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.
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?
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?
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.
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.
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?
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
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.
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.
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?
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.
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.
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:
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.
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.
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?)
Therefore the fields in the circuit are conservative, aren't they?
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.
We should remember that the electrostatic potential V is a conservative field; the magnetic scalar potential Vm is not a conservative field.
If B is not a function of time, (5) and (6) evidently reduce to the electrostatic equations
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.Quote(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.QuoteTherefore 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_PotentialQuoteYou 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,QuoteWe 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),QuoteIf 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?
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).
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.
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)
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:QuoteIn 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,QuoteIn 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?QuoteThe 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.
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)
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,QuoteThe 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.
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
Therefore the fields in the circuit are conservative, aren't they?
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.