of the conservative field potential that does not vary with the path and the voltage that varies with the path already clear?

Anyone can use them according to their needs. Do you think that electrical engineers only use the definition of electric potential and never the definition of voltage? Kirk T. McDonald is a physicist. In his article, he also uses the concept of potential to analyze electromagnetic circuits to express something. Are people going to keep pestering and saying what's the inconvenience and disadvantage of this because it doesn't directly correspond to the power loss in the resistor, etc.?

I suggest that if you think that there is a violation of the laws of physics that leads to wrong results, please point it out directly, it will help focus the discussion.

My aim right now is simply to compare the conventions (as the scalar potential is new to me, but you've convinced me that it has some legitimacy) to see what they actually correspond to - especially in this distributed case, as the analysis with negligible length resistors hides some glaring differences. I really wanted to see what you thought specifically - even though we agree on the fundamental physics, you've said repeatedly that a voltmeter measurement from the outside, or with a T-shaped loop on the inside, hides or 'masks' the voltage in the wire. Do you still claim that, or can we say definitively now that if you want to correctly measure scalar potential, use Mabilde's setup, and if you want to correctly measure path voltage, measure from the outside along a specific path?

Also scalar potential isn't as 'path independent' as you may regard it. That's only the case if the wires and resistors are fixed in place. The moment you move a resistor (of non-neglible length) closer or farther form the source, or rotate it with respect to the induced field, its scalar potential can change dramatically. That doesn't happen with path voltage, which only changes when you move the path to the other side of the solenoid - I have another post coming showing that lol. (This for me is the interesting physics discussion we could be having, not arguing about fundamentals with people who don't want to learn.)

(I included a slightly modified diagram, showing both of Lewin's corresponding paths, and Mabilde's.)

 

[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.

 

[alan123hk]

you've said repeatedly that a voltmeter measurement from the outside, or with a T-shaped loop on the inside, hides or 'masks' the voltage in the wire.

I don't remember if I ever said that, and if I did say that was an inaccurate statement, I apologize, I meant of course that if the leads of a voltmeter are disturbed by the induced electric field from changing magnetic field, it cannot be accurate measures the scalar potential between two points on the ring circuit. But I must have mentioned repeatedly that I was describing the scalar potential generated by the charge, because this is focus of controversy.

especially in this distributed case, as the analysis with negligible length resistors hides some glaring differences

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?

Also scalar potential isn't as 'path independent' as you may regard it. That's only the case if the wires and resistors are fixed in place. The moment you move a resistor (of non-neglible length) closer or farther form the source, or rotate it with respect to the induced field, its scalar potential can change dramatically.

Even if a pure electrostatic system does not contain induced electric fields caused by changing magnetic fields, when you move a charged capacitor, resistor along a wire, it changes the potential distribution in space, so I can say that the electric potential produced by pure electrostatics is not Path independent?

That doesn't happen with path voltage, which only changes when you move the path to the other side of the solenoid - I have another post coming showing that lol. (This for me is the interesting physics discussion we could be having,

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

 

[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]

I don't remember if I ever said that, and if I did say that was an inaccurate statement.... But I must have mentioned repeatedly that I was describing the scalar potential generated by the charge, because this is focus of controversy.

We were going back and forth on this point, that's why I wrote that "test leads" post. Maybe I misinterpreted but I thought you claimed afterwards that the outside voltmeter reading doesn't even measure path voltage correctly as opposed to scalar potential (which it doesn't). Doesn't matter, we're on the same page now.

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?

It's not that any calculations are wrong, if you know what convention you're using. It's just that in the case of negligible length, there's very little induced voltage, so scalar potential is virtually equal to the path voltage, and it's hard to see much difference - the voltage across a resistor is still equal to IR regardless of convention. But If you have a distributed length, there's now a very noticeable difference between the two. Path voltage is the V in V=IR, scalar potential subtracts the induced voltage from this. Now consider a lay audience - youtube viewers, maybe with some physics interest and knowledge but not a deep understanding or even awareness of these conventions. If they're told the 'voltage' is 9V when IR is 12V, they might think..."hmm that's not the voltage I'm familiar with, something weird is going on". They might actually investigate further and learn that there's more than one thing called voltage, and Lewin is actually correct according to the more commonly understood convention. You can see the potential for confusion here, which explains a lot of the reason why so many people argue with each other over this problem. That's the main reason I've been posting here - to figure out why the hell people are disagreeing over something that should be absolute.

Even if a pure electrostatic system does not contain induced electric fields caused by changing magnetic fields, when you move a charged capacitor, resistor along a wire, it changes the potential distribution in space, so I can say that the electric potential produced by pure electrostatics is not Path independent?

I take your point. But most people who have worked with voltage at all, even in simple circuits, aren't used to values changing as soon as you nudge a wire or turn a resistor. It's taken as a given that the location of circuit elements doesn't matter, just how things are connected. You can imagine how weird is to see scalar potential values change even with small movements, when path voltage is stable (path voltage only changes when you put a wire on the other side of a solenoid). Again, thinking of a lay audience, they might realize that - despite path dependence - path voltage locally performs the way that they're used to thinking about. If the goal is to resolve this debate, (not just here but in a wider audience), people should have a clear understanding about what these different conventions do differently. I'll put up some diagrams on this last point.

 

[Muu9]

Just thought I'd drop some links:
http://www.phy.pmf.unizg.hr/~npoljak/files/clanci/guias.pdf
https://intuitivephysics.me/validity-of-kirchhoffs-law
https://www.eevblog.com/forum/chat/kirchhoff_s-loop-rule-is-for-the-birds/

 

[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.