What is the relationship between voltage and flux in Faraday's Law?

[alan123hk]

Just to make sure this part is clear, the fellow represented the changing magnetic field of Professorr Lewin's apparatus with batteries as the EMF's. There is a significant DC current in the fellow's center loop, but this DC current makes only a small and constant magnetic field. The difference in the two voltmeters is due to the EMF's and resistors, directly analogous to Professor Lewin's changing magnetic field. Your explanation is very good, other than I think it would be much better and very exact if you deleted this phrase. It only can refer to Professor Lewin's closed loops.

You are right, I understand that he want to explain something, but the DC circuit he used in the demonstration and the electromagnetic induction circuit used by the professor are essentially different.

 

[Charles Link]

I am sorry but this video is silly. It sets up a "straw man" man and knocks it down. Prof. `Lewin was showing that two ordinary voltmeters connected to the same two points in a circuit can give wildly differing readings. This does not require some "Uri Geller" rigging of stage props as presented here.
Lewin presented a cautionary tale because such circumstances can often occur in real circuits particularly if any appreciable currents exist. Slavish devotion to Kirchhoff's laws and the lumped element model can leave you scratching your head muttering "WTF?". This is particularly true if there are ground loops. The fact that this became controversial is simply a measure of lack of understanding of the physics involved by folks who should know better.

That was very much my first interpretation of it, but please read my "Edit 3" of post 56. I do think the fellow put much careful thought in designing his apparatus.

 

[vanhees71]

It's an analogy, not more nor less. In both cases you have EMFs and no potential differences. Maybe the analogy helps to understand the important difference between the 2.

The one thing I don't understand in the video is the statement that Kirchhoff's Laws were violated, which of course is not the case. Kirchhoff's Laws follow directly from the Maxwell equations under the usual assumptions of the quasistationary approximation and both Lewin's example and the present video's one are analyzable using Kirchhoff's Laws.

 

[Charles Link]

@alan123hk Please see also post 52, which I think might have been overlooked, and adds a detail or two to the "Faraday effect" concepts.

The transformer that you mentioned back in post 46 contains some basic physics involving the Faraday effect. I think you might also find the transformer laminations of post 52 of interest. That seems to be one feature they don't emphasize enough in many of the write-ups involving transformers.

 

[hutchphd]

That was very much my first interpretation of it, but please read my "Edit 3" of post 56. I do think the fellow put much careful thought in designing his apparatus.

Yes but it is just silly. From what I see his analysis is correct and if you somehow mistook batteries for wires you in your circuit you might be surprised. But even then you could not recreate Lewin exactly. I do not see the point, and personally find it another vexing attempt to "explain" discrepancies that flow from lack of understanding. At least he didn't talk about "the Electric Field" between the plates of the batteries.

 

[vanhees71]

This argument I also didn't understand in the movie, namely the claim that the two voltmeters are attached to the same two points in the circuit by just ignoring the batteries, i.e., the batteries were interpreted just as ideal wires, which is of course precisely nonsense, because they are EMFs. The explanation thereafter was correct though. The claim that you cannot use Kirchhoff's laws is, however, not true, because of course you can use Kirchhoff's laws to analyze the circuit correctly and in fact that's what the movie maker does himself when saying that the voltages add/subtract to approx. 6V.

 

[Charles Link]

I think the fellow was probably imitating Professor Lewin, who I think claims Kirchhoff's laws don't work. What really is the case, and that's what Professor's Lewin's paradox is all about, is that $E_{induced}$ is a non-conservative field, i.e. $\oint E_{induced} \cdot ds \neq 0$. Thereby, when there exist non-zero $E_{induced}$, we cannot assign voltages to points in the circuit, and expect two voltmeters in different orientations to have the same reading.

 

[alan123hk]

The claim that you cannot use Kirchhoff's laws is, however, not true, because of course you can use Kirchhoff's laws to analyze the circuit correctly and in fact that's what the movie maker does himself when saying that the voltages add/subtract to approx. 6V

I guess this may be related to the following short film, please start listening at 16:40.

In any case, I believe that this is just a way for the professor to express it from a certain angle, and the purpose is to inspire our thinking. I personally believe that the Kirchhoff's laws can of course be applied to the analysis of any circuit.

The transformer that you mentioned back in post 46 contains some basic physics involving the Faraday effect. I think you might also find the transformer laminations of post 52 of interest. That seems to be one feature they don't emphasize enough in many of the write-ups involving transformers

https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ posts

This core Loss calculation techniques seems very useful. I will study it when I have time. Thank you for sharing.
But to be honest, as an electronic technician of consumer products, in the actual working environment, I rarely analyze core losses in depth, because the information about core loss of magnetic materials is usually provided by the manufacturer.

 

[Charles Link]

I guess this may be related to the following short film, please start listening at 16:40.

In any case, I believe that this is just a way for the professor to express it from a certain angle, and the purpose is to inspire our thinking. I personally believe that the Kirchhoff's laws can of course be applied to the analysis of any circuit.
https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ posts

This core Loss calculation techniques seems very useful. I will study it when I have time. Thank you for sharing.
But be honestly, as an electronic technician of consumer products, I rarely use these things in the actual working environment, because the information about core loss of magnetic materials is usually provided by the manufacturer.

I recommend that you learn and understand why it works. Did you ever look at the side of a transformer and see the laminations=layers? These are very necessary at preventing core loss because of the eddy currents created in the core by EMF's (by Faraday's law) from the ac magnetic field in the core. The insulation between the layers blocks the eddy currents, that basically would circulate in a direction opposite that of the current in the primary coil. Without these laminated layers, the core losses would be huge.

 

[alan123hk]

I recommend that you learn and understand why it works. Did you ever look at the side of a transformer and see the laminations=layers? These are very necessary at preventing core loss because of the eddy currents created in the core by EMF's from the ac magnetic field in the core.

I have seen this laminated structure, and I know it is very important to prevent the core loss because the eddy currents created in the core by EMF. But I usually use ready-made magnetic cores (EE, EI, etc.), and the manufacturers often provide detailed information, such as
https://product.tdk.com/info/en/catalog/datasheets/ferrite_mn-zn_material_characteristics_en.pdf

 

[Charles Link]

I have seen this laminated structure, and I know it is very important to prevent the core loss because the eddy currents created in the core by EMF. But I usually use ready-made magnetic cores (EE, EI, etc.), and the manufacturers often provide detailed information, such as
https://product.tdk.com/info/en/catalog/datasheets/ferrite_mn-zn_material_characteristics_en.pdf

A google of ferrites shows that they are electrically non-conductive, unlike iron, and thereby laminations may not be necessary for this material. Perhaps you can supply some additional detail. I may not be up-to-date on the current state of the art.

 

[alan123hk]

A google of ferrites shows that they are electrically non-conductive, unlike iron, and thereby laminations may not be necessary for this material. Perhaps you can supply some additional detail. I may not be up-to-date on the current state of the art.

Actually I don't know much about the current state of magnetic cores, but in my memory, I don’t seem to have seen a laminated structure method for iron powder core and ferrite core. However, I believe they also have eddy current loss, but relatively speaking, they are much less.

 

[Charles Link]

See https://en.wikipedia.org/wiki/Magne...cores are,or nonconductive cores like ferrite.
Looks like ferrites are only used on occasion.
and a quote from the article: "The current flowing through the resistance of the metal heats it by Joule heating, causing significant power losses. Therefore, solid iron cores are not used in transformers or inductors, they are replaced by laminated or powdered iron cores, or nonconductive cores like ferrite."

 

[alan123hk]

I continued to study the following electromagnetic induction circuit and found a very interesting thing. If I am not mistaken, according to my inference, the real voltage between point A and point B should be +0.4V. The voltage measured by the voltmeters on the left and right sides is different, that is, 0.9V on the right side and -0.1V on the left side. The main reason is actually caused by the induced voltage of the 4 connecting wires (L1, L2, L3, L4). of the two voltmeters (VM2 and VM3).

 

[Charles Link]

If you are referring to the electrostatic integral $V_{AB}=\int\limits_{A}^{B} E_s \cdot ds=+.4 \, volts$, I think I agree with that result.

 

[alan123hk]

If you are referring to the electrostatic integral VAB=∫Es⋅ds=+.4volts, I think I agree with that result.

Yes, it should be the electrostatic line integral between A and B = + 0.4V

 

[Charles Link]

Perhaps it is worthwhile to show our work for this:
On the path from top to bottom on the right side $\int E_{total} \cdot ds=\frac{\mathcal{E}_{total}}{2}+V_{AB}=.9 =IR_1$.
For the path from bottom to top on the left side $\int E_{total} \cdot ds=\frac{\mathcal{E}_{total}}{2}-V_{AB}=.1 =IR_2$.
Meanwhile $\mathcal{E}_{total}=1.0$, and the .9 and .1 come from the loop equation $\mathcal{E}_{total}=IR_1+IR_2$.
Solving, we get $V_{AB}=+.4$ volts, where $V_{AB}=\int\limits_{A}^{B} E_s \cdot ds$, with $E_s$ the electrostatic component of the electric field.

Because the $E_{induced}$ is in the circular direction in this problem, it appears that a voltmeter whose resistor is placed along the line from A to B passing through the center of the circle, (with only the electrostatic component along the line from A to B), would indeed read +.4 volts, as I think @alan123hk was pointing out in post 74.

 

[arydberg]

Thanks for your reply.

My idea is that from the front view, there are two loops connected to the voltmeter, the EMF induced on the right side seems to be +0.5V, and the EMF induced on the left side seems to be -0.5V. There seems to be a contradiction, maybe there is no contradiction at all. I think it is possible to add two values. So the answer should be that the maximum chance is 0V.

Is my inference correct?

I agree

 

[rude man]

I agree

Take one half of the ring, say the right side if current is clockwise. Call the points A (top) and B (bottom). Then $\int_A^B \bf E \cdot \bf dl$ = +0.5V.
But that is not the whole story.
The meter circuit includes (is pierced by) flux $= -(1/2) d\phi/dt = -emf/2$ = -0.5V. The area is half the coil's area.
So the emf along the ring half cancels the emf induced in the meter loop. You should convince yourself that the signs of $\int_A^B$ and $d\phi/dt$ are opposite, not additive.
You can do this with either half of the ring and the same thing applies: $\int_A^B dl = emf/2 = -1/2 d\phi/dt = 0.5V$.