#562 – Electroboom!

[HuronKing]

Ok, then in three small sentences or less explain where the non-conservative field is in Lewin's ring.

\${\nabla \times E = -\frac{\partial B}{\partial t}}\$
Hint: left side of the equation.

Yes, but the right part of the equation makes it conservative again!  The only way it will be non-conservative is if you do this:

\${\nabla \times E = 0}\$

Ah, no. The right part doesn't "make" the left part conservative. The right part just says "there is a time-varying magnetic flux" and the left part says "there's an electric field with a curl", and the curl is what makes the left part non-conservative. As to answer the "where", the "curled" electric field is in a plain perpendicular to the magnetic flux vector.

Yes it does!  That is why KVL do work.

Emphasis mine. Is there a typo here? Are you saying that the curl of E being 0 means the field is non-conservative?

Have you taken a course in vector calculus or are you self-taught on this subject?

[jesuscf]

Ok, then in three small sentences or less explain where the non-conservative field is in Lewin's ring.

\${\nabla \times E = -\frac{\partial B}{\partial t}}\$
Hint: left side of the equation.

Yes, but the right part of the equation makes it conservative again!  The only way it will be non-conservative is if you do this:

\${\nabla \times E = 0}\$

Ah, no. The right part doesn't "make" the left part conservative. The right part just says "there is a time-varying magnetic flux" and the left part says "there's an electric field with a curl", and the curl is what makes the left part non-conservative. As to answer the "where", the "curled" electric field is in a plain perpendicular to the magnetic flux vector.

Yes it does!  That is why KVL do work.

Emphasis mine. Is there a typo here? Are you saying that the curl of E being 0 means the field is non-conservative?

Have you taken a course in vector calculus or are you self-taught on this subject?

Yes, the old good "Straw Man Fallacy".  Here, read it from Hayt yourself, from the attached pdf.

[thinkfat]

With regards to whether KVL holds in that circuit or not - it doesn't, in the circuit Lewin chose for his experiment. It may very well hold in the circuit you created, but it is a different circuit, by your own words, because the electric fields are observed along different paths. It is also very, very peculiar and so full of uncertainties that it will be very hard for you to claim that the results you obtained are "exact" and "true". Your measurements were one or two percent off from your calculation as I recall, but can you say where that error came from? Probing? Calculation? Assumptions? Resistor tolerance? Volt meter error? That's why I called it a Nothing-Burger.

The main source of error, I believe, is probing.  One has to be very careful for the varying magnetic field not to affect the measuring instruments.  I find it interesting that you are willing to dismiss my experiment because I got a less than 2% discrepancy with the theoretical KVL calculation, but you are willing to immediately accept Lewin's results when he doesn't even provide any actual numerical measurements from his experiment.

Also, have you done the experiment yourself?  Do you have any measurements?

I don't dismiss your experiment because of those 2%, but because you cannot say where they come from. You "believe" the error is from probing. That means you're absolutely sure that your calculation (and thus assumptions) are correct. Why? Your approach to modeling was trivially assuming complete uniformity in distribution of the scalar PD. How can you be sure of that?

I can surely repeat the experiment as soon as I've found my stash of ring cores that is hidden somewhere in the pile of boxes from the last move. But I'm reasonably sure that I'll just observe what others have already found.

[HuronKing]

Ok, then in three small sentences or less explain where the non-conservative field is in Lewin's ring.

\${\nabla \times E = -\frac{\partial B}{\partial t}}\$
Hint: left side of the equation.

Yes, but the right part of the equation makes it conservative again!  The only way it will be non-conservative is if you do this:

\${\nabla \times E = 0}\$

Ah, no. The right part doesn't "make" the left part conservative. The right part just says "there is a time-varying magnetic flux" and the left part says "there's an electric field with a curl", and the curl is what makes the left part non-conservative. As to answer the "where", the "curled" electric field is in a plain perpendicular to the magnetic flux vector.

Yes it does!  That is why KVL do work.

Emphasis mine. Is there a typo here? Are you saying that the curl of E being 0 means the field is non-conservative?

Have you taken a course in vector calculus or are you self-taught on this subject?

Yes, the old good "Straw Man Fallacy".  Here, read it from Hayt yourself, from the attached pdf.

Have you read it? No strawman here. I seriously have no idea WTF you're talking about because on the very page you cited, Hayt writes:
"Any field that satisfies an equation of the form of Eq. (20), (i.e., where the closed
line integral of the field is zero) is said to be a conservative field."

You wrote,
"The only way it will be non-conservative is if you do this:

∇×E=0"

Which is the opposite of what Hayt wrote. Is this a typo? Or have you self-taught yourself vector calculus? Those are the only two possibilities I can think of for such a statement.

[HuronKing]

Hayt goes on to write,
"The integral is zero if ρ1 = 1, 2, 3,... , etc., but it is not zero for other values of ρ1,
or for most other closed paths, and the given field is not conservative. A conservative
field must yield a zero value for the line integral around every possible closed path."

There are paths in the Romer-Lewin ring whose line integrals are non-zero. Proof by counterexample then is that the field is non-conservative, KVL doesn't hold (as it requires 0 line integral around every path, as Feynman writes), and voltage is non-unique in the presence of non-conservative fields.

How is this even a debate?

[HuronKing]

I should mention that, strictly speaking, a vector field with zero curl CAN be non-conservative. That is, zero curl does not imply the field is conservative.

https://mathinsight.org/path_dependent_zero_curl

This is why I, personally, much prefer the integral representation of Maxwell's Eqs than the differential form.

I wish I had this website when I went to college. Their examples and details are quite nice:
https://mathinsight.org/conservative_vector_field_determine

[jesuscf]

Ok, then in three small sentences or less explain where the non-conservative field is in Lewin's ring.

\${\nabla \times E = -\frac{\partial B}{\partial t}}\$
Hint: left side of the equation.

Yes, but the right part of the equation makes it conservative again!  The only way it will be non-conservative is if you do this:

\${\nabla \times E = 0}\$

Ah, no. The right part doesn't "make" the left part conservative. The right part just says "there is a time-varying magnetic flux" and the left part says "there's an electric field with a curl", and the curl is what makes the left part non-conservative. As to answer the "where", the "curled" electric field is in a plain perpendicular to the magnetic flux vector.

Yes it does!  That is why KVL do work.

Emphasis mine. Is there a typo here? Are you saying that the curl of E being 0 means the field is non-conservative?

Have you taken a course in vector calculus or are you self-taught on this subject?

Yes, the old good "Straw Man Fallacy".  Here, read it from Hayt yourself, from the attached pdf.

Have you read it? No strawman here. I seriously have no idea WTF you're talking about because on the very page you cited, Hayt writes:
"Any field that satisfies an equation of the form of Eq. (20), (i.e., where the closed
line integral of the field is zero) is said to be a conservative field."

You wrote,
"The only way it will be non-conservative is if you do this:

∇×E=0"

Which is the opposite of what Hayt wrote. Is this a typo? Or have you self-taught yourself vector calculus? Those are the only two possibilities I can think of for such a statement.

What I mean is that, over the influence of an external magnetic field, if you add the voltage drops from the resistors you'll get a number that is not zero.  That is where the induced emf comes into play.  The sum of the voltage drops in the resistors is equal to the induced emf.  Energy in is equal to energy out.  The circuit is conservative and KVL works.  Is that clear now?

[jesuscf]

Hayt goes on to write,
"The integral is zero if ρ1 = 1, 2, 3,... , etc., but it is not zero for other values of ρ1,
or for most other closed paths, and the given field is not conservative. A conservative
field must yield a zero value for the line integral around every possible closed path."

There are paths in the Romer-Lewin ring whose line integrals are non-zero. Proof by counterexample then is that the field is non-conservative, KVL doesn't hold (as it requires 0 line integral around every path, as Feynman writes), and voltage is non-unique in the presence of non-conservative fields.

How is this even a debate?

The only path allowed is the path of the circuit composed of wires and resistors.  The integral does not apply to any other arbitrary path.

[jesuscf]

I should mention that, strictly speaking, a vector field with zero curl CAN be non-conservative. That is, zero curl does not imply the field is conservative.

https://mathinsight.org/path_dependent_zero_curl

This is why I, personally, much prefer the integral representation of Maxwell's Eqs than the differential form.

I wish I had this website when I went to college. Their examples and details are quite nice:
https://mathinsight.org/conservative_vector_field_determine

I agree.  I also prefer the integral representation.

[jesuscf]

With regards to whether KVL holds in that circuit or not - it doesn't, in the circuit Lewin chose for his experiment. It may very well hold in the circuit you created, but it is a different circuit, by your own words, because the electric fields are observed along different paths. It is also very, very peculiar and so full of uncertainties that it will be very hard for you to claim that the results you obtained are "exact" and "true". Your measurements were one or two percent off from your calculation as I recall, but can you say where that error came from? Probing? Calculation? Assumptions? Resistor tolerance? Volt meter error? That's why I called it a Nothing-Burger.

The main source of error, I believe, is probing.  One has to be very careful for the varying magnetic field not to affect the measuring instruments.  I find it interesting that you are willing to dismiss my experiment because I got a less than 2% discrepancy with the theoretical KVL calculation, but you are willing to immediately accept Lewin's results when he doesn't even provide any actual numerical measurements from his experiment.

Also, have you done the experiment yourself?  Do you have any measurements?

I don't dismiss your experiment because of those 2%, but because you cannot say where they come from. You "believe" the error is from probing. That means you're absolutely sure that your calculation (and thus assumptions) are correct. Why? Your approach to modeling was trivially assuming complete uniformity in distribution of the scalar PD. How can you be sure of that?

I can surely repeat the experiment as soon as I've found my stash of ring cores that is hidden somewhere in the pile of boxes from the last move. But I'm reasonably sure that I'll just observe what others have already found.

Wait a minute!  Are you admitting that any arbitrary part of the circuit under the influence of an external varying magnetic field has a measurable induced emf?  Can you please explain that to Sredni?

[thinkfat]

Hayt goes on to write,
"The integral is zero if ρ1 = 1, 2, 3,... , etc., but it is not zero for other values of ρ1,
or for most other closed paths, and the given field is not conservative. A conservative
field must yield a zero value for the line integral around every possible closed path."

There are paths in the Romer-Lewin ring whose line integrals are non-zero. Proof by counterexample then is that the field is non-conservative, KVL doesn't hold (as it requires 0 line integral around every path, as Feynman writes), and voltage is non-unique in the presence of non-conservative fields.

How is this even a debate?

The only path allowed is the path of the circuit composed of wires and resistors.  The integral does not apply to any other arbitrary path.

For Dr. Lewins experiment, it obviously needs to be pointed out that also the paths involving the measurement equipment are important. People seem to forget that, but it is key to understanding it. And of course the equation applies to every possible path through the circuit. How could it be any different.

[HuronKing]

What I mean is that, over the influence of an external magnetic field, if you add the voltage drops from the resistors you'll get a number that is not zero.  That is where the induced emf comes into play.  The sum of the voltage drops in the resistors is equal to the induced emf.  Energy in is equal to energy out.  The circuit is conservative and KVL works.  Is that clear now?

This is not what conservative fields mean. Path-dependence is the key. I linked this to you pages and pages ago.
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/8-2-conservative-and-non-conservative-forces/

Don't know how much simpler it can get.

And also why I asked if you were self-taught on this subject - you are confusing terminology and making an inconsistent theory.

The only path allowed is the path of the circuit composed of wires and resistors.  The integral does not apply to any other arbitrary path.

This is very explicitly not what Hayt or any of the other published authors on this subject have written (when time-varying fields are present inside the contour of the loop).
KVL predicts that going clockwise around the loop or counterclockwise around the loop doesn't matter - the voltage should be the same, i.e. the line integral should be zero around ANY path of the wires and no work is done.

That doesn't just mean ANY configuration of wires (which you're admitting here your theory won't work for any arbitrary arrangement of wires, so KVL is dead?). Romer is perfectly okay predicting what happens when the wires move, that's what Figure 6 of his paper shows. But also the DIRECTION of the path matters. Of course it should - it's an integration of vectors, which have direction. Romer makes note of this in his Figure 3 and copious amounts of the mathematical literature on line integration discusses this.

This isn't at all surprising in a complete interpretation of Faraday's Law. The periodic output waveform from Romer's Figure 5 makes perfect sense - when the direction of the magnetic flux changes the sign of the voltages across the resistors also changes.

[bdunham7]

For Dr. Lewins experiment, it obviously needs to be pointed out that also the paths involving the measurement equipment are important. People seem to forget that, but it is key to understanding it. And of course the equation applies to every possible path through the circuit. How could it be any different.

Not pointing it out and hoping people missed it--perhaps expecting the curled e-field resulting from the changing magnetic flux to somehow stop at the perimeter of the inner ring--was the sole basis for this being a memorable, 'miind-blowing' demonstration. 

[jesuscf]

What I mean is that, over the influence of an external magnetic field, if you add the voltage drops from the resistors you'll get a number that is not zero.  That is where the induced emf comes into play.  The sum of the voltage drops in the resistors is equal to the induced emf.  Energy in is equal to energy out.  The circuit is conservative and KVL works.  Is that clear now?

This is not what conservative fields mean. Path-dependence is the key. I linked this to you pages and pages ago.
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/8-2-conservative-and-non-conservative-forces/

Don't know how much simpler it can get.

And also why I asked if you were self-taught on this subject - you are confusing terminology and making an inconsistent theory.

In Lewin's ring can you please calculate the power added to the circuit by the external varying magnetic field (via Faraday's law) and compare it to the power consumed by the resistors?  Is energy conserved in the circuit or not?

[HuronKing]

What I mean is that, over the influence of an external magnetic field, if you add the voltage drops from the resistors you'll get a number that is not zero.  That is where the induced emf comes into play.  The sum of the voltage drops in the resistors is equal to the induced emf.  Energy in is equal to energy out.  The circuit is conservative and KVL works.  Is that clear now?

This is not what conservative fields mean. Path-dependence is the key. I linked this to you pages and pages ago.
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/8-2-conservative-and-non-conservative-forces/

Don't know how much simpler it can get.

And also why I asked if you were self-taught on this subject - you are confusing terminology and making an inconsistent theory.

In Lewin's ring can you please calculate the power added to the circuit by the external varying magnetic field (via Faraday's law) and compare it to the power consumed by the resistors?  Is energy conserved in the circuit or not?

Your question is nonsensical in the context of this discussion. I also sent you this link.
https://courses.lumenlearning.com/physics/chapter/7-5-nonconservative-forces/

If you'd read this, or had any formal training in mathematics or physics (I'm happy to admit I received formal education, have you? I really am interested to know if you've ever passed exams on these topics), you'd see how ridiculous this question is when we are talking about non-conservative forces and fields. From the link above:

"An important characteristic is that the work done by a nonconservative force adds or removes mechanical energy from a system. Friction, for example, creates thermal energy that dissipates, removing energy from the system. Furthermore, even if the thermal energy is retained or captured, it cannot be fully converted back to work, so it is lost or not recoverable in that sense as well."

Are you asking if the energy consumed in the resistors to the Romer system is the same as the energy supplied by the field? Yes... obviously.

But that has diddly-squat to do with whether the work done in one path versus another path with the same start and end point is the same. Figure 1 in the link above make this point as simple as possible - probably too simple since friction can't supply energy and doesn't circulate like EM-fields can but alas this is the baseline I see.

Hayt makes this same distinction in the pages you posted earlier for Figure 4.4. And that's the question you should be asking. Is the work done on a charge the same in every possible path (which includes direction!) in the Romer ring? Romer's analysis shows that it is not... and so the voltage between A-B depends on the path... which is the whole reason KVL dies in this experiment.

I think you ought to read Hayt's Chapter 9. And if you can't make it that far, read Hayt's Chapter 7.6.
"Although we shall consider the scalar magnetic potential to a much greater extent in Chapter 8, when we introduce magnetic materials and discuss the magnetic circuit, one difference between V and Vm should be pointed out now: Vm is not a single-valued function of position. The electric potential V is single-valued; once a zero reference is assigned, there is only one value of V associated with each point in space. Such is not the case with Vm."

And Hayt goes on to prove it using a coaxial cable and ties the whole thing together beautifully on P.212.

I don't know how much more explicitly you want it. In Romer's ring, since the voltage is the contribution of both the electric potential and the magnetic potential (whose contribution is a current flow from induction), and the magnetic potential is multi-valued depending on the path of integration, then there is no unique, single-valued voltage in the network.

Let KVL die. Let yourself graduate from Hayt's Chapter 4 and embrace the entirety of Applied EM.

*Edit*
Corrected an ambiguity in my phrasing to make it clear that the magnetic potential is in amperes and gives rise to a current. Covid and flu booster can make one miss some things!

[jesuscf]

Your question is nonsensical in the context of this discussion. I also sent you this link.
https://courses.lumenlearning.com/physics/chapter/7-5-nonconservative-forces/

If you'd read this, or had any formal training in mathematics or physics (I'm happy to admit I received formal education, have you? I really am interested to know if you've ever passed exams on these topics), you'd see how ridiculous this question is when we are talking about non-conservative forces and fields. From the link above:

It makes perfect sense, because the derivation of Kirchhoff's circuital laws are based in the principle of conservation of energy.  And yes, I read the link you posted, and it completely agrees with what I am saying.

(Yes, I have received formal education: 5 years undergrad, 2 years masters, and 4 years PhD.  What about you?)

Are you asking if the energy consumed in the resistors to the Romer system is the same as the energy supplied by the field? Yes... obviously.

Therefore the fields in the circuit are conservative, aren't they?

I don't know how much more explicitly you want it. In Romer's ring, since the voltage is the contribution of both the electric potential and the magnetic potential (whose contribution is a current flow from induction), and the magnetic potential is multi-valued depending on the path of integration, then there is no unique, single-valued voltage in the network.

You are mixing magnetic circuits with electric circuits here.  In the wire ring with two resistors we don't have a magnetic circuit.  But since you are at it, please add a thermal circuit, and why not, a hydraulic circuit as well!  All of those can be solved with laws equivalent to Ohms law, KVL, and KCL.

[HuronKing]

It makes perfect sense, because the derivation of Kirchhoff's circuital laws are based in the principle of conservation of energy.  And yes, I read the link you posted, and it completely agrees with what I am saying.

The KVL law predicts that voltage in a closed path must be zero regardless of path. That's not true in the presence of non-conservative fields, period.

(Yes, I have received formal education: 5 years undergrad, 2 years masters, and 4 years PhD.  What about you?)

MSEE and 2 years Ph.D in progress and P.E. license.

Therefore the fields in the circuit are conservative, aren't they?

Go directly to jail. Do not pass GO. Do not collect $200.

How can you read Hayt, have a formal education (I assume in EE), and come away with that conclusion about magnetic fields? It's astounding to me.

Here, try this maybe?
https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Book%3A_Electricity_and_Magnetism_(Tatum)/09%3A_Magnetic_Potential/9.01%3A_Introduction_to_Magnetic_Potential

You are mixing magnetic circuits with electric circuits here.  In the wire ring with two resistors we don't have a magnetic circuit.  But since you are at it, please add a thermal circuit, and why not, a hydraulic circuit as well!  All of those can be solved with laws equivalent to Ohms law, KVL, and KCL.

So the contribution of the magnetic field does not matter to you? No wonder you can't see the non-conservative interactions at play.
Hayt is using magnetic circuits as part of his progression in getting to the complete form of Faraday's Law. Chapter 4 is about electrostatic fields, where all the fields are conservative, work done on any path is zero, and KVL can happily be applied.

But trouble starts to happen when the magnetic fields come in and start to wobble a little. Hayt writes on P.212,

We should remember that the electrostatic potential V is a conservative field; the magnetic scalar potential Vm is not a conservative field.

That is achieved by simply taking the closed loop line integrals of H (H and B are related directly by permeability constant so its trivial to do the unit conversions) and seeing that they no longer have unique values.

So what happens when we take the derivative of those B-fields and those derivative are not zero and then we integrate them?

By Chapter 9, KVL is gone. Hayt writes (P.280),

If B is not a function of time, (5) and (6) evidently reduce to the electrostatic equations

Which are just the KVL equations of Chapter 4. If B IS a function of time, then those equations can't reduce - KVL doesn't exist here, the closed loop contours are now entirely path dependent.

See, the trick is in seeing that magnetic forces on a charge in a magnetic field are not, by their nature, conservative. But, if they don't vary with time - everything is cool. Their derivative is zero and they disappear. Yet, if they DO vary with time - then all the nasty complications of path-dependency analyzed in Chapter 7 MUST be taken into account. Chapter 9 does this - but KVL cannot live in Chapter 9 because KVL is said, repeatedly by Hayt, to be path independent.

Faraday's Law is not. And can never be. Voltages in the presence of time-varying fields are not unique.

[jesuscf]

It makes perfect sense, because the derivation of Kirchhoff's circuital laws are based in the principle of conservation of energy.  And yes, I read the link you posted, and it completely agrees with what I am saying.

The KVL law predicts that voltage in a closed path must be zero regardless of path. That's not true in the presence of non-conservative fields, period.

(Yes, I have received formal education: 5 years undergrad, 2 years masters, and 4 years PhD.  What about you?)

MSEE and 2 years Ph.D in progress and P.E. license.

Therefore the fields in the circuit are conservative, aren't they?

Go directly to jail. Do not pass GO. Do not collect $200.

How can you read Hayt, have a formal education (I assume in EE), and come away with that conclusion about magnetic fields? It's astounding to me.

Here, try this maybe?
https://phys.libretexts.org/Bookshelves/Electricity_and_Magnetism/Book%3A_Electricity_and_Magnetism_(Tatum)/09%3A_Magnetic_Potential/9.01%3A_Introduction_to_Magnetic_Potential

You are mixing magnetic circuits with electric circuits here.  In the wire ring with two resistors we don't have a magnetic circuit.  But since you are at it, please add a thermal circuit, and why not, a hydraulic circuit as well!  All of those can be solved with laws equivalent to Ohms law, KVL, and KCL.

So the contribution of the magnetic field does not matter to you? No wonder you can't see the non-conservative interactions at play.
Hayt is using magnetic circuits as part of his progression in getting to the complete form of Faraday's Law. Chapter 4 is about electrostatic fields, where all the fields are conservative, work done on any path is zero, and KVL can happily be applied.

But trouble starts to happen when the magnetic fields come in and start to wobble a little. Hayt writes on P.212,

We should remember that the electrostatic potential V is a conservative field; the magnetic scalar potential Vm is not a conservative field.

That is achieved by simply taking the closed loop line integrals of H (H and B are related directly by permeability constant so its trivial to do the unit conversions) and seeing that they no longer have unique values.

So what happens when we take the derivative of those B-fields and those derivative are not zero and then we integrate them?

By Chapter 9, KVL is gone. Hayt writes (P.280),

If B is not a function of time, (5) and (6) evidently reduce to the electrostatic equations

Which are just the KVL equations of Chapter 4. If B IS a function of time, then those equations can't reduce - KVL doesn't exist here, the closed loop contours are now entirely path dependent.

See, the trick is in seeing that magnetic forces on a charge in a magnetic field are not, by their nature, conservative. But, if they don't vary with time - everything is cool. Their derivative is zero and they disappear. Yet, if they DO vary with time - then all the nasty complications of path-dependency analyzed in Chapter 7 MUST be taken into account. Chapter 9 does this - but KVL cannot live in Chapter 9 because KVL is said, repeatedly by Hayt, to be path independent.

Faraday's Law is not. And can never be. Voltages in the presence of time-varying fields are not unique.

You are very confused by what you are reading.  For instance, can you provide a guesstimate of the reluctance of the ring circuit?  Assume the wiring is either copper or aluminum.   How would that affect the electric behavior of the ring circuit?

[HuronKing]

You are very confused by what you are reading.  For instance, can you provide a guesstimate of the reluctance of the ring circuit?  Assume the wiring is either copper or aluminum.   How would that affect the electric behavior of the ring circuit?

And phhhewwwwww away we go. It's funny you should ask this though as if it's some kind of gotcha question. Hayt has a discussion of this too:

In an electric circuit, the voltage source is a part of the closed path; in the magnetic circuit, the current-carrying coil will surround or link the magnetic circuit. In tracing a magnetic circuit, we will not be able to identify a pair of terminals at which the magnetomotive force is applied. The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero).

So, the MMF of equation 44 on P.257 is entirely dependent on the number of loops of the coil enclosing the circuit. In other words, the value of the MMF changes based on how many times you circumscribe the closed loop path.
Hayt defines on P.256 that the Reluctance of a circuit is a function of the magnetic scalar potential (which we saw from Chapter 7.6 is path dependent). The units are A-t/Wb. So, the Reluctance is also path-dependent.

And I like how Hayt tosses in, for good measure, a reminder that closed line integral of E is NOT zero in this circuit or the circuits of Chapter 9 which is also explicitly the definition of a non-conservative field. I think he mentions this because on P.256 he writes after showing the closed loop integral of E dot dl,

In other words, Kirchhoff’s voltage law states that the rise in potential through the source is exactly equal to the fall in potential through the load.

Once more for the people in the back? What does he write on the very next page of P.257?

The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero)

Can we stop with this fiction that Hayt somehow agrees with the cockamamie proposition that KVL holds in all cases? He doesn't. None of the published authors do - because it's wrong.

[jesuscf]

You are very confused by what you are reading.  For instance, can you provide a guesstimate of the reluctance of the ring circuit?  Assume the wiring is either copper or aluminum.   How would that affect the electric behavior of the ring circuit?

And phhhewwwwww away we go. It's funny you should ask this though as if it's some kind of gotcha question. Hayt has a discussion of this too:

In an electric circuit, the voltage source is a part of the closed path; in the magnetic circuit, the current-carrying coil will surround or link the magnetic circuit. In tracing a magnetic circuit, we will not be able to identify a pair of terminals at which the magnetomotive force is applied. The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero).

So, the MMF of equation 44 on P.257 is entirely dependent on the number of loops of the coil enclosing the circuit. In other words, the value of the MMF changes based on how many times you circumscribe the closed loop path.
Hayt defines on P.256 that the Reluctance of a circuit is a function of the magnetic scalar potential (which we saw from Chapter 7.6 is path dependent). The units are A-t/Wb. So, the Reluctance is also path-dependent.

And I like how Hayt tosses in, for good measure, a reminder that closed line integral of E is NOT zero in this circuit or the circuits of Chapter 9 which is also explicitly the definition of a non-conservative field. I think he mentions this because on P.256 he writes after showing the closed loop integral of E dot dl,

In other words, Kirchhoff’s voltage law states that the rise in potential through the source is exactly equal to the fall in potential through the load.

Once more for the people in the back? What does he write on the very next page of P.257?

The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero)

Can we stop with this fiction that Hayt somehow agrees with the cockamamie proposition that KVL holds in all cases? He doesn't. None of the published authors do - because it's wrong.

Once again, we are not dealing with a magnetic circuit here.

[HuronKing]

Once again, we are not dealing with a magnetic circuit here.

That is why you fail. I've taken you, step-by-step, through Hayt's progression from electrostatics, into magnetic circuits (where the non-conservative properties of magnetism are explored), and how they directly lead to the conclusions of Chapter 9, which Hayt teased in Chapter 8, and I quote again,

The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero)

Your answer is the equivalent of saying,
"Nuh uh!"

A Ph.D. is telling me that the properties of magnetic fields do not apply to the Romer ring. I guess Hayt was just navel-gazing in Chapters 7 and 8 and none of that stuff has anything to do with Chapter 9...

I'm truly astounded (and I guess I now understand Lewin's comments in his lecture about being accused of cheating on the experiment). Unless you have something substantive, I am not replying to your comments anymore.

[jesuscf]

Once again, we are not dealing with a magnetic circuit here.

That is why you fail. I've taken you, step-by-step, through Hayt's progression from electrostatics, into magnetic circuits (where the non-conservative properties of magnetism are explored), and how they directly lead to the conclusions of Chapter 9, which Hayt teased in Chapter 8, and I quote again,

The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero)

Your answer is the equivalent of saying,
"Nuh uh!"

A Ph.D. is telling me that the properties of magnetic fields do not apply to the Romer ring. I guess Hayt was just navel-gazing in Chapters 7 and 8 and none of that stuff has anything to do with Chapter 9...

I'm truly astounded (and I guess I now understand Lewin's comments in his lecture about being accused of cheating on the experiment). Unless you have something substantive, I am not replying to your comments anymore.

Great, from now on I will try not reply to your messages either as I am tired of your fallacious diatribe.

[HuronKing]

Great, from now on I will try not reply to your messages either as I am tired of your fallacious diatribe.

LoL. You're the one who trotted out Hayt. You don't even know what's in it as I took you page by page. As for fallacies, you never actually said anything about what I wrote is fallacious. However, there is absolutely nothing more fallacious than this statement, here once more in all its glory:

Yes, but the right part of the equation makes it conservative again!  The only way it will be non-conservative is if you do this:

∇×E=0

When asked for clarification, you told me this is not a typo.

The statement that this equation is non-conservative is so astoundingly, blisteringly, wrong that it entirely sums up the position of the KVL-Always-Holder.
After all this, you still can't tell the difference between a conservative electrostatic field and a non-conservative magnetostatic field - it's hopeless to try to go beyond this and analyze what happens if one takes the derivative of this non-conservative field.

And this one,

Therefore the fields in the circuit are conservative, aren't they?

And you're a Ph.D? Of electrical engineering (I assume?)?!?!

[bsfeechannel]

Not pointing it out and hoping people missed it--perhaps expecting the curled e-field resulting from the changing magnetic flux to somehow stop at the perimeter of the inner ring--was the sole basis for this being a memorable, 'miind-blowing' demonstration.

The sole basis for the memorable mind blowing demonstration was to be sure that people would not pay attention when he demonstrated in his very lecture 16 @27:00 that the electric field outside the solenoid has no influence whatsoever over the probes (In the previous lecture 15, he went in more detail about this). And that's because inside any closed path that does not include the solenoid the induced electric field is, wait for it, wait for it, conservative.

And it is conservative because the magnetic field outside the solenoid is precisely dick.

Remember curl E = - dB/dt. If B = 0, curlE = 0.

What I like about Lewin's super demo is how he covered all the bases. He left no escape route for those who were not prepared. After his demo you either grow a pair and learn this thing properly or you'll be immediately identified as a dilettante.

[thinkfat]

The funny thing about this whole "energy conservative" stuff is, there has to be a source of energy in the first place, for the circuit to consume it. This source of energy can be either lumped or not. If the physical circuit allows for it, you can lump the energy source into a two-terminal component and put it actually into the circuit. In other circuits, like in Dr. Lewins experiment, you can not do it. You're forced to admit that the energy source is external to the circuit and has no discrete source, and this is where KVL stops working.

The reason is simply because such circuit is not energy conservative. The energy source is not IN the circuit. This breaks the very foundation on which KVL is based.

PS: before anyone gets any funny ideas: No, Scalar PD is not the savior of KVL here. You can calculate a theoretical "half-way" voltage between the two points where the Oscilloscopes are connected in Dr. Lewins circuit, but you cannot replace the EMF with a voltage source connected to those points without disturbing the results.