Undergrad Walter Lewin Demo/Paradox: Electromagnetic Induction Lecture 16

[tedward]

@tedward it seems to me you still believe that the pie shaped leads @mabilde uses to sweep around the single turn loop contribute to the voltmeter reading on their own. This is not so. Once you can understand this I think you will understand the broader picture of what is going on in many places of this so called paradox.
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You say the pie shaped wires in the @mabilde video form a loop and imply that this alone is what causes the reading. If we were to extend the wires in this loop out radially alot farther before we form the loop, what do you suppose would happen? In other words, we have a large pie shaped loop but it is not part of the round loop that was previously swept around. If the round part of the new loop is significantly far away the solenoid the pie shaped loop will generate a very insignificant voltage. The flux from the solenoid is not strong enough to CUT the now larger part of the pie loop. It's too far away. Remember the violin strings/bow scenario I described? The radial wires inside OR outside the solenoid will contribute nothing.

Ok, this tells me that you have a fundamental misconception about how induced emf works. You need to learn it the right way if we're going avoid talking past eachother. Take your example from earlier of putting voltmeter leads around a solenoid, we can learn a lot from this. Starting with just the voltmeter by itself (no solenoid), shorting the leads gives you a reading of zero, as they should. No flux, no emf. Now take the leads and put them around this very tall (say infinitely tall) solenoid and short them on the other side. Now there is a changing magnetic flux in your measurement loop. Think of the imaginary surface formed by your loop, like the soapy film on a plastic wand for blowing bubbles. The flux lines penetrate through this surface perpendicularly. The emf is determined by the rate of change of the total amount of flux penetrating this surface. That's Faraday's law.

The voltmeter will now read this emf. If you move the wires around, as long as the total flux in your loop stays the same, the emf reading will not change. Say you make your voltmeter loop much bigger, but the same solenoid is still inside. The total emf WILL NOT CHANGE, because the amount of flux through your loop has not changed. (Now, in the real world, the solenoid is not infinitely tall, and the field lines will circle back back around again, passing through your loop, canceling out the flux. But that's just a problem with this particular setup, there are ways of avoiding a return path and you can make your loop as big as you want, like confining the magnetic field to a torus shaped iron core).

Here's a another thought experiment. Take your voltmeter leads, and short them so it reads zero. Hold your loop right next to the solenoid, but just make sure the solenoid remains outside your loop. The voltmeter will still read zero. This is because there is no flux through your loop. No flux, no emf. We see that flux can only cause an EMF in a LOOP. It doesn't cause an emf in a section of a wire. Now you can talk about induced ELECTRIC FIELD in a wire, certainly.

The only way an emf can manifest itself is with an electric field, which is why considering the field is crucial. Say this particular loop (with the solenoid near, but outside), is long and thin, so two long wires connected by very short segments, with the solenoid outside next to a long wire. The solenoid will try to exert an induced field on both the long wires, but in the loop these are opposite directions, so they cancel eachother out, and there is no net electric field in any of the wires - i.e. no emf. The voltage drop across any section is therefore zero. (I think it's slightly more complicated than that as I believe some static charge will reposition themselves, but the end result is still the same).

So in your idea of extending the pie shaped loop - in the ideal sense, with the solenoid infinitely tall (so no return path), it would make no difference. The reading would't change. In this real setup, once the loop gets big enough you will catch the returning flux and it will cancel out to zero. But this has nothing to do with distance, just a limitation of the practical setup. If the solenoid was wrapped around an iron core in a rectangular donut type shape, with the other end of the rectangle outside the loop, we won't have this problem and you can make your loop as large as you like.

Now here's another experiment to convince you of @mabilde's error. Say he's measuring a section of conducting wire in his set up. The real voltage is zero, but he's reading an emf. This is due SOLELY to the flux from the solenoid, not the scalar potential (which is non-physical on its own). Now, change the area of the loop. keep the contacts on the wire ring where they were, but press the lead wires together, so the loop becomes a closed T-shape with no area. Now there is no flux from the solenoid through this loop, and you will read zero volts, as you should. This is why his setup is so carefully made - to get precise measurements from the emf that exactly line up with the derived value for scalar potential. It occurred to me thinking it about it this morning that he's doing this intentionally to pretend he's measuring the scalar potential. It's a bullshit measurement. More on this - I have a rant coming....

[And by the way, your mental model of the" violin bows" of the flux cutting across the wire is simply wrong. I used to imagine something similar until I realized that the magnetic flux can be set up to exist only in the center, nowhere near the circuit wires, and you still get an emf. Read up on Faraday's law!!]

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

(warning: Long response - even for me).

After some thread/post cleanup work, the thread will be reopened. After a PM conversation with @tedward he will try to be more brief in his posts, and support his points using more math.

Thanks for your patience.

 

[Averagesupernova]

The below animation shows all that is needed to know about induction to prove my point about cutting the lines is the only thing that sets up a current in a conductor in a magnetic field. The action starts in the video at about 32 seconds. If we cannot nail this part down and agree then this discussion is doomed. Keep in mind that the motion only needs to be relative. An increasing or decreasing field with the right orientation of the conductors relative to the flux will cause a current in the conductor. Conversely, an increasing or decreasing field with the conductors oriented in a different way will not cause a current in the conductor.

https://www.pengky.cn/zz-generator-...lternator-principle/alternator-principle.html

 

[tedward]

Ahh, now I see what you meant. Thanks for posting this. This view of induction IS correct when you have a FIXED (unchanging in magnitude or direction) magnetic field, and MOVING wire components in the B-field. The emf is caused by the magnetic part of the Lorentz force:

F = qv x B

Essentially, free charge moving through the fixed B-field experiences a magnetic force which drives current. The emf is therefore caused by magnetic force, there is no E-field directly involved (I believe). The effect is consistent with Faraday's law:

emf = -d(Phi)/dt

where Phi is the magnetic flux through the loop as it moves.

Now the situation with a transformer, or our single loop with a flux, is different. It is a FIXED wire with a CHANGING magnetic field confined to a small region inside the loop, which does not touch the wire itself (it doesn't have to). In other words, outside the solenoid, and at the wires in particular, the magnetic field is zero. Now it's the changing flux through the loop that creates an INDUCED electric field circulating through space, and in particular at the wire loop. This is induced electric field is described by Maxwell's version of Faraday's law:

curl (E) = -d(B)/dt

When integrated, this becomes the same emf equation written above. So in this second case (our situation), there is no 'violin string bowing.' The electric field is simply created by the changing flux, per the Maxwell-Faraday equation.

Oddly, there are these two different physical scenarios both give rise to the same equation. Feynman talks about this in his EM lectures, I highly recommend you read them as he gives a great treatment (link below):

"We know of no other place in physics where such a simple and accurate general principle requires for its real understanding an analysis in terms of two different phenomena. Usually such a beautiful generalization is found to stem from a single deep underlying principle. Nevertheless, in this case there does not appear to be any such profound implication. We have to understand the “rule” as the combined effects of two quite separate phenomena." - Richard Feynman

https://www.feynmanlectures.caltech.edu/II_17.html

 

[Averagesupernova]

This view of induction IS correct when you have a FIXED (unchanging in magnitude or direction) magnetic field, and MOVING wire components in the B-field.

It makes no difference which is fixed and which is not. It also makes no difference whether the flux lines are cutting the wire due to an increase or decrease in field strength or the magnet is moving. All scenarios generate a current in the wire that obeys the right hand rule. Let's keep this dumbed down because all we are interested in is determining whether the pie shaped wires can be oriented inside the solenoid and not contribute to a voltmeter reading simply by their presence.

 

[tedward]

It makes no difference which is fixed and which is not. It also makes no difference whether the flux lines are cutting the wire due to an increase or decrease in field strength or the magnet is moving. All scenarios generate a current in the wire that obeys the right hand rule. Let's keep this dumbed down because all we are interested in is determining whether the pie shaped wires can be oriented inside the solenoid and not contribute to a voltmeter reading simply by their presence.

You're correct that both scenarios generate current / emf in the wire. But you made an argument a few posts back that if the pie-shaped wire is far away from the solenoid, eventually the magnetic field lines aren't strong enough to 'cut' the wire (as you put it). We can keep this dumbed down - but, if you make an argument that is based on an incorrect understanding, we're not going to agree on anything.

The fact is, the only thing that determines how much emf is induced in a loop of any size, at any distance, is the rate of change of the total amount of flux that penetrates the AREA enclosed by the loop. This is a mathematical equivalence that is based on Stoke's Theorem (usually taught in Calc III).

No magnetic field has to touch the wires at all. If you read just a part of the Feynman lecture I linked to, this is what he refers to as 'flux rule'. It's about as basic as Faraday's law gets, and anybody who's interested in this circuit problem should understand it, as it's highly relevant to what we're discussing. So yes, moving the pie shaped wires farther outside of the solenoid region increases the area of the pie slice, but not the amount of flux that penetrates it, so the emf would remain the same (practical considerations of the negative returning flux aside).

Now consider changing the area of the pie slice INSIDE the solenoid where it already is. Keep the ends of the two leads attached to the same two points on the conducting wire ring, but move the voltmeter lead wires themselves around to make the area they form larger or smaller. This changes the amount of magnetic flux passing through the loop, which changes the induced emf, and the reading on the voltmeter. If you make the area of this loop zero (with the leads still in place, making a T-shape), you will have zero flux, therefore zero emf, and a zero reading. With no flux in the loop to throw off the readings, this means that there is ZERO VOLTAGE DROP across the two points on the conducting wire ring.

 

[Averagesupernova]

Ok so let's look at this another way. We agree (I think) that a completely shorted ring will read zero volts no matter how the pie shaped probes are positioned. A ring that is not shorted shows a voltage that varies depending on the position. Your claim is that in the case of the varying voltage, we are not reading a voltage that is present from point A to point B. We are reading a voltage generated by the loop that is formed by the pie shaped probes and a portion of the ring. Suppose I give you the benefit of the doubt here and agree to that. So how can you explain the zero volts scenario with a shorted ring? You can't have it both ways. We still have the pie loop and still have the same field and it is NOT generating a voltage.

 

[alan123hk]

I think the test method used by mabilde is correct. When we measure voltage with a voltmeter, the DUT has to supply energy, so the Lewin circuit has to supply energy, so the voltmeter loop has to somehow let the changing magnetic flux through to get the energy, otherwise it won't be able to measure anything thing.

Of course, according to Faraday's law, the test result is the induced EMF in the pie-shaped area due to the change of the passing magnetic flux. But this is not enough to prove that it cannot represent the scalar potential difference generated between the ring arcs at the same time.

I think whether you agree or disagree, both parties should try to come up with some stronger arguments to support it. The method you said of pressing the leads together so that the loop becomes a closed T-shape with no area is problematic because the leads of the voltmeter will be affected by the induced electric field. When you are trying to measure the potential difference created by a pure charge, do not subject the voltmeter and its leads to the induced electric field, this may cause errors in the measurement results.

I reiterate my opinion again, I agree with mabilde's method is correct.

 

[tedward]

Ok so let's look at this another way. We agree (I think) that a completely shorted ring will read zero volts no matter how the pie shaped probes are positioned. A ring that is not shorted shows a voltage that varies depending on the position. Your claim is that in the case of the varying voltage, we are not reading a voltage that is present from point A to point B. We are reading a voltage generated by the loop that is formed by the pie shaped probes and a portion of the ring. Suppose I give you the benefit of the doubt here and agree to that. So how can you explain the zero volts scenario with a shorted ring? You can't have it both ways. We still have the pie loop and still have the same field and it is NOT generating a voltage.

Help me out here - I'm not quite sure what you mean by a shorted ring vs. a non-shorted ring?

 

[Averagesupernova]

A non shorted ring is one similar to the Lewin experiment and @mabilde with the pie shaped leads that show a voltage. A shorted ring would be just that. A shorted ring is demonstrated in the @mabilde video at 32:10. Lewin never demonstrated a shorted ring.

 

[tedward]

I think the test method used by mabilde is correct. When we measure voltage with a voltmeter, the DUT has to supply energy, so the Lewin circuit has to supply energy, so the voltmeter loop has to somehow let the changing magnetic flux through to get the energy, otherwise it won't be able to measure anything thing.

Of course, according to Faraday's law, the test result is the induced EMF in the pie-shaped area due to the change of the passing magnetic flux. But this is not enough to prove that it cannot represent the scalar potential difference generated between the ring arcs at the same time.

I think whether you agree or disagree, both parties should try to come up with some stronger arguments to support it. The method you said of pressing the leads together so that the loop becomes a closed T-shape with no area is problematic because the leads of the voltmeter will be affected by the induced electric field. When you're trying to measure the potential difference created by a pure charge, you can't subject the voltmeter and its leads to an induced electric field.

I reiterate my opinion again, I agree with mabilde's method is correct.

[Sorry, what is the DUT?]

Ok, so you agree that the pie shaped measurement loop is penetrated by the corresponding part of the solenoid flux, and this emf is what the voltmeter reads per Faraday's law. I think where our thinking differs, is what the intended measurement is. As we've discussed, we have two different voltage conventions:

Path voltage = Int(E . dl) (between two points on a specified path.)

Scalar potential = Int(Es . dl)

where I'm using E as the total electric field (which charge actually responds to), and Es is the electric field from the only the static charge distribution, or the static field. I'm going to stick to this terminology to keep the two conventions seperate. In the case of a battery providing the emf, there's no difference. With induced emf, these are different quantities.

With direct measurement (no flux in the measurement loop), a voltmeter can only measure path voltage, as it is the actual energy difference per unit charge between two points, or looked at another way, the work per unit charge done by an outside force moving a test charge along the path. Put a voltmeter across a resistor, you get the V in V = IR. Put one across conducting wire, you get zero.

So if @mabilde is using the scalar potential convention for voltage (he never explicitly says so but he does cite the McDonald paper so I can only assume that's what he means), he cannot measure it directly. Scalar potential (as I understand it) is a mathematically derived value, not an observable physical phenomenon, as you cannot extract the static field from the net field outside of an equation.

Now I agree that if the resistors are of negligible length, and in this precise circular setup with the leads positioned exactly as they are, that the scalar potential difference across a section of conducting wire is numerically equivalent to the induced emf through his measurement loop. So if the flux emf is .25V, we can say the difference in scalar potential is the same.

My problem is that he doesn't mention scalar potential, discuss the difference in voltage conventions, or acknowledge that he's really just measuring the flux through his loop. As @rude man said in one of his posts, @mabilde is 'simulating' the scalar potential measurement, not measuring it directly. Well I can 'simulate' rain by pissing on your leg. And if say "please enjoy this rain simulation", there's no problem. But if I tell you it's raining, most people would call that lying. Either way it creates a mess. The point is that he's convinced people that there is a measurable energy drop that exists between two points of conducting wire, which is completely bogus. (I'll answer the second part about the voltmeter leads canceling the 'emf' in a separate reply).

 

[Averagesupernova]

[Sorry, what is the DUT?]

Device Under Test.

 

[Averagesupernova]

My problem is that he doesn't mention scalar potential, discuss the difference in voltage conventions, or acknowledge that he's really just measuring the flux through his loop.

Doesn't the fact that he reads zero volts with the pie shaped leads in a shorted loop condition imply he's not measuring flux? If he were measuring flux a reading would show up here. As I said before, you can't have it both ways.

 

[tedward]

Thanks for the clarification on the shorted ring. We could take that to be a string of identical resistors as in the video, a ring of uniform resistance, or single loop of conducting wire (with no other lumped resistors). I certainly do NOT agree that the correct measured voltage (meaning path voltage, IR drop, etc) is zero between any two points, as the emf is now evenly distributed across the ring, as opposed to conducting wire with a lumped resistor in it, where all the voltage drop is across the resistor.

However Mabilde DOES measure zero in his flux-influenced setup, as the resistor drop is canceled by the flux emf through his leads at all points. Now the scalar potential, the mathematical concept he's trying to measure, IS essentially zero between any two points as static charge does not build up anywhere. But he can't measure it directly so he uses the flux through his loop to give him the same thing.

So what should the 'path voltage' drop (as in V = IR) be? First setup your voltmeter correctly. Best to put it outside the wire circle in such a way that no flux enters your measurement loop. Put your leads around say, a 90 degree section, and you'll measure 1/4 of the total emf. This represents the physical energy drop per unit charge along this shorter, direct path. If you use P = IV this measurement will correspond to the real, measurable power dissipation (heat) in this quarter section of the ring.

But why doesn't this give you the 3/4 voltage drop instead? How does the voltmeter 'know' which path you want to measure? Because THIS loop is flux-free. If you analyze the path along the long way, going through the 3/4 loop section, it's obvious that this loop has the ENTIRE flux/emf passing through the loop, which affects the measurements accordingly. Working out the math, you'll see that the 3/4 voltage drop, MINUS the emf induced by the flux (and a sign flip accounting for lead placement), gives you the same 1/4 drop:

-(3/4 emf -1 emf) = 1/4 emf

Both analysis paths are completely consistent. If you want to measure the 3/4 drop directly, just cross your voltmeter on the other side of the flux. Now you have a flux-free loop with the leads around the 3/4 section, and you'll measure 3/4 of the total emf, which corresponds to the V = IR in this section.

This is the non-0V, paradox-free solution I promised you. Voltmeters only measure the section of the ring corresponding to a flux-fee loop. If you can understand this thought experiment, you will understand Lewin's circuit completely, as it's pretty much the same thing. Faraday's law always wins.

 

[Averagesupernova]

I certainly do NOT agree that the correct measured voltage (meaning path voltage, IR drop, etc) is zero between any two points, as the emf is now evenly distributed across the ring.

You seem to contradict yourself. You say in one place you cannot have a voltage across a wire. But in the above quote....
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You cannot have a voltage measured across the ring anywhere as long as the conductivity all the way around is the same. The voltage is lost across the internal resistance of the source and the source is the ring. The source and the load occupy the same space. The double A batteries in a loop with resistors example I gave above is not meant to be just some puzzle I use to make me look smart. It is supposed to illustrate that the same thing is happening with AC induced into the ring.

 

[tedward]

Ok - first, there's no contradiction. A conducting wire loop with one lumped resistor in it concentrates all the E-field (and hence voltage drop) in the resistor, so there's no field (and hence no voltage drop), anywhere else. The resistance in the wire is effectively zero by comparison with the lumped resistor. Just like a voltage divider with a 1-ohm and a 1k resistor in series - the 1k resistor hogs all the voltage. That's what I meant by no voltage drop in conducting wire, as in Mabilde's setup.

If you ONLY have conducting wire, and no lumped resistors (like your 'shorted wire') you essentially have a uniform, non-zero resistance across the wire. So now all the voltage drop is distributed evenly, with every section dissipating energy at the same rate. It doesn't matter if it's conductive material or highly resistive material all the way around - it's disitributed evenly. Imagine a string of 1-ohm resistors - they're small but they all share the voltage equally.

 

[tedward]

You cannot have a voltage measured across the ring anywhere as long as the conductivity all the way around is the same. The voltage is lost across the internal resistance of the source and the source is the ring. The source and the load occupy the same space. The double A batteries in a loop with resistors example I gave above is not meant to be just some puzzle I use to make me look smart. It is supposed to illustrate that the same thing is happening with AC induced into the ring.

Again, your model of alternating batteries and resistors is a great thought experiment for understanding the DIFFERENCE between induced voltage and the DC/battery variety. You HAVE to think about electric field. In a resistor, electric field points from high potential to low potential, in the direction of current. In the standard model of a battery, electric field points from high potential to low potential INSIDE the battery, which is OPPOSITE the direction of current. So in a simple circuit with a battery and few resistors in series, the sum of electric field (read sum of voltage drops) is zero.

V_battery + sum(V_resistors) = 0

where the voltage across the battery is positive and the resistors are negative. In terms of electric field, integrating the E-field in the direction of current:

Int(E_battery) + Int(E_resistors) = 0

where the battery E-field is counted as negative, and the E-field in any resistor is positive.

The same exact thing happens in your alternating battery/resistor circuit, but it's spread out - each battery's E-field cancels with the E-field of the resistor right next to it - they're in opposite directions. Put a voltmeter between any two points in the circuit and the voltage is effectively zero, +/- one tiny battery.

Now with an INDUCED emf, the situation is very, very different. THERE ARE NO BATTERIES!! All you have are resistors. Energy is supplied from the outside of the system, via the changing magnetic flux. The e-field is only in one direction, the direction of current. The 'scalar potential' is certainly zero, because there is no charge build up at any single point. But the voltage drop, or path voltage, in any section depends on that fraction of the loop. You'll get something like this:

V_AB = (Length_AB / Length_circuit)*emf.

You can measure this voltage with a voltmeter, provided the measurement loop is flux-free.
Of course this also means the sum of the total voltage around the loop equals the induced emf:

Int(E.dl) = -d(phi)/dt.

This right here is Faraday's law, in integral form. It says it all right there - the integral of the loop's E-field is non-zero, meaning it is NON-CONSERVATIVE. This means, if we use path voltage as our convention:

Sum(V) around loop = -d(phi)/dt

Lewin shouldn't even have to defend his case, because Faraday says it for him. That's what Faraday's law means!

So now you have a mental model for your shorted (or uniform-resistance) ring: start with your battery / resistor ring, which sums to zero, and just delete all the batteries.

 

[Averagesupernova]

Imagine a string of 1-ohm resistors - they're small but they all share the voltage equally.

But they can't in a closed loop being driven inductively by a solenoid as in the @mabilde experiment. They will heat but you will never measure a voltage across them. Yes it seems impossible but its true. You may measure a little since individual wires that connect them are not the same as a resistor. But if you use resistor wire you will never measure anything. If you do it is due to voltmeter lead routing.
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Maybe an experiment could be devised using resistor wire. Measure the voltage all you want from any point to any other. By watching the heat with an infrared camera you should be able to tell if the voltage you measure is real. I guarantee all of the resistor wire will be heating evenly and that implies any voltage you measure at all can't be real because unless you are measuring 180 apart on the ring, one side has to be heating more than the other if the voltage measured is real.

 

[Averagesupernova]

@tedward I think we may be talking past each other up until the lead placement of the meter. You waste a great deal of time explaining things in a very roundabout way that I understand and have for many years.

 

[tedward]

If I over explain things you already understand, I apologize, but it's only because I need to state basics to make sure we're on the same page. Otherwise I assume we're disagreeing on fundamentals.

 

[tedward]

Maybe an experiment could be devised using resistor wire. Measure the voltage all you want from any point to any other. By watching the heat with an infrared camera you should be able to tell if the voltage you measure is real. I guarantee all of the resistor wire will be heating evenly and that implies any voltage you measure at all can't be real because unless you are measuring 180 apart on the ring, one side has to be heating more than the other if the voltage measured is real.

Cool experiment. Use resistor wire, put a solenoid in the center. Set up your infrared camera / heat sensor and calculate the power dissipation in any section, and calculate the honest-to-goodness voltage from
P = IV.
Take a voltmeter, measure any section you like of any size. But measure the section from the outside of the ring, so NO FLUX penetrates your loop (unlike Mabilde who intentionally does the precise opposite). I guarantee that you will measure the same voltage that you measure from heat loss.

 

[Averagesupernova]

I guarantee that you will measure the same voltage that you measure from heat loss.

You can't. The positive and negative field occupy the same space. They cancel just like they do in the case of a AA battery next to a resistor. There is still a current but the field around the section you are trying to measure is canceled by the field of the opposite polarity caused by the induction.
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Imagine hooking many voltmeters around the outside. They would all have to add up and by the time you get to the last part of the ring you would have to have all the voltage across that space. You won't. If you get readings on all those voltmeters it is currents induced into the leads in that case.
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The basis of our disagreement is the placement of the leads. I happen to believe the opposite of what you do concerning this.

 

[tedward]

You can't. The positive and negative field occupy the same space. They cancel just like they do in the case of a AA battery next to a resistor. There is still a current but the field around the section you are trying to measure is canceled by the field of the opposite polarity caused by the induction.
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Imagine hooking many voltmeters around the outside. They would all have to add up and by the time you get to the last part of the ring you would have to have all the voltage across that space. You won't. If you get readings on all those voltmeters it is currents induced into the leads in that case.
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The basis of our disagreement is the placement of the leads. I happen to believe the opposite of what you do concerning this.

Try this - let's start with where we absolutely agree. We see, in the video, mabilde measure a sector of a resistor ring, with voltmeter lead wires pointing radially outward from the center, directly and symmetrically over the solenoid in a pie-slice shape. He measures zero volts - right? Right.

I claim that, by careful and purposeful arrangement, he's effectively measuring the scalar potential, which SHOULD be zero in this situation, by indirect means (i.e. measuring the flux and subtracting from the true voltage drop so it cancels out). But whatever we call this measurement, it is definitely NOT the real voltage-drop value we're looking for, corresponding to heat dissipation, which of course has to be non-zero, and theoretically measurable by other means for comparison (or we could simply calculate this voltage from V = IR for this section of the resistance).

Let's pretend this voltmeter is tiny, located at the center, and the leads are flexible and are easily movable. Keeping the leads attached to the same points in the resistor loop, move the voltmeter to a position outside the circle. The shape or area formed by the leads no longer matters. This is because (can we agree?) there is no flux passing through this new measurement loop, outside the solenoid.

Now in this new position, what will we measure? If you claim it's STILL zero, you are in fact claiming that his measurement loop, (which we all agree is part of a circuit and subject to the same physical rules as everything else), is completely UNAFFECTED by flux passing through it. In other words, with flux in the loop, we measure zero. Without flux in the loop, we still measure zero because the flux has no effect. If you believe that to be true, than you are tossing Faraday's law in the trash. If that's true, induction is not possible in the first place.

I claim that Faraday's law works in all cases. The difference in flux will show up in the measurements. With no more flux/emf to cancel out true voltage drop, we will get an accurate reading of what we're looking for - the voltage corresponding to the power dissipation in that section of the ring, or the V from V = IR. There is no paradox. The complimentary section of the ring cannot be measured directly in this position, because the measurement loop corresponding to this section DOES have the flux through it. If you want to measure THAT section, move the voltmeter to the other side, CROSSING the area of the solenoid, while keeping the leads attached. This new position is topologically distinct from the first position. With this new flux-free loop, we will measure the true voltage of the complimentary section.

 

[Averagesupernova]

@tedward our fundamental disagreement is the placement of voltmeter leads. I've tried to explain my position. I say the arrangement of the leads in the @mabilde video is the only position for the leads to not be influenced by the field and you disagree. I've given up convincing you of this for now even though I still stand with @mabilde.
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I've tried to explain my position from a different angle and that is the shorted ring vs a ring with one or several discrete resistors. Your claim is that a shorted ring with evenly distributed resistance such as a wire will develop voltage across part of it due to induced current through it. My claim is that it will not. I'm not sure if you understand the reason for this or not. I have tried explaining it.
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So, let's take the AA battery scenario one more time. Assuming identical batteries, wire them all up in series in a loop. No resistors. Yep, short circuit. Measure the voltage anywhere you like. You will get zero. They will heat. I watched several Lewin videos lately and he actually short circuited batteries and yes, the one he held shorted for long enough to verify the heating. But, there is no voltage across it. And there won't be any voltage across any of them in a multi battery loop. The voltage is dropped across the internal resistance of each battery. This isn't some little resistor we can get at by cracking open the case of any of the AAs. But it is real and will heat. The DC setup I have described is immune to lead placement errors. This is why I always come back to it. The AC setup will act the same but it is NOT immune to voltmeter lead placement. So my position is any reading you get when measuring such a ring with distributed resistance has to be false. This false reading is what is causing ALL of the discrepancy in the Lewin experiment. Distributed resistance or not, the lead placement is throwing the readings.
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Faraday is not wrong, nor is Kirchoff. You believe that I am tossing Faraday away saying it is wrong when I am not. I am saying it is misapplied. I believe that Lewin and you are tossing Kirchoff away saying it is wrong while it is not. You are misapplying it.

 

[hutchphd]

You believe that I am tossing Faraday away saying it is wrong when I am not. I am saying it is misapplied. I believe that Lewin and you are tossing Kirchoff away saying it is wrong while it is not. You are misapplying it.

SHORT AND SWEET

1. Ohms "Law" is remarkably useful approximation for many but not all materials.
2. Faraday's Law is one of Maxwell's equations and is (so far as is known) always correct
3. Kirchhoffs Circuit "Laws" can be used with care for lumped circuit analysis. They are very useful within limits.

Attempts to legalize Kirchhof's circuit laws by redefining Maxwell Fields and Potentials are misguided and often foolishly complicated as witnessed herein.

Interestingly Kirchhoff's approximations in scattering theory seem to provoke similarly disparate opinions as to applicability.

Can we be finished?

 

[Averagesupernova]

Can we be finished?

Lol. I've been 'finished' in my ideas on this for a few years. I would like to find someplace on the net if not here, someone who has taken Lewin's side and had it explained to them and realized the problems with it. All I ever find is us vs them. No one ever says: 'Oh NOW I get it. Yes, I can now see it. It certainly has been a lesson for me!" I never see that, maybe I need to look harder.

 

[tedward]

I promise I will address the lead placement 'error' in a different reply to keep these from getting too long. And your shorted battery loop example, as you describe it, makes perfect sense to me. No disagreement. But for now,

LET'S TALK ABOUT ENERGY

I'm sure you know this but I will spell it out so we can make comparisons. A battery is a device that stores chemical energy. This energy comes from ... chemical reactions ... (my chem is not great). But I do know that this stored energy does work to separate charge, from neutral to a split positive / negative pair at the terminals. That's about as far as I need to understand the internal workings. So the battery does the work of separating charge, depositing this charge it's terminals. It now acts like a capacitor, but one whose charge, and therefore potential difference never changes. I think of this as a replenishing capacitor. It's important to remember the electric field INSIDE the battery, just like a capacitor, points from positive to negative - this negative field is a record of the work the battery had to do to separate the charge.

The battery has now converted chemical energy to electrical energy, or emf (voltage). The emf of the battery is the sum of the electric field throughout the rest of circuit, from positive terminal to negative terminal, in the direction of current. Attached to a resistive wire, this emf is delivered to the wire uniformly, with uniform electric field everywhere. If the resistors are lumped, charge distributions form at the resistor ends which redistribute the electric field, concentrating the field ONLY IN THE RESISTORS and leaving conducting wire with no field. The electric field in the resistors represent a record of the energy they receive from the battery. This energy is then dissipated, but we don't count this twice - we only count what the resistors receive. So battery does the hard of work of mining chemical energy, and gives this energy to the freeloading resistors to spend how they like. The zero-sum electric field is a record of this internal transaction. And the transaction will continue for as long as the battery has stored energy.

So now we have good old fashioned Kirchoff's law, in both voltage and field form:

Sum(V) around loop = 0

Int(E.dl) = 0

Here energy was both created and spent WITHIN the loop. Again it doesn't matter if we use lumped or distributed resistance.

Now let's talk about the induced emf case. The changing flux of the solenoid CREATES an electric field throughout space, which interacts with the free charge in the circuit. The integral of this field in the circuit loop is the emf. Just like the battery, the electric field is concentrated only in the resistors, or uniformly spread in the case of resistive wire. How does this field simply arise from nothing? The same way you can turn on your TV with a remote - it's an electromagnetic wave that propagates from the solenoid radially outward until it hits something, traveling at the speed of light. How does it deliver energy? The same way you feel warm sunlight coming from a source 93 million miles away. The field contains energy and can transfer it from one place to another. From the point of view of the loop, this is FREE ENERGY.

The electric field in the resistors still point in the direction of current, but there is no record of any work done (like a battery). There is no negative field to cancel out the positive. This loop must worship the solenoid like a god, as all it does is give out free energy. It never has to do any work, and its balance sheet (the electric field, or some of voltage drops) only shows income. Where does the energy REALLY come from? Lot's of sources, but imagine some guy pushing a bar magnet back and forth through the solenoid. He feels a force - not just the inertial force but a magnetic force on his bar, so he's doing work the whole time until he gets tired and runs out of his stored energy. He has replaced the battery, and he is certainly not part of the loop. So conservation of energy works out, but now we need to account for it differently:

Int(E.dl) = -d(phi)/dt

or Sum(V) around loop = -d(phi)/dt.

Another way to think of Faraday's law:

Sum(V) = work done by the guy pushing the magnet, per coloumb.

 

[tedward]

Lol. I've been 'finished' in my ideas on this for a few years. I would like to find someplace on the net if not here, someone who has taken Lewin's side and had it explained to them and realized the problems with it. All I ever find is us vs them. No one ever says: 'Oh NOW I get it. Yes, I can now see it. It certainly has been a lesson for me!" I never see that, maybe I need to look harder.

I mean, I could say the same exact thing from the other side. We're both confident, and one of us is certainly wrong. I am trying to understand your arguments and 'get inside your head' as it were, as I find the discussion useful even if we don't agree (I've really learned from thinking about this intently). So if we don't agree, no problem. See what you think of my energy argument :)

 

[Averagesupernova]

@tedward you can't say it applies for DC and not for AC. Source resistance works always. If you think it cannot apply then you are essentially saying that for AC circuits the field polarity doesn't add up to zero around the circuit. You have to know that can't be true. Transformer secondaries have a real source resistance that is the resistance of the copper in the winding. They output a voltage, it's placed across the load, those voltages cancel to zero around the loop.
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Take a transformer with a center tapped secondary and put it in parallel with another identical transformer but hook them so they fight each other. Now you have two secondaries fighting each other and four nodes you can hook voltmeter leads to. All but a very small fraction of the voltage will be lost in the copper resistance in the secondary windings. You will not measure very much no matter where you stick the probes. It's no different than the single turn winding previously discussed.