I Walter Lewin Demo/Paradox: Electromagnetic Induction Lecture 16

[tedward]

AC vs DC is not really the issue. Batteries are DC, but you could also have a non-induced AC source that is in the loop, at least ideally. We used to model use these models in circuit analysis class all the time. Similarly, our transformer does not have to be AC. If you manipulate the external magnetic field so that it is increasing linearly, you essentially have a constant, non-oscillating DC emf, which is often a simpler model for these kinds of problems. With math,

B = kt (linearly increasing magnetic field)

dB/dt = k (constant derivatie)

emf = -k*area (constant flux rate in a single direction everywhere)

So this induced DC emf can certainly add up to non-zero around the loop. It's always in, say, the clockwise direction. As for AC, I think you mean that the field polarity adds up to zero OVER TIME. But freeze time at any moment where the field peaks, and it's all in one direction, just like the DC case. Every point in the loop at a single moment has a clockwise emf, half a period later it's counter-clockwise.

Transformers do have a small resistance, as there is a lot of copper, but it is usually negligible compared to the load resistance. I promise you - when we talk about the voltage output of a transformer, we are not talking about the negligible ohmic voltage drop across the transformer windings. This is a classic source of confusion. Certainly in the ideal case, the ohmic voltage drop across the windings is neglected completely without affecting anything.

At some point in every E/M physics class, a bright student will ask the question, how can a transformer (or inductor) have a voltage drop across it if it's resistance is effectively zero? You will usually get a hand-wavey response unless the professor knows what they are talking about. The answer is certainly NOT the ohmic drop through the windings, which is zero. The real answer is at the HEART of this entire problem. The answer is path dependence due to the non-conservative electric field. Measure the path outside the terminals, you get the emf voltage. Measure a path through the windings (as I've described several times) and you get zero. Two different paths, two different values. That's becasuse the loop that connects them has a changing flux, the Faraday term on the right side of his law, -d(phi)/dt. The sum around this loop is not zero, it is equal to the emf. Hence the path difference. This path dependence is usually hidden in a coiled up transformer, but is exposed when you unwrap the transformer into a single loop, which is the ENTIRE POINT of Lewin's demonstration.

 

[Averagesupernova]

The answer is certainly NOT the ohmic drop through the windings, which is zero.

You may want to think twice about that statement before you dig yourself in any deeper.
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Admittedly, there can be and often are losses ahead of the transformer when short circuiting but there will ALWAYS be losses in the secondary when it's shorted. If there is a current in a shorted secondary then there is loss. We're not talking about superconductors. You are approaching the point where you are saying 1=2.

 

[tedward]

Don't know if I 100% understand you - you mean take a secondary coil, put a changing flux through it (from any source - primary, whatever), and short the output with no resistors / load? Then yes, of course you will get a large drop across the coil, because there is hardly any other resistance in the circuit outside of the coils. Copper of course has a small resistivity , and there is a lot more length of wire in the coils than in the shorting wire, so it will get the lions share of the drop. Put any significant resistance / load in the circuit, and most of the drop will be on the load. And yeah, maybe it's a minor practical consideration that often gets neglected in an ideal treatment, as we usually model the small resistance with another part of the circuit, as part of a lumped resistance. But if you think the ohmic voltage drop across the coil in the circuits we're considering is the same as its output, you've missed the entire point.

 

[Averagesupernova]

you mean take a secondary coil, put a changing flux through it (from any source - primary, whatever), and short the output with no resistors / load?

Well of course that's what I mean! Hasn't that been the central point of the discussion? Just because the secondary winding in the transformer example I gave a few posts back is more than one turn makes no difference. The whole thing is exposed to the flux and the whole I*R voltage is lost across the resistance of the copper. Just how difficult are you trying to make this?

 

[tedward]

The bottom line is Faraday's law, when you break it down, is simply a statement of conservation of energy.

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

This simply means all the energy delivered to the loop = time rate of change of flux. Energy in from the outside equals energy spent by the resistors. The loop sum is non-zero, or E-field is non-conservative, since energy comes from outside the loop.

If there is no changing flux anywhere, then you get Kirchoff's law

Int(E.dl) = Sum(V) = 0

Any energy spent by a resistor comes from a battery in the loop. The E-field is conservative.

Period.

 

[tedward]

Well of course that's what I mean! Hasn't that been the central point of the discussion? Just because the secondary winding in the transformer example I gave a few posts back is more than one turn makes no difference. The whole thing is exposed to the flux and the whole I*R voltage is lost across the resistance of the copper. Just how difficult are you trying to make this?

Apologies, I misunderstood what you were describing. I was thinking transformer with a load, not the uniform ring. In the last part of #91 I was describing a typical LR circuit where the winding resistance is neglected.

 

[Averagesupernova]

So I have no idea where you are at on this now. Nothing has changed for me. It is my position that a shorted single ring as a secondary you cannot measure a voltage between any two points on the ring/loop. Getting a voltmeter reading when doing this means the voltmeter leads are placed improperly and contributing to the reading.

 

[Averagesupernova]

I don't understand why we keep drifting off into places that distract from the point at hand. The whole point of what I have introduced into this is to prove it is not possible to measure a voltage between points on a shorted ring. After that we can discuss non-distributed resistance on the ring. An agreement with my position causes the requirement to use specific orientation of test leads on other configurations of the resistance on the ring.

 

[tedward]

The whole point of what I have introduced into this is to prove it is not possible to measure a voltage between points on a shorted ring.... An agreement with my position causes the requirement to use specific orientation of test leads on other configurations of the resistance on the ring.

So I'm writing a post on the lead argument / voltage masking debate right now, because it's a question that keeps coming up. But to this point - when you say the specific orientation of test leads are you referring to Mabilde's pie-shaped leads over the solenoid?

 

[Averagesupernova]

But to this point - when you say the specific orientation of test leads are you referring to Mabilde's pie-shaped leads over the solenoid?

ANY placement. It's critical since a pair of leads for a voltmeter have the exact same properties as the DUT. They are not an insignificant part of the setup.

 

[tedward]

THE TEST LEAD QUESTION

An important question that keeps coming up is whether or not a voltmeter measurement in a circuit with an induced emf, placed either across a resistor or a section of conducting wire, 'masks' the intended measurement because it is subject to the same induced field as the section being measured. It's a valid concern, provided there is actually something there to measure. The answer to this question depends only on your choice of voltage convention.

Certainly we don't have this problem when we measure a resistor in a regular DC circuit. A voltmeter is a just a resistor placed in parallel, so it's true voltage (and reading) is exactly the same as the measured resistor. But isn't it subject to the battery's emf as well? Of course it is. You can look at it either way - "measuring" the resistor's voltage, or reacting to the emf in exactly the same way as the resistor. It's the same thing said two different ways. Does it make a difference if we're talking about induced emf?

Let's say we're trying to measure the 'scalar potential' in a section of conducting wire in an induced-emf circuit, just wire around a solenoid with one lumped resistor. The scalar potential here is the integral of the Electrostatic field only:

$V_s = \int{\vec E_s \cdot\vec dl}$

In this case $E_s$ can be thought of as the negative of the induced field that would be present if it wasn't canceled out, or as it would appear in free space. Some people refer to this as the 'emf' in this section of the wire (a term I object to, but I digress).

I place my voltmeter in parallel with the wire section. The wire has a presumably non-zero scalar potential between the two points of measurement, and for the exact same reason my voltmeter leads have the same scalar potential as long as it closely follows the DUT. The voltmeter should read zero as the 'potentials' in the wires cancel each other out. It certainly does, so it seems like this argument works in this case. We now know that voltmeters can't measure scalar potential independently. (Keep in mind the scalar potential is just a mathematically derived quantity anyway, charges have no way of responding to it independently of induced field).

Now say we want to measure the 'path voltage' between two points on a wire. This the standard (in my opinion) definition of voltage that is measurable, and corresponds to the integral of the total E-field along a path:

$V = \int{\vec E \cdot\vec dl}$

This definition is the same as the voltage drop across a resistor from Ohm's law, and actually represents the work done on a test charge between two points along a given path. So let's hypothesize that there is an actual non-zero path voltage that doesn't disappear in steady-state. Well the same non-zero path voltage would occur in the lead wires, as the argument goes, so the zero reading (same one we got before) doesn't tell us anything.

Since the measurement itself is suspect, we have to simply find another way to determine what the path voltage should be, in order to determine if voltmeter measurements are valid. Luckily, as path voltage corresponds to net electric field, this is easy to do. Net electric field, the only field present that actually exerts a force on a charge, is the sum of induced field and electro-static field:

$E = E_i + E_s$

In the steady-state case (after charges have reached equilibrium, virtually instantly), there can be no net force on a free charge in a region with no (or negligible) resistance, or you would have arbitrarily large / infinite current. Since the net force on any free charge is zero, then the net electric field must be zero, since electric field determines the force felt by a charge according to:

$\vec F = q\vec E$

So we can rest assured that the net electric field in a section of conducting wire between resistors is zero. Since the definition of path voltage is the integral of electric field, this means that the voltage drop - the same voltage drop we use with Ohm's law, is zero, in perfect agreement with our voltmeter. This also means a voltmeter measurement across a resistor only measures the resistor's drop, and we don't need to worry about 'missing' the voltage in the wire, because there is none.

This makes perfect sense. If the (completely valid) argument is that voltmeters are not immune to physical effects of the circuit, then they must be affected by both induced electric field and static field, just like the section they are measuring. If these effects cancel out in a conducting wire - and they do - than they cancel out in the leads on a voltmeter, resolving the issue.

When does your voltmeter give incorrect readings? Whenever you have changing magnetic flux passing through your measurement loop, due to the induced emf, per Faraday's law.

CONCLUSION: You can't use a voltmeter to measure the mathematical concept of scalar potential (regardless of whether it cancels out in your measurement loop, or is simply non-physical). A voltmeter ONLY measures path voltage. If you're using the standard path voltage convention, (and there's no flux in your loop), you can trust your voltmeter.

 

[Averagesupernova]

@tedward I've read through your last post and I want to take some time to absorb what you've said and implied. I'll try to have a reply later today.

 

[Averagesupernova]

I still keep coming back to the same thing. The measurements @mabilde makes from the center are correct. I've explained how I think this is possible. I used the rubbing of violin strings vs cutting through as an analogy. You didn't buy it.
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You've implied that magnetic fields do not act on a wire, they act on a loop. Well, that's a bit of a BS thing to say considering we cannot measure it unless we form a loop. So it's pointless to introduce such ridiculous statements. Flux cutting through any tiny small portion of the wire will cause induction.
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Unless I have misunderstood, you have also implied that a loop outside of the solenoid is immune to the changing flux. This is false. It can be proven by sorting voltmeter leads together and orienting them around on the outside.
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Now for some specifics. You claim that the pie shaped affair inside the solenoid is just a pickup loop. That is an understandable position, but wrong. My position is that the pie shaped wires are not cut by the flux due the orientation with the field. I back this up by pointing out if it were just forming a pickup loop it would not read zero in the case of a shorted loop. Your position on that argument is that it is just canceling. That argument is not valid since the current in the pie shaped loop will be in the same direction as the large loop. There was talk about the pie shaped loop being a simulation of the real loop. This is an understandable thing to buy into until we disconnect the pie from the main loop and and give it its own stand alone arc at which time it will read the same as when the pie was directly connected to the main loop when the main loop was not shorted.
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We need to come to an agreement about how the conductors behave in the B field while moving in various ways relative to said field. I have my doubts if we will.
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I'm also not certain you have agreed that a shorted secondary ring in this setup can not have any voltage measured anywhere on it when the probes are placed correctly.
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All that being said, I think your last post is simply a more confusing way of saying all of the things you've already said previously in this thread.
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If anywhere in this post I have misunderstood your position on something I am sorry and invite you to clarify. Although I doubt I have misunderstood anything.

 

[alan123hk]

The test method used by @mabilde is undoubtedly correct. Single-turn transformers and multi-turn transformer work exactly the same. There is a scalar potential difference between any two points on the transformer winding wires. This not limited to the open circuit output of the two endpoints, there is a potential difference between any two tap outputs on any segment of the winding.

My personal idea is that there is no such thing of "simulate measurement", unless you manipulated it intentionally and improperly, or it was just simulated on the computer. It is important that the measurements are valid and accurate. If we insist that we are measuring something that doesn't actually exist, it's an invalid and inaccurate measurement. That is, if you say that this scalar field and potential generated by electric charges simply does not exist in electromagnetic devices.

But if that doesn't exist, how does that explain the countless secondary multi-tap output transformers people use every day. I think they're all using the scalar potential part (at least approximately) of the transformer. That potential difference is not only measurable, but also provide stable stream of energy. In addition, the magnitude of the potential difference is exactly equal to the EMF of the corresponding winding.

https://pressbooks.bccampus.ca/singlephasetransformers/chapter/multi-tap-transformers/

 

[tedward]

I'm pretty sure I understand your point. If so, this is (here at least) a discussion of convention and terminology, basically semantics. From our other discussions, WE at least (I believe) agree on the fundamental physics: what magnetic flux is, what Faraday's law says, how the various components of electric field combine, etc. We seem to disagree mostly on 'voltage' convention, i.e. what's a useful measurement and what's not, and what to include when analyzing a circuit.

I'm not an EE, so it's interesting to me to learn how EE's treat things differently from physicists. My background is ME, so my electrical background goes up to phsyics for engineers and circuit analysis, plus what other else I've learned/taught myself over the years (I've also been tutoring math and physics for well over 20 years, hence my continued interest in learning physics and making sure my understanding is rock solid). So I'm certainly willing to learn and/or give some ground on semantics.

Obviously transformers work, including the multi-tap variety. My model for this is just an externally controlled magnetic field through a coiled wire / single loop (with low/negligible resistance) connected to a circuit with a high resistance load, so I'm not really thinking too much about the primary. You can certainly take a voltmeter, measure the terminals (or wherever the tap points are), and get the 'voltage' output. If you put a black box around the transformer, you couldn't differentiate it from an 'ideal' AC voltage source (the AC version of a battery, if that's a thing).

One way to think about this output voltage is simply considering the electric field, through space, connecting the terminals on a path outside the coils, as Feynman does for an inductor in his published lectures. With this view, if I analyzed a circuit loop though this path, I would consider this an $\int E \, dl$ voltage, or path voltage, and Kirchoff's law applies. You could fairly call this (As Feynman does) a potential difference, as the field is locally conservative outside the coils. If I analyzed a complete circuit loop through the coils themselves, I wold simply call this the emf of the circuit (as Lewin prefers to), as there is a changing flux penetrating the surface the loop bounds (my rotini pasta) per Faraday's law.

In our discussions of the single loop transformer (the various versions of the Lewin circuit), in a section of conducting wire (with at least one lumped resistor in the circuit) there is always an electrostatic field to cancel the induced field, resulting in zero net E-field. Since the scalar potential is $\int E_s \, dl$ and the induced voltage is $\int E_i \, dl$, it seems the terminology of 'scalar potential' is equivalent to 'induced voltage', at least within a negative sign. So I don't necessarily see a conflict here, what I call emf you might call scalar potential. As I think on it now, this is related to much of the confusion.

From a concrete physics perspective, and to avoid all sources of potential (no pun intended) confusion, I've taken an absolutist perspective, and have been trying to only consider what I would consider 'real', or physically measurable quantities. This means I exclusively use 'path voltage', or $\int E \, dl$ where E is the net electric field. I treat the net field as the only field that is physically 'present', as it is the only field that is responsible for exerting a net force, or doing actual work, on charge, without considering any other field. It's the only thing that charges 'feel'. If a charge is in a section of conducting wire, and induced field is balanced by static field pushing back, it feels zero force, and no work is done on it as it moves. The induced field that would be there in free space is no longer there, canceled by the force of electrostatic charge. So whichever term you use for it, scalar potential, induced voltage, or emf, it has no independent effect. That is my guiding star.

Hope that clears up my view. I'm also very curious what you think of the test lead question post. (more on Mabilde's setup in a separate post).

 

[tedward]

The test method used by @mabilde is undoubtedly correct...

My personal idea is that there is no such thing of "simulate measurement", unless you manipulated it intentionally and improperly, or it was just simulated on the computer.

I get your point about 'simulating' a measurement. If I want to measure loss of energy due to friction of an object moving on a surface, I can measure the kinetic energy at two points and subtract them - even if I can't measure the heat dissipated in the air directly - and report it as friction loss. So part of my issue might be philosophical, but a lot of it is certainly about intent. Again choice of convention plays a big role here.

Mabilde is an EE professor (iirc), and likely subscribes to the convention (which is as far as I can tell unique to that field) that 'voltage' refers to 'scalar potential' only - though he never states this. He references Kirk McDonald's paper in his analysis, which defines scalar potential strictly as the electrostatic potential between points of accumulated charge at the ends of the resistors. Defined this way, this potential certainly adds up to zero around the loop, as the electrostatic field, on it's own, is conservative. This is apparently what he's trying to measure around the circuit.

As I laid out in my last post, the scalar potential between two points, at least in a section of conducting wire, corresponds to the induced voltage that would be felt between two points in free space. But we're not in free space anymore, this is conducting wire with a lumped resistance in the active circuit, so the net E-field here is in fact zero. The induced voltage / scalar potential in this region now ONLY exists as math, because charge cannot respond to it independently (all the electric field is concentrated in the resistors). But fine, let's pretend it's there. We're essentially asking what would the induced voltage be in this the section of the wire, either in free space or before the fields had reached an equilibrium.

You can't measure this quantity with a direct voltmeter measurement, simply based on the argument that has been raised repeatedly: that voltmeter leads cancel the thing you're trying to measure, since scalar potential, as it's defined, can exist in conducting wire. But one way to do it is to set up your voltmeter leads so that it feels the exact same amount of flux - and therefore emf - that correspond radially to this very symmetric circuit. This subtracts the portion of the emf through the loop from his measurement, giving him a non-zero number that of course changes with the angle of his pie-slice: both area of he slice and the arc-length are proportional to area.

Now if he's trying to measure this abstract quantity, and he describes his intent, his process, and how he intends to measure it, I have no problem. But consider what he states he's measuring. I'll have to watch the video again, but as I recall he never mentions the words scalar potential (tell me if I'm wrong), only 'voltage'. He certainly never discusses different voltage conventions (scalar potential vs. path voltage). So he's assuming everyone watching subscribes to the same definition that he's been trained, as apparently EE's are, to use. So, by using his set up, he sure makes it look like he's measuring something real in this copper ring - that charges actually gain/lose energy as they move across this conducting wire, even though absolutely no work is done on them as there is precisely zero net field there. He then proceeds to show, that the energy gained in the conducting wire is lost in the resistors. This is only true from the scalar potential convention, not from the common understanding of voltage (the true net work done on a charge per coulomb), as the net work done on a charge around the loop by the electric field is most definitely not zero.

Now consider his audience, which includes anyone who watches youtube who's interested in physics: certainly high schoolers, college students, teachers and other academics, and the casual science buff. They're convinced, as they saw with their own eyes, that there is a measurable difference in voltage / energy between two points of zero resistance conducintg wire. So of course, when they do a voltage sum, thinking they're using the more common path voltage convention, they have to take this into account, and the loop sum must be zero!! But most of the audience is not familiar with the technical differences in convention, and most assume we're talking about the standard type of voltage - the type that Lewin is using. So Lewin must be wrong!!

In reality, the emf provided by the flux term is ALREADY TAKEN INTO ACCOUNT in the resistors, and factoring in the scalar potential in the copper ring just subtracts this sum to get zero. So without understanding this strict convention, the audience has been forced to accept scalar potential as a convention unwittingly. What do they learn? That an induced emf circuit works exactly like a DC battery circuit, and the sum of the 'voltages' around the loop is zero. Therefore Lewin is simply confused, and path independence (a fundamental physical idea) is nonsense.

 

[alan123hk]

I'm also very curious what you think of the test lead question post.

The thick blue line outside is a section of the ring circuit, and the thick red line is the lead wire of the voltmeter pressed into a T shape. Both are conductors, so charges accumulate on their surfaces to cancel the induced electric field. Since the induced electric fields inside them are almost equal, the potentials generated by the accumulated charges on their surfaces are also almost exactly equal.

Obviously, the potential difference generated by the charge measured by the voltmeter now moves from the two points a-b to the two points c-d, which should be roughly equal to the arc length between points c and d multiplied by the induced electric field.

Also, of course I admit that Farady's law is the basis of everything and is the king, because it is always correct.

The distance between the associated thick red and blue lines is approximately zero. I've separated them slightly for easier viewing.

 

[Averagesupernova]

@tedward you really need to lose the assumption that is acceptable to include whatever is induced into the voltmeter leads that causes changing voltage readings based on position. This path dependency view is nonsense. By coming to an agreement on how the leads can be positioned so the they are not contributing to the reading and then placing them there shows that everything make sense. It is my view that @mabilde has done this. There are other ways of doing this but in the single loop scenario I believe he has chosen the best method.
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Real electrical engineers always work with the simplest accepted laws (which admittedly are often shortcuts of something more complex) to obtain the desired end result.
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One of the most arrogant people I ever met once told me that when something is not making sense the first thing that should happen is to ask yourself what you are doing wrong. I was rather impressed with a statement like that coming from such an arrogant individual. He obviously applied that to himself. It's a trait many people would do well to adopt. Had Lewin done this, or whoever came up with this prior to him, I wonder where we would be.

 

[hutchphd]

One of the most arrogant people I ever met once told me that when something is not making sense the first thing that should happen is to ask yourself what you are doing wrong. I was rather impressed with a statement like that coming from such an arrogant individual. He obviously applied that to himself. It's a trait many people would do well to adopt.

I agree with your friend absolutely.
The point (I believe?) you are missing is that Prof. Lewin was perfectly aware of what he was doing. He was not confused by the result. Nor should be anyone else who understands Maxwell's Equations.

 

[alan123hk]

Only In the case of an electrostatic field, the voltage is equal to the potential difference. Due to the conservative nature of the static field, the voltage does not depend on the integration path between any two points. In the case of time-varying electromagnetic fields, voltage and potential difference are not the same. The potential difference between two points is unique, while the voltage and induced emf between two points depends on the integration path.

For Lewin's circuit paradox, two points in a circuit cannot be at different potentials just because the voltmeters are on different sides of the circuit. This is a probing problem. We can think of the voltmeter as measuring the voltage produced across the source impedance of the probe wire as the current flows through it, which is why the voltages on both sides of the voltmeter are different. So, stubbornness and arguments may be because everyone has a slightly different idea of definitions, conventions, and terminology.

 

[hutchphd]

For Lewin's circuit paradox,

It is neither a paradox nor a surprise to Prof Lewin. Jeez.

 

[alan123hk]

It is neither a paradox nor a surprise to Prof Lewin.

I sincerely believe this.
(I mean I belive that Prof. Lewin was perfectly aware of what he was doing. He was certainly not confused by the result.)

 

[hutchphd]

You are then sincerely mistaken.

 

[alan123hk]

Right - this is the new definition of Ohm's law that accommodates scalar potential. But the good old V = IR that everyone actually uses and is measured by a voltmeter is the path voltage, Int(E.dl).

So it is not inconsistent with Ohm's law, because the current and power loss in a resistor is calculated in terms of the voltage , not potential difference.

But when it is different from the case of electrostatic field, we have to change the expression from j=c*Ec to j=c*(Ec+Ei), where j = current density, c = conductivity, Ec+Ei = total field

 

[Averagesupernova]

The point (I believe?) you are missing is that Prof. Lewin was perfectly aware of what he was doing. He was not confused by the result.

I can't see how that can be when he said Kirchoff is wrong. I thought we about had this resolved. I said in an earlier post that the setup did not match the schematic. Had the setup been represented correctly on paper then transformer secondaries would have been drawn in and he would not have been able to claim he was probing the same point with both voltmeters.

 

[hutchphd]

He said that Kirchhoff was wrong when blindly used in the situation he presented. Not "Kirchhoff" (the man) but "Kirchhoff" (the Law) when carelessly applied. Lewin was not confused about either Kirchhoff's circuit law nor Faraday's Law .

 

[Averagesupernova]

He said that Kirchhoff was wrong when blindly used in the situation he presented. Not "Kirchhoff" (the man) but "Kirchhoff" (the Law) when carelessly applied. Lewin was not confused about either Kirchhoff's circuit law nor Faraday's Law .

That's a stretch. Taking that approach and to put it the way he did and not explain what's really going on is irresponsible. Especially for someone in his position.

 

[hutchphd]

He was teaching the sophomore EM course at MIT. He explains it in great detail in previous and subsequent lecturees. I do not understand the vitriol it engenders: none is appropriate. He did not ascribe it to voodoo.
He was warning his students not to blindly apply Kirchhoff by using a vivid and effective lecture demo. More power to him.

 

[Averagesupernova]

Fine, whatever. I can't say I've ever fallen into the blindly following Kirchoff trap or whatever. I still think it's a silly thing to do. I could say the same thing about ohm. Incandescent bulbs don't follow ohms law when it is misapplied. E * I doesn't give us Watts when we misapply and ignore current being out of phase with volts.

 

[tedward]

I still keep coming back to the same thing. The measurements @mabilde makes from the center are correct. I've explained how I think this is possible. I used the rubbing of violin strings vs cutting through as an analogy. You didn't buy it.

Flux cutting through any tiny small portion of the wire will cause induction.

So now that you've collected all our disagreements in one place, It's easy to see where the common thread lies that runs through all of this. At first I wondered why you were ignoring or disagreeing with any argument that dealt with flux passing through a loop. It's painfully obvious now (from your description) that you simply do not understand what magnetic flux is, or how it affects a circuit. I don't know if you're a high-schooler, an electrical engineer, or have a Ph.D. I don't care. Whatever model or rule-of-thumb you've learned, and however you learned it, it is wrong. It does not help the discussion to ignore your clear misunderstanding that you have laid out above for anyone to see. If you want to speak intelligently on these topics, you have to know the fundamentals first. Now that I can clearly see where the fence between our yards lies, it's probably a waste of time discussing further. But this is a physics forum where people come to learn/discus physics, and the teacher in me wants you to learn this correctly, so that something fruitful comes from this discussion. You can open an A.P. Physics textbook, watch Lewin's fantastic lectures, or even watch some Kahn Academy videos. But learn it right. I don't care if we agree on the Lewin 'paradox' anymore, I just want you to (re)learn the basics for your own benefit. I'll break it down for you (or anyone else who wants to learn) here and respond to some of your other points separately.

THE SOLENOID'S FIELD

First let's get our picture right. In our solenoid, current through the windings generates a magnetic field. The field points vertically (oscillating up and down) along the axis, is uniform, and only exists inside the solenoid volume. I'm using the common assumption of an infinitely tall solenoid to avoid worrying about the return path flux, which is a practical consideration but not relevant to the ideal case. The point is that there is no flux between the solenoid and the loop circuit, even if there is considerable distance from the solenoid to the loop wire.

MAGNETIC FLUX

What is magnetic flux? It's the amount of magnetic field, summed over a defined area, that passes through that area. The calculus version is written like this: $$\Phi_B = \iint \vec B \cdot d\vec A $$
In a simple situation like ours, it reduces to a simple formula: $$\Phi_B = BA_s$$ where $B$ is the magnitude of the field in the solenoid, and $A_s$ is the cross section area of the solenoid only, regardless of the size of our circuit. We only include the solenoid area as there is no magnetic outside the solenoid to contribute to the sum.

FARADAY'S LAW

How does it affect the circuit? Faraday's law (applied to our situation) basically says two things. First, if the flux through the solenoid changes in time, it creates an electric field in space that surrounds it, always encircling the solenoid in one direction. It also says that that the circulation of this electric field, meaning the total sum of the field on any circular path around the solenoid, equals the time rate-of-change of this magnetic flux: $$\oint \vec E \cdot d\vec L = -\frac {d \Phi_B}{dt} $$ In space, you can picture the electric field as clockwise arrows circling the solenoid, where the field strength decreases with radius but the total sum of any circular (or any path) is always the same. This is an important point - the electric field strength decreases with distance, but the total circulation is the same no matter the radius. And that circulation is always non-zero, as long as there is a changing flux. That's what non-conservative means. The negative sign is just there as a nod to Lenz' law, which says the direction of the induced field opposes changes in the flux.

EMF

When our circuit loop is placed around the solenoid, this electric field interacts with free charge in the loop, pushing charge around and creating a current. Inside the circuit, we now refer to the circulation of the field an electromotive force, or emf. $$emf = \oint \vec E \cdot d\vec L $$ This emf is a property of the entire loop itself, and is also called the induced voltage of the circuit. It's important to remember that emf is not some new mysterious physical quantity. It's just the sum of the induced electric field over the length of the loop. The ONLY manifestation of induction here is via electric field.

As consequence of Faraday's law, we can also say that any closed path that has NO flux penetrating it, has zero emf. This is really handy when analyzing voltmeter loops to make sure that they are unaffected by unwanted magnetic flux. In fact, you can apply Faraday's law to the entire circuit loop, smaller loops in a network, loops that other have other loops inside them, paths through free space, or any combination of circuit and free path. It always works.

Now let's talk about your model for a second. You have spoken several times of your 'violin string bowing' picture. The best I can tell is you got this from an induction generator picture (like the video you linked to), which involves a fixed field and moving conductor. In this case, there is no electric field directly responsible for moving charge - it is simply the magnetic force (one part of the Lorentz force): $$F_B = q \vec v \times \vec B$$ Interestingly, this different physical phenomenon gives rise to exact same equation - Faraday's law. The Feynman lecture I linked to (did you bother reading it?) discusses this ambiguity.

So how do I know your model is wrong? Because the magnetic flux, as we said, only exists in the solenoid. It does not have to come in contact with the loop. The loop could theoretically be at any distance, with no magnetic field of consequence in between. So this bowing picture with flux interacting with the loop, on it's face, falls flat. Now I don't want to confuse the issue, but the purist in me needs to mention that the way the induced electric field is created in the first place is via an electromagnetic wave, assuming the flux is oscillating like an AC source. The 'M' in this EM wave plays no part in our analysis, as it does not contribute to flux or do work on any charge in our circuit. And if the flux in the solenoid is increasing linearly, as many examples treat it, there is no magnetic field at the loop at all.

Obviously, in Mabilde's setup, he does put his copper ring as close as possible to the solenoid. This is to avoid the practical issue of the returning path of the solenoid field (comes out the 'pipe' at the top and turns around to re-enter at the bottom) which has much weaker field strength but could certainly affect measurement loops.

So now you should be caught up. That's what we mean by flux. We do not speak of flux 'cutting through wires', that is meaningless. Certainly induced electric fields have a physical effect on wire sections, either alone or part of a circuit, but I'll address that in another post as well as some of your other points.