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

[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: $\int_A^B dl = emf/2 = -1/2 d\phi/dt = 0.5V$.