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• mabilde
Gold Member
The only law that must be obeyed in physics territory is Maxwell. Period.

(And tax Law)
All laws must be obeyed within the constraints of their definitions. This is of course drifting into territory that is a different thread than this one here on PF by (some) different posters. But the point still stands.

alan123hk
The only law that must be obeyed in physics territory is Maxwell. Period.
(And tax Law)
Not necessarily the case, tax laws are often violated, so you know from the news, if it is not legal tax avoidance, those people will be convicted and punished.

tedward
The only law that must be obeyed in physics territory is Maxwell. Period.

(And tax Law)
Relative to the ground is not relevant. Coil relative to the pie shaped is relevant.

Yes the flux lines are precisely parallel with bore of the solenoid, or as you call them, straight up. But, as the field grows and shrinks with the current in the solenoid the vertical flux lines move in and out from the exact center of the bore. This does NOT cause the flux lines to cut across the short wires that make up the pie shape at an angle that is non parallel to the to those short pie shaped conductors. The flux lines simply travel parallel with those conductors. This does NOT cause a current to flow.
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Think of the flux lines as actual strings or wires. The only way current is induced is if the flux 'strings' are to move in such a way as to tangle with the wire. Sliding parallel like a violin bow along the strings is a good example to compare to. The bow moving across the strings or the strings sliding across the bow will get a sound out of a violin but the same comparison with induction will not yield a current.

Well, naturally that is reflected by the voltage generated but I'm not sure this is how you meant it.
Your description of the changing flux tells me that you have don't have a solid understanding of Stoke's theorem and/or Faraday's law, and probably have a bad mental picture of the magnetic flux itself. The magnetic field, B, generated by the current in a solenoid, whether constant or changing, points along the axis of the solenoid (vertically in this case), and is spread uniformly throughout the interior, just like the velocity field of water flowing through a pipe. If the current oscillates, so will the strength of the B-field, but the field lines themselves don't move - they just oscillate their direction, positive or negative, or in this case up and down. The B-field does not point radially outward until you leave the solenoid (the end of the pipe so to speak), where it bends outward around the top and completes the loop outside and back through the bottom, though it's much weaker here.

Check out @mabilde's field diagram at 7:19 in the video for a picture, or google image search magnetic field of a solenoid. That uniform field in the center changes strength and sign only, not location. Since the field is perpendicular to the circular area of the single loop, the net magnetic flux through it just the the field strength times the area, and the emf is the rate-of-change of this quantity.

Stokes' Theorem says that the line integral of a field along the boundary of a surface is mathematically equal to the integral of the curl of the field over the area of the surface. Maxwell's equations says the curl of the electric field is -dB/dt. Put them together and you have Faraday's law - the circulation of electric field along a loop is equal to the rate of change of the magnetic flux through the surface bounded by the loop. The magnetic field doesn't even have to touch the loop at any point - the E-field created by the changing flux spreads through out space, circling the solenoid. The emf is the line integral of this e-field.

Mabilde's voltmeter leads, together with the particular section of the single loop he's measuring, forms a sector - an imaginary pie-wedge shaped surface, just like in a pie chart. When a changing magnetic flux penetrates this surface, it creates an E-field along the this measurement loop itself, and the integral of this e-field around the loop is the induced emf. This quantity is the only thing he's measuring apart from the drops in the resistors. There is no 'voltage drop' in the wire section of the loop he's measuring as it's a conductor. When there's no resistor in his measurement loop, all he has is a flux measuring device. He gets great results too (his setup is marvelous) - very clean numbers predicted exactly by Faraday's law. He just completely misinterprets what he's reading.

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Gold Member
There is no 'voltage drop' in the wire section of the loop he's measuring as it's a conductor.
I've thought of this many times and it's intuitive for there not to be. Take a look at a thermocouple. There is a voltage based on the temperature differential between the open and closed ends and it is most certainly a conductor. I know it seems like I'm trying to cloud the issue. All I want to point out with this is that just because something is a conductor doesn't mean it cannot have a voltage across it.

tedward
So I had to review thermocouples, but if I understand the basic principle, the Seebeck effect describes how a temperature difference provides an emf in some conducting material. Connect two different materials that have a different material properties (voltage to temperature gradient), and you get a measurable voltage at the ends. The idea (I believe) is similar to an induced emf, but now it's temperature differences that create the emf. But the response of free charge is the same - charges respond to the temperature-induced emf by creating a very brief current, until they pile up at the two ends - positive charge on one end and negative on the other.

Once the system has reached equilibrium (on extremely small timescales), current stops flowing. The charge distribution creates its own static field in the opposing direction, which cancels out the temperature-induced field. So now you have a potential difference /voltage between the ends, and zero net field inside the conductor. This has to be the case - if there was any lingering NET E-field, free charge would flow until was canceled out, that's just how conductors work. The same thing happens in conductors where there is an induced emf - charge piles up at the resistors, or at the ends of an open circuit, creating a static potential that cancels out the induced field. This is just like two unconnected wires sticking out the ends of the terminals of a DC battery. You can measure the potential difference between the ends, but there's no NET e-field in the wires themselves.

Now, there ARE exceptions to this rule. Commonly cited are induced fields with very high frequencies (like terahertz) or extremely large circuits like transmission lines, where the emf changes too fast for the charge to keep and you can have temporary fields that don't equalize. But a better exception which illustrates the point is this: Take a conducting wire, - not a circuit, just a rod of copper say. Position the rod perpendicular to a magnetic field. Say the field is pointing vertically upward, and the rod is horizontal. Now move the rod perpendicular to its length and to the field at some velocity. The magnetic force (the qv x B component of the Lorentz force) pushes positive charge to one side and negative to the other. Again, the charge builds up at the ends until an equilibrium is reached, with a measurable voltage at the ends, and an opposing electrostatic field is created by the charges. But the here's the cool part - the magnetic force is just that, a force.

It's not from an electric field (at least in the stationary frame of the original magnetic field we're moving in). So the electrostatic field from the charges exists, and acts to cancel out the net FORCE on any free charge. Basically the electric force on a charge and the magnetic force cancel eachother out, but the net field is still present. So you have a conductor with a sustained non-zero electric field, but the net force on any free charge is zero. This can only happen if a conductor is moving in a B-field, which isn't happening in any of our examples. But it does hammer in the point that you can't have net FORCE on a charge in a conductor once it's reached equilibrium. And for stationary circuits, this means you cannot have a net FIELD in the conductor. (Feynman discusses this in his E/M lectures).

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alan123hk
What I meant by saying the scalar potential ignores the induced emf, is that this definition of voltage focuses on the scalar potential and disregards the induced emf as a voltage, even though yes, the scalar potential is influenced by the induced field. Let's imagine this experiment: A good diagram would be to think of a circuit as a clock, with single resistor in a circular loop with a magnetic flux at the center, so the loop IS the transformer secondary, similar to the Lewin circuit. We place the tiny resistor at the top of the loop, at the 12 o'clock position. Say the emf is 12V, and the resistor is 12 ohms, giving a current of 1A. You draw the net E-field (induced plus electrostatic) along the circuit. It's zero everywhere except for inside the resistor (which is negligible length for now), and it points to the right. This field is large, and when multiplied by the length of the resistor gives 12V (same as the emf).
You then separate the diagram into two components - one for the induced field only, one for the electrostatic field. From symmetry, the induced field diagram is uniform around the clock, pointing clockwise, just as if it would be in free space, from 0 (midnight) to 12 noon. Now the static field diagram must cancel out the induced field diagram everywhere but in the resistor, so this diagram is also uniformly spread in the COUNTER-clockwise direction - starting at 12 noon and going back to 0, or midnight. The other difference about the static field is that it now just in the resistor, it points to the right, or clockwise. Now since the static field sums up to zero around the clock, we can define a number that we'll call scalar potential. We disregard the induced field diagram and only focus on the potentials on the static field diagram. Taking midnight to be ground (0V), the scalar potential increases in the clockwise direction, against the static field. One hour is one volt, so going clockwise to noon gives you your 12V. But in the tiny space between noon and midnight (say this is one-second on the clock), the potential suddenly drops back down to zero.
Now overlay the actual clock circuit over the diagram, and write down the potential values around the diagram (same as the clock numbers in volts). Pick any two points, say 3:00 and 5:00. The scalar potential predicts a difference of 2V. Hook up a voltmeter to these points, and you measure zero. This is because the voltmeter only measures integral of NET E-field, it has no way of discerning static from induced. Now, pick up a small test charge and place it in the loop (this part is just a thought experiment of course). As you move this charge between 3:00 and 5:00, you will find it takes no work to move it in either direction, no matter the size of the charge, even though the scalar potential difference is 2V. Now move the test charge from one side of the resistor to the other, through the resistor. If you go left from just after midnight to noon, it will take you 12V times your charge worth of work, as the scalar potential predicts. (If you go the other direction, the field pushes your charge for you and the work you do is -12V * q.) But now push your charge from say 2:00 to 10:00 through the resistor. The scalar potential difference is 8V. But the work you do is still 12V*q, just as when you moved it from midnight to noon. What we're seeing is that you cannot do work calculations using the scalar potential only. This is what I mean when I say that scalar potential is unrelated to the work done on a charge. Without including the induced field, the notion of potential difference as work - something that is foundational for thinking about voltage - no longer holds. The best you can say, as you mentioned, is that the TOTAL of the scalar potential in the solenoid is equal to the amount of work done on the load. The individual values along the path are meaningless. (Edit: see below)

What about Ohm's Law? As long as the resistor is negligible length, there's no problem. But now say our 12 ohm resistor extends uniformly from 6:00 to 12 noon clockwise, so on the left side of the clock. Emf and current are still the same, 12V and 1A respectively. So by Ohm's law, the Int(E.dl) voltage (I'll call this path voltage) across this resistor is 12V. But considering the induced emf ONLY, by symmetry, this must be 6V, spread uniformly. Therefore, the electrostatic potential across the resistor is also 6V, as they add up to 12V. But the scalar potential, 6V, doesn't give you IR which is 12v. This is what I mean when I say scalar potential doesn't conform to Ohm's law. You can still apply Ohm's law, but not without including the induced field, per the equation you linked to.

It seems to me that using this definition of voltage is bending over backwards to satisfy a single requirement: "Sum of voltage around a loop is always zero." But what are we giving up? You have to redefine Ohm's law to include a separate term, the integral of the induced field. You give up the notion of potential difference as work, unless you include that same term. You can't measure it with a voltmeter. And it does nothing to change the physics - the fact that the net E-field is still non-conservative and therefore path dependent. On top of that, imagine a circuit that has no symmetry, just a messy loop with an irregular field. Trying to calculate the static potential at every point means you have to somehow integrate the induced field up to any point. It's a damn near impossible task except for the most symmetric circuit. And the minute you do work out all the scalar potentials, as soon you move wires around everything is different. The net E-field voltage doesn't have this problem.
It's far more useful to define 'path voltage' as the integral of E.dl, as it is the same voltage that V=IR uses, is measurable by a voltmeter, and naturally corresponds to work done on a test-charge along a given path. Yes, you now have to give up path independence and the idea of a fixed potential, but that's a direct consequence of the Maxwell-Faraday equation, and should not be surprising.
(In your transformer diagram, all choosing different taps does is change the amount of emf in your circuit, equivalent to increasing the number of windings or strength of d(phi)/dt.) EDIT: Ok I understand why you included this diagram, the scalar potential markings indicate the amount of emf you'll get. But that still only applies if you use a path between the terminal points that flux penetrates. If you add up the E.dl voltages along the wire itself, between ANY two points, or as in my clock circuit above, you get zero.
I think I see exactly what you mean. I really admire your insight. You now ask the most central and interesting question about this discussion. I actually started thinking about these things that confuse me a long time ago. I hope I can reply soon but I'm out shopping and lunch right now and when I get back I'll share some of my thoughts on these issues although I'm not too sure if I'm thinking right.

tedward
I think I see exactly what you mean. I really admire your insight. You now ask the most central and interesting question about this discussion. I actually started thinking about these things that confuse me a long time ago. I hope I can reply soon but I'm out shopping and lunch right now and when I get back I'll share some of my thoughts on these issues although I'm not too sure if I'm thinking right.

Thanks - for sure, I'm enjoying the conversation. Glad I joined this, it's good to be able to bounce ideas off people who know what they're talking about rather than arguing in YouTube comments haha.

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alan123hk
What we're seeing is that you cannot do work calculations using the scalar potential only. This is what I mean when I say that scalar potential is unrelated to the work done on a charge. Without including the induced field, the notion of potential difference as work - something that is foundational for thinking about voltage - no longer holds. The best you can say, as you mentioned, is that the TOTAL of the scalar potential in the solenoid is equal to the amount of work done on the load. The individual values along the path are meaningless. (Edit: see below)
You seem to describe the problem in a complicated way. I try to understand it simply. It is to divide a whole into two parts, the first part is the scalar potential and the scalar field, and the second part is the induction field. For example, when a positive test charge moves in the scalar field from a place of low potential to a place of high potential, it increases the energy of the scalar potential. You can do the calculations independently, since you already know that the induced electric field energizes this positive test charge.

Then you can imagine that the test charge actually moves in the induced electric field at the same time and in the same path. The magnitude of the induced electric field is the same as that of the scalar field, but in the opposite direction, so the net field in the conductor is zero. The direction of motion of the test charge is the same as that of the induced electric field, so we can deduce that the energy provided by the induced electric field is exactly equal to the energy added by the scalar potential energy.

I think the key here is whether you can convince yourself that when a test charge moves in a conductor with a net electric field of zero, it actually corresponds to the energy conversion process described above. If you are not satisfied or agree with the above reasoning process, you can try to analyze it with Poynting vector, then you will get a complete energy flow diagram.

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Homework Helper
2022 Award
I believe you. But what you do not seem to understand is:
1. This is not the ordinary and customary description
2. I believe most physicists find this "dual field" complication unnecessary and fraught
The pictures you carry around in your head are entirely up to you, but please realiize there may be a reason why they are not mainstream physics.

I must admit to being completely mystified by the negative response to Prof Lewin's lecture. I recall my perhaps incorrect inference from him that the most vehement attacks on the lecture demonstration were from outside the physics department. These included charges that he had used sleight of hand somehow!
So I guess I will remain mystified and give it up. Seems obvious to me.

DaveE
tedward
I believe you. But what you do not seem to understand is:
1. This is not the ordinary and customary description
2. I believe most physicists find this "dual field" complication unnecessary and fraught
The pictures you carry around in your head are entirely up to you, but please realiize there may be a reason why they are not mainstream physics.

I must admit to being completely mystified by the negative response to Prof Lewin's lecture. I recall my perhaps incorrect inference from him that the most vehement attacks on the lecture demonstration were from outside the physics department. These included charges that he had used sleight of hand somehow!
So I guess I will remain mystified and give it up. Seems obvious to me.
What's funny, is that when you hear people's reasons for assuming Kirchoff to always be true, they almost never mention scalar potential. Check out RSD Academy on youtube, who's done multiple videos on this problem arguing against Lewin. (Mabilde does the same analysis in one of his videos). He never (iirc) mentions scalar potential, or identifies two opposing electric fields, or conflicting conventions for defining voltage. If he did, he would recognize that this is just a difference in conventions rather than saying Lewin was just flat out wrong. If you follow his analysis, he treats voltage the way we usually think about it - the total drop across a resistor or IR, integral of net electric field, etc.

He simply inserts the emf in the circuit, attributing it to different sections of the wire, without calling it a field. Then he shows that emf - sum(IR) = 0. Which is of course true as a conservation of energy equation, but not as a loop equation: sum(IR) = emf. So he's not considering scalar potential, he just gets the physics wrong. Then he points to McDonald's paper and says, "see, I'm right, there's a potential difference in the wire that you just can't measure". It's laughable.

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hutchphd
Gold Member
Sigh...
There is no doubt that Walter Lewin is an excellent teacher. As I said in my first post in this thread, I don't understand the controversy.
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It seems to me a number of folks are not following rules. Firstly is Walter Lewin. Kirchoff holds. The resistors form a series circuit as far as the test setup is concerned. Where the broken rule comes in is that Lewin is not accounting for the fact that the connecting wires are forming the secondary winding of a transformer. Actually two secondaries. One is the node A and the other node D. Or I should say the wires between the resistors where said nodes are. So technically, the schematic form of this setup is not matching the real world setup. It doesn't matter that "they're just wires", how they are functioning needs to be accounted for.
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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.
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All that being said, the @mabilde setup somewhat resembles a directional coupler. Yes, that stance is a bit of a stretch. But instead of using a primary solenoid and the single loop as a secondary, we could replace one resistor with a signal generator and watch the scope in the @mabilde setup while flipping the test leads from one side to the other and notice the change. Directional couplers for RF are often laid out on a circuit board with very few if any components soldered to the board. They may not look like it but they essentially combine a voltage signal with a signal from a current transformer.
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There is nothing magical about this. Some folks are just misinterpreting the results or are not seeing the setup and schematic for what they are. So, in a way, Lewin did perform a sleight of hand. But I don't think he intended to, and a lot of folks have fallen for it. Truly a perfect storm. If the lecture had been presented in a different manner I wonder where things would be today.

tedward
Sigh...
There is no doubt that Walter Lewin is an excellent teacher. As I said in my first post in this thread, I don't understand the controversy.
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It seems to me a number of folks are not following rules. Firstly is Walter Lewin. Kirchoff holds. The resistors form a series circuit as far as the test setup is concerned. Where the broken rule comes in is that Lewin is not accounting for the fact that the connecting wires are forming the secondary winding of a transformer. Actually two secondaries. One is the node A and the other node D. Or I should say the wires between the resistors where said nodes are. So technically, the schematic form of this setup is not matching the real world setup. It doesn't matter that "they're just wires", how they are functioning needs to be accounted for.
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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.
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All that being said, the @mabilde setup somewhat resembles a directional coupler. Yes, that stance is a bit of a stretch. But instead of using a primary solenoid and the single loop as a secondary, we could replace one resistor with a signal generator and watch the scope in the @mabilde setup while flipping the test leads from one side to the other and notice the change. Directional couplers for RF are often laid out on a circuit board with very few if any components soldered to the board. They may not look like it but they essentially combine a voltage signal with a signal from a current transformer.
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There is nothing magical about this. Some folks are just misinterpreting the results or are not seeing the setup and schematic for they are. So, in a way, Lewin did perform a sleight of hand. But I don't think he intended to, and a lot of folks have fallen for it. Truly a perfect storm. If the lecture had been presented in a different manner I wonder where things would be today.
So I've thought about the "just model the wires as a transformer" argument a lot. Understanding exactly what this means gets to the heart of the debate, and no one seems to talk about it, though it's now quite clear to me. It depends on what we mean by the 'voltage' of a secondary, or in general an inductor. It all comes down to path difference.

In an transformer, you can measure the electric field along two paths - Path 1 is THROUGH the coiled wire itself, which we almost never do. Path 2 is the the nearly straight path connecting the terminals OUTSIDE the coil (as we usually do with a voltmeter). .For the first path, imagine taking a voltmeter and measuring points on the coil very close together, and add up the measurements as you reposition the voltmeter point by point: Vab + Vbc + Vcd... etc. all the way down, following the coil in a corkscrew manner.

These path measurements are different - why? Imagine the loop formed by these two paths - coiling down the wire (1), and straight up outside (2). This loop has a changing magnetic flux passing through the surface it bounds. As I mentioned in an earlier post, the surface is complicated, but it looks like a rotini pasta or a the film formed when a corkscrew is dipped in soapy water. You absolutely need to understand Stoke's theorem to follow this, but basically the rate of change of magnetic flux that penetrates this surface equals the integral of electric field along this whole loop (Faraday's law).

If there's a changing flux passing through, the total loop voltage is NON-ZERO. In fact the loop voltage is equal to the emf, -d(phi)/dt. If the loop sum is non-zero, then these two paths have to have different 'voltages'. The outside path measures the electric field on the outside, and therefore is a true voltage. It equals the emf of the transformer. The coiled path measures zero, as there is zero-field in the conductor. The total sum of the loop is the emf. This is path dependence, a consequence of the non-conservative induced field. If you put this transformer in a circuit with a single resistor, you can consider two loops - resistor + coil (Loop 1), or resistor + outside straight path (Loop 2). Loop 2 is how we usually analyze the circuit with Kirchoff's law. If we use Loop 1, we have to use Faraday's law, including the emf as a separate term.

We almost never observe this path dependence because who would ever bother taking those cumbersome voltmeter measurements on a tight coil? So how do we investigate it? We unwrap the transformer. We make it a single loop, and strengthen the flux to get a good emf. Now we can show the circuit in a plane. We can even put the flux in the center of the circuit, so the entire loop is the transformer. But we still have two paths/loops we can consider. Loop 1 is the resistor plus the (uncoiled) wire - it is essentially the path of the circuit itself under consideration. Loop 2 has the resistor, but leaves the wire, goes around the flux at the center to avoid it (through free space), and returns to the wire at some point.

You can analyze Loop 2 with Kirchoff's laws as there is no flux passing through it. This loop treats the circuit itself as a transformer and gives it a 'voltage', so the loop sums to zero. It's the part of the path through free space that has the electric field associated with the emf, NOT the conducting wire. That's 'modeling the wires as a transformer'. But Loop 1 does not sum up to zero, it only has a resistor. It DOES have the flux passing through it, which is accounted for with Faraday's law. Two DIFFERENT loops, two DIFFERENT loop measurements, SAME final accounting:

Vr = emf (Loop 1, Faraday's law)

or

Vr - emf = 0 (Loop 2, Kirchoff's law)

Now attach voltmeters to either side - one across the single resistor on the left, one between two points of the wire on the right. But connect them at the same two points on the circuit, just along different paths. The first one will measure the emf, as an IR drop across the resistor. The second will record zero, as there is no drop in a conducting wire. How do the meters know which path they're measuring? From which side of the flux they're on. Take the left meter, lift it up over the flux (over the top of the solenoid) to the other side, and you will see the number gradually change as the voltmeter lead crosses over the flux area, as we are changing the flux through the measurement path, until it becomes Path 1. This is exactly what @mabilde shows in his video, rather beautifully.

So we see that "modeling the loop as a transformer" really means "change the path of analysis so it doesn't include the flux". That defeats the purpose of understanding the underlying physics of non-conservative fields. Lewin has deliberately, intentionally deconstructed the transformer to understand what is really going on. Wrapping it up again and modeling it as a coil avoids the ambiguity but simply hides the physics. As you said, there is nothing magical about this.

(I have some diagrams for this, I'll put them in another post. My intention is to make some videos on this as well that makes this really clear).

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alan123hk
I have to mention again that this treatment of describing the electric fields generated by the charges on the surface of conductors in an electromagnetic induction system, and further analyzing the interaction of the electric fields generated by these charges with the induced electric fields, is certainly not my own idea or invention. I found a lot of this material on the Internet, Dr. Lewin's lectures on his circuit paradoxes were also one of my study objects.

Note that Ei is the induced electric field and Ech is the electric field generated by the surface charge, Q+ Q- are positive and negative charges respectively

tedward
I have to mention again that this treatment of describing the electric fields generated by the charges on the surface of conductors in an electromagnetic induction system, and further analyzing the interaction of the electric fields generated by these charges with the induced electric fields, is certainly not my own idea or invention. I found a lot of this material on the Internet, Dr. Lewin's lectures on his circuit paradoxes were also one of my study objects.

View attachment 321873
Note that Ei is the induced electric field and Ech is the electric field generated by the surface charge, Q+ Q- are positive and negative charges respectively
What's the timestamp on this video? I want to understand your thinking here.

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Gold Member
@tedward all this talk of electric fields seems to cloud what's really at hand. You might question how the subject can be discussed by ignoring this. I'll tell you how. The same way we do basic circuit analysis without ever worrying about the lowly electron. It simply isn't needed.
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The turns ratio of primary to secondary in a transformer does not fail, ever, as long as copper losses and loading is accounted for. So, yes, there is X number of volts dropped across each turn and to delve farther into that, X volts per unit of length within a turn. And yes, it's measurable. There is nothing to say there needs to be more than one turn total. I've gotten mixed signals from you concerning this. It's been my perception that your opinion is that turn per turn will never add up in a coil to the total voltage.
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In the Lewin and @mabilde setup, all we really need to worry about is that we have a secondary coil with two discrete resistances wired across it. We need not worry about fields until we start trying to measure voltages on portions of the coil so as to not introduce errors into the measurements. How this is done I've already covered.
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One last exercise for our brains:
Suppose we have a transformer that has two secondary coils wound in such a way so they supply the same voltage. Many transformers are configured like this. Hook them in series so they are additive. For sake of discussion let's say each secondary is putting out 10 volts. We don't know how many turns there are and we don't care. Put in series they would together put out 20 volts. Now take and put a resistor across the 20 volts output of the two secondaries in series. Next break the connection made between the two secondaries and insert another resistor. We now have the same setup as the Lewin experiment only we have more turns. Somehow all the readings in this setup are accepted and I assume it is because it is easy for people to visualize that there is a transformer. The way of thinking changes. Also people can be sloppy with voltmeter lead placement with no consequences in the measurements.

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alan123hk
What's the timestamp on this video? I want to understand your thinking here
For those interested, you can refer to the video link I added in #12 of this discussion thread. I personally think that watching it from 33 minutes will be the most exciting part, but this may not let you understand the whole interesting story.

Frankly, I don't really understand some of the debates raised here, such as what is mainstream and what isn't, leading students away from the most important subjects of study, etc. I am an engineering graduate and have been engaged in engineering for decades. I don't know anything about the academic world of physics, but I am very interested in physics. What I care about is whether the method used is reasonable, whether it will find the right answer, and most importantly, whether it will violate the laws of physics. If there is a simple and sound concept and method that can show correct results, why are some people so negative about discussing and using it?
This is just one of the options available, I don't think anyone is trying to lead others not to learn from the mainstream and the best.

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tedward
@tedward all this talk of electric fields seems to cloud what's really at hand. You might question how the subject can be discussed by ignoring this. I'll tell you how. The same way we do basic circuit analysis without ever worrying about the lowly electron. It simply isn't needed.
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The turns ratio of primary to secondary in a transformer does not fail, ever, as long as copper losses and loading is accounted for. So, yes, there is X number of volts dropped across each turn and to delve farther into that, X volts per unit of length within a turn. And yes, it's measurable. There is nothing to say there needs to be more than one turn total. I've gotten mixed signals from you concerning this. It's been my perception that your opinion is that turn per turn will never add up in a coil to the total voltage.
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In the Lewin and @mabilde setup, all we really need to worry about is that we have a secondary coil with two discrete resistances wired across it. We need not worry about fields until we start trying to measure voltages on portions of the coil so as to not introduce errors into the measurements. How this is done I've already covered.
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One last exercise for our brains:
Suppose we have a transformer that has two secondary coils wound in such a way so they supply the same voltage. Many transformers are configured like this. Hook them in series so they are additive. For sake of discussion let's say each secondary is putting out 10 volts. We don't know how many turns there are and we don't care. Put in series they would together put out 20 volts. Now take and put a resistor across the 20 volts output of the two secondaries in series. Next break the connection made between the two secondaries and insert another resistor. We now have the same setup as the Lewin experiment only we have more turns. Somehow all the readings in this setup are accepted and I assume it is because it is easy for people to visualize that there is a transformer. The way of thinking changes. Also people can be sloppy with lead placement with no consequences.
(warning: Long response - even for me). A lot of the talk about electric fields was to explain how there can be no field in a conducting wire, and hence no voltage between two points along a wire. This comes up especially when trying to talk about the 'scalar potential' argument, a voltage convention that DOES sum to zero around any loop, and is cited as a justification of Kirchoff's law always holding. But I don't think you're interested in this convention. We both agree (I hope) by what we mean by voltage. V = IR across a resistor, it's measured by a voltmeter, it's the work on a charge between two points, etc. But to understand how a transformer really works, you have to discuss electric field just a bit. It is, after all, central to Faraday's law, as is the magnetic field, and shouldn't be ignored. But I'll try to only reference the field when I absolutely have to.

Apologies if I sound patronizing, but just to make sure we're on the same page, electric field is the force per unit charge on a small test charge. It's a vector that points from a higher potential to a lower potential, i.e always in the direction of DECREASING potential. This is just like the gravitational field here on Earth (as in g = 9.81 N/kg), which points toward the ground, in the direction of decreasing potential energy. The connection between voltage / potential and electric field is that a potential difference is the integral, or sum, of an electric field over the length between two points. To keep things simple, if electric field is constant, then potential difference or voltage is just the electric field times the length of the path, so V = Ed. So if one point has a potential of 8V, and another at 5V, then there must be a path between them whose electric field points towards the 5v point, and whose sum over the length is 3V. The average field here is 3V divided by whatever the length is.

So for a DC battery - measure the terminals with a voltmeter, and you are measuring the sum of the electric field between those points along the path of the voltmeter leads. Move the wires around with the ends fixed, and nothing happens, as the path changes but the voltage doesn't. It's very cool when you think about it - the field could vary over space but the sum is always the same along any path.

Enter the transformer - say it has many turns, don't care how many. When you measure the terminals with a voltmeter, you are measuring the sum of the electric field between an imaginary line between those terminals, running OUTSIDE the coil, along the path of the voltmeter leads. This is what we refer to as the voltage of the transformer, or the output. You can define the voltage of an inductor the same way, but we don't care about self-inductance for now. What causes this voltage? The changing magnetic flux inside the coil volume, and Faraday's law. An induced electric field is created in a circular pattern, which pushes charge through the coils. The more coils, the more wire exposed to the induced electric field, and greater emf, or voltage measured at the terminals.

But here's the thing - if the transformer isn't hooked up to anything, the charge has nowhere to go and piles up at the open terminals (almost instantly), one positive and one negative. These accumulated charges generate their own field, pointing from positive to negative, in the OPPOSITE direction through the coils. The charge builds up at the ends exactly enough so that the induced field is canceled out in the conducting wire itself. It has to, otherwise more charge would flow until it did balance. This happens even if the transformer is hooked up to a load - the charge just builds up at the load resistor but current still flows, the same way a crowd forms when many people are trying to enter through the same narrow doorway. Either way, the electric field is zero IN the conducting coils. So if voltage is measured THROUGH this wire path, the sum should be zero. So we've got a conflict in measurements.

Now the way you measure this non-voltage very important. If you put the voltmeter on adjacent turns, you will measure one turn worth of emf. If you measure 5 turns apart, measuring on a path outside, you'll get 5 turns worth of emf, like you described. But to get an accurate measurement along this path, you would have to measure a small angle on the same turn - say a quarter turn, with the voltmeter outside. Your voltmeter should say zero. Measure 4 of these quarter turns in succesion, moving the leads as you go, and you'll still get zero. But as we said, measuring from point to the same position on the next turn, and you WILL get an emf. Why? Path dependence. You can't avoid it.

Think of a wire loop with a break in it. If you put your leads on opposite ends of the break, your measurement circuit, even with twisted leads, has a circular area for flux to pass through and you'll get an emf. But if you put the leads almost next to eachother on the same side of the break, no flux passes through your loop (remember the voltmeter and leads are outside the loop), so no emf. Adding up many of these small measurements around the circle, and you get zero. This happens on the whole coil as well. Measure top to bottom, you get the full emf, as the flux passes through every turn. Measure adjacent points on the same turn one at a time, in such a way that no flux gets through, and the sum of all the readings will be zero. It's an awkward and cumbersome way to measure the voltage (which is why you will rarely see anyone try it) but that's what we mean when we say the voltage of a path THROUGH the coils.

It's a lot easier to perform / visualize with only one loop, that doesn't cross itself, lying in a plane. It's basically the Omega shape - a circle with two connected feet that connect to the rest of the circuit. If there's a changing flux through the circular part, and you hook up a voltmeter underneath, connecting the feet, completing the circle, you will measure the emf. This is because your measurement path includes a full circle for flux to pass through. But now hold your voltmeter above the circular part, in the same plane, connected to the same two points. Now your measurement loop is a crescent shape, and it lies OUTSIDE the circle. No flux passes through this crescent, so you will measure no emf. That's path dependence. The difference in measured voltage is due to the fact that the Induced e-field is non-conservative: loops don't have to sum to zero, and different paths can give different measurements.

Lewin's circuit is basically the same idea, but now the entire circuit is the single loop transformer. Now we have the resistors as a load, whose total drop is equivalent to the induced emf. If you measure the wire path, making sure your voltmeter loops never allow any flux to pass through, You'll measure only the resistor's drop which is non-zero, and equal to the emf passing through the entire circuit. Now, draw a weird path that goes through the resistors around the loop, but doubles back around the flux in the middle , returns to the wire and completes this C-shaped loop, the voltages will add to zero, as this new section of the path subtracts out the emf from the flux. Essentially measuring the voltages across the left and right sides does the same thing - different paths, different voltages.

For your proposed circuit model with the two transformers, it works out to the same thing. Say one's on top and one's on the bottom. Let's replace them with single loop transformers, like the omega shape, and make the emf whatever you like. The top one is an omega, the bottom one is an upside down omega, and they each have changing magnetic flux passing through them. Any loop around the whole circuit that connects the feet of the omegas will measure the transformer voltages, and everything sums up to zero. This is because this loop does not include any of the flux. But measure a loop that follows the actual wire path, and you will measure the total emf. This is because the entire loop has flux through it, and therefore has a net emf. Again this is better to see in a diagram, I will try to post some when I get a chance.

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Gold Member
@tedward I read your last post and absorbed some of it. I also watched Lewin's video that @alan123hk linked to in post 12. I have many times considered the closed loop with resistor wire that was discussed. However, I never considered using two different wires. I think that complicates it. One wire in a single loop can illustrate it just as well. Turn the solenoid on and stick the probes at 90° apart. Using one voltmeter there is already a perceived paradox. Which path do you choose? The long or the short?
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Here's the answer: The only way there can be no paradox is if the voltage reading is zero. Guess what? Go to the video @mabilde linked to in the first post of this thread. This is exactly what happens at time 31:10 when the voltmeter leads are routed correctly. Admittedly this is crude substitute for resistor wire. So, it would seem that at 31:10 Kirchoff holds. Again.
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The Lewin video in post 12 shows what appears to be a very upset Walter Lewin. And I honestly feel bad for him. At the end of his video he talks about some electrical engineer somewhere who said the results of Lewin's experiments are caused by bad probing. I would say this is true. Again, because of the routing of the voltmeter leads. They are behaving exactly how I would expect. Why should they behave any differently than the other wires in the circuit that are basically routed the same way?

tedward
I think by the second wire, you mean the imaginary path I described. I didn't mean a wire - Kirchoff's and/or Faraday's laws work for any loop in space, it doesn't have to lie along a wire.

Your example with the resistor wire loop (uniform resistance) is a classic example. Does a voltmeter give you 1/4 of the total emf or 3/4 ? This exact problem used to trip me up back in college - 20 years before I ever heard of Walter Lewin :). I never got a good answer on how to a loop sum should work, or what a voltmeter should measure (and never tried it myself). Lewin's demo finally helped me figure it out. Ohm's law says that the 'voltage drop' across each section should give you the corresponding IR. Say the emf is 4V, total resistance is 4 ohms, and current is 1 amp. the short section SHOULD have 1V across, and the longer section 3V. This voltage drop is a real thing that corresponds to heat dissipated, so it should be measurable.

Place 2 identical voltmeters, connected at the same 90 degree points, where both voltmeters are placed OUTSIDE the circle. This is to ensure that NO MAGNETIC FLUX passes through the measurement loops, formed by the voltmeter leads and the section of the circle they're measuring. One should measure 1V, the other should measure 3V, as expected (I know, I know, @mabilde got zero, I'll get there). The question is, how do we tell which value a voltmeter will show?

Take the meter showing 1V. It's connected to the short resistor section, and this section along with the voltmeter leads bound an area that has no flux through it. Faraday's law says no flux, no induced emf. So the the thing we're measuring is just the resistor drop as intended. Now without moving the voltmeter at all, simply examine the OTHER loop that is formed by the same voltmeter, going around the long way. It consists of the same voltmeter leads, and the 3/4 length section. That section has -3V across it from the perspective of the lead placement (it would be positive if we flipped the leads). But this LOOP has the total magnetic flux passing through it!!

Faraday's law says that this flux will affect your measurements, by including a 4V emf. I believe it was you who asked if I thought a voltmeter shorted around a solenoid would record an emf? Of course it would! The same thing happens here. So instead of measuring 3V through the long resistor, you get -3V + 4V = 1V. So either way you measure it, using either loop, you get 1V. One is the short resistor voltage with no emf, the other is the longer resistor voltage WITH an emf.

Now do the same thing with the other voltmeter. If the colored leads are placed the same as the first voltmeter, it should read -3V. This is the no-flux loop, which is an accurate measurement. Now analyze the other path going all the way around the circle. It measures 1V - 4V = -3V. Again the difference in sign on the emf from before has to do with the lead placement. The point is either path you choose to analyze, you get the same result. If it didn't, we'd really have a problem.

So what happens if you take one voltmeter and move it to the other side? Well there's no way to move the voltmeter to the other side - while keeping the leads attached to the same points - without passing the voltmeter and its wires across the area where the flux is. As soon as you cross to the other side, you've essentially moved the flux from one loop to the other. So the no-flux loop as a different value. 1V becomes -3V, or the -3V becomes 1V if you move back. And it doesn't flip like a switch - if the solenoid as a large effective area, you will see the voltmeter measurement change gradually as it passes through the flux region, until the meter is completely on one side. It's almost like magic - but @mabilde does this beautifully early in his video. Of course, he cites this as the kind of "probe error" he accuses Lewin of [facepalm]. I really wish Lewin had done this as well, though you would need the AC version of the experiment rather than the pulse.

So what about @mabilde's readings on his 'continuous resistor'? It's actually a very good approximation to the uniform ring. Again, I think his experimental setup is fantastic. But of course he reads zero volts no matter where he places the leads, so you probably think all my analysis so far is complete bunk. But think about what he's doing - he has twisted voltmeter wires up until the center of the loop. So far, so good. Then his leads spread out in that pie-slice shape, connected to a section of the resistor ring. So in his measurement loop, he as a section of the ring (let's say it's 1/4 of the circle). His flux emf is 1V if I recall. So there should be a 0.25V drop in the resistors, 1/4 of the total emf. But what about the flux through his pie slice?? His leads are placed DIRECTLY OVER THE SOLENOID, in this wonderful precise setup he has. So he's getting exactly 1/4 of the emf passing through his measurement loop! It cancels out the voltage drop completely! This is why to get an accurate reading the voltmeters in my above experiment had to be outside, so they got not flux.

Compare this to just a bit earlier in the video, when he uses the same setup to measure the conducting ring. At one point he's measuring conducting wire with no resistors in between, so he SHOLD measure zero, as there are other lumped resistors elsewhere in the circuit. But he does measure something, and it changes as he moves his leads around the circle. But it has nothing to do with the conducting wire, there is no voltage drop or 'emf along the wire' as he claims. He's just measuring the flux through his pie-slices!! As I said in that other post, he created a very precise flux-meter. Talk about bad probing!

Now @mabilde is not an idiot (I think I read somewhere he teaches electrical engineering?). If you watch his whole video, there are parts where he seems to understand perfectly that flux through your loop adds/subtracts an emf to your measurement. So what gives? I think, as he cites the McDonald paper, that he's using the 'scalar potential' convention for voltage (which actually would line up with his results), but he never explicitly says so. Sometimes he's treating voltage like a scalar potential, sometimes in the usual way we think about it. But if so it's very disingenuous. I can't wrap my head around how he could completely ignore the flux through his loops and claim he's actually getting a real measurement.

As for Lewin, if someone with a Masters in EE - who obviously doesn't understand basic theory - claimed I had used bad probing, when Faraday's law predicts my results exactly, I'd be pretty pissy too. We can talk about the voltmeter wire mirroring the circuit thing another time, this is getting long again.

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Gold Member
@tedward you've got a lot for me to absorb in your last post at this time of the day. Getting sleepy. The second wire I refer to is, as I understand, two resistor wires not having the same resistance per foot in series on the loop. Why Lewin thinks he needs both of them to show there's a paradox (which there really isn't) I have no idea. He can use a single wire with X ohms per unit length. The perceived paradox will show up because there is the same current in the entire loop, which is true, and depending on which way you look, the voltmeter is across different resistances. But, nature doesn't like paradoxes. So as I said there's only one voltage that satisfies everyone and prevents the paradox. It's zero volts.
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Instead of using a loop with an induced current let's make one out of double A batteries all wired in series with resistors all of the same value also in the loop. Battery, resistor, battery, resistor, all the way around the loop. Pick as many as you like. 1,000 or 1,000,000, whatever. Through basic circuit analysis you will find that no matter where you probe with a DC voltmeter you will either measure 1.5 volts or zero. Not satisfied? Understandable. No problem. Change the battery voltage to much smaller. Add more to the loop to come up with the same open circuit voltage if you were to break the loop. Add resistors so you always have the same number of batteries and resistors. Now stick the probes anywhere you want. The most you will read is the voltage of a single battery. Continue this long enough and the voltmeter will read zero no matter what.
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Apparently Walter Lewin is unable to grasp this and I find it sad. The perceived paradox is there with batteries and resistors measuring DC as well. I suspect folks that read this post might actually build the double A resistor test setup as it's a lot less hassle than the inductive loop and no errors introduced onto the voltmeter leads. We'll talk more tomorrow.

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tedward
Happy to talk more tomorrow - but I'll leave you with this. Yes the single uniform resistor is easier to understand, and the 'paradox' occurs with any non-symmetric measurement. And you're right, nature doesn't like paradoxes. But there is a solution other than the 0V-at-any-point solution - read my last post once or twice.
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Your DC battery / resistor daisy-chain example is great! I remember thinking of the same thing at one time when I was first wrestling with this. And you're absolutely right, you will measure either a single battery or 0V, no matter where you put the leads, and in the small-battery limiting case it will be essentially zero between any two points. But that exact line of thinking will help you understand the difference between induced voltage and voltage from a source like a battery. DC or AC doesn't matter, you can replicate either scenario with both. It's all about whether the source is in the loop or from the outside. More on this later.

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tedward
You seem to describe the problem in a complicated way. I try to understand it simply. It is to divide a whole into two parts, the first part is the scalar potential and the scalar field, and the second part is the induction field. For example, when a positive test charge moves in the scalar field from a place of low potential to a place of high potential, it increases the energy of the scalar potential. You can do the calculations independently, since you already know that the induced electric field energizes this positive test charge.

Then you can imagine that the test charge actually moves in the induced electric field at the same time and in the same path. The magnitude of the induced electric field is the same as that of the scalar field, but in the opposite direction, so the net field in the conductor is zero. The direction of motion of the test charge is the same as that of the induced electric field, so we can deduce that the energy provided by the induced electric field is exactly equal to the energy added by the scalar potential energy.

I think the key here is whether you can convince yourself that when a test charge moves in a conductor with a net electric field of zero, it actually corresponds to the energy conversion process described above. If you are not satisfied or agree with the above reasoning process, you can try to analyze it with Poynting vector, then you will get a complete energy flow diagram.
I'm not 100% sure I get you, but it seems that you're saying the scalar potential increase really describes the energy added by the induced field. I'm not sure I agree with that. If you look at the combined field, the only place energy is added (non-zero e-field) is in the resistors. This energy is of course needed to overcome the resistance, so the kinetic energy of the charges don't increase. Kind of like pushing a box along a very smooth path - you don't need to push it with much force, just the bare minimum to keep it moving. But once you hit a very rough surface, like cement, you need to supply energy to get it past the rough part and overcome friction. So I don't see any energy interactions until the charge hits the resistor, where the energy supplied by the net field is dissipated as heat. I could see the total of the scalar field over a loop (neglecting the drop in the resistors) as being equivalent to the total of the induced field, but I don't see how you can tie that to certain positions along the path, where really the net field is zero.

Gold Member
@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.

alan123hk
I'm not 100% sure I get you, but it seems that you're saying the scalar potential increase really describes the energy added by the induced field. I'm not sure I agree with that. If you look at the combined field, the only place energy is added (non-zero e-field) is in the resistors.
Yes, because a resistor is connected in series, then the potential energy does not increase, and the energy supplied by the induced electric field is immediately lost on the resistor, but if the resistor is replaced by a capacitor, the charge stored in the capacitor will increase, and the corresponding potential energy also increased.

Gold Member
Yes, because a resistor is connected in series, then the potential energy does not increase, and the energy supplied by the induced electric field is immediately lost on the resistor
This is correct. In really really dumbed down layman's terms, the battery/transformer secondary/etc. does something and the resistor immediately undoes it. All of this ties together. The internal resistance of the lowly AA battery is what 'undoes' when the AA is shorted. The resistor wire loop that has been discussed in the single turn coil measures zero because the 'doing' and 'undoing' are happening in the exact same space. When the voltage source is separated away from the load, the voltages and currents we are all familiar with make more sense. Only because it's familiar.

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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!!]

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Mentor

Averagesupernova and hutchphd
Mentor
(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.

renormalize, tedward and hutchphd
Gold Member
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

Gold Member
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.

Gold Member
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.

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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?