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.