Walter Lewin Demo/Paradox: Electromagnetic Induction Lecture 16

[Averagesupernova]

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

 

[tedward]

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

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

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

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

[Sorry, what is the DUT?]

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

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

Scalar potential = Int(Es . dl)

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

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

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

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

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

 

[Averagesupernova]

[Sorry, what is the DUT?]

Device Under Test.

 

[Averagesupernova]

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

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

 

[tedward]

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

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

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

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

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

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

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

 

[Averagesupernova]

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

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

 

[tedward]

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

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

 

[tedward]

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

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

V_battery + sum(V_resistors) = 0

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

Int(E_battery) + Int(E_resistors) = 0

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

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

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

V_AB = (Length_AB / Length_circuit)*emf.

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

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

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

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

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

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

 

[Averagesupernova]

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

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

 

[Averagesupernova]

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

 

[tedward]

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

 

[tedward]

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

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

 

[Averagesupernova]

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

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

 

[tedward]

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

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

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

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

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

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

 

[Averagesupernova]

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

 

[hutchphd]

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

SHORT AND SWEET

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

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

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

Can we be finished?

 

[Averagesupernova]

Can we be finished?

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

 

[tedward]

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

LET'S TALK ABOUT ENERGY

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

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

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

Sum(V) around loop = 0

Int(E.dl) = 0

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

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

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

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

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

Another way to think of Faraday's law:

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

 

[tedward]

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

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

 

[Averagesupernova]

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

 

[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).