Motional EMF in Faraday disc, co-rotating magnet axial mean flux

[FusionJim]

So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday disc.



The Faraday disc voltage is an integral of the area swept by the rotor conductor in one full revolution. I understand the magnetic mean in the azimuthal direction to mean whether there is a net magnetic flux in a single direction within this area that is swept by the rotor conductor.
For example, if we had two stationary half ring magnets each having an opposite pole , then the disc would generate zero EMF as the two EMF's would create only eddy currents and cancel the total integral.


If we rotate this two pole magnet together with the disc then there are no eddy currents and also no EMF. Now there is no EMF not because the two opposite EMF contributions cancel but because the total field is equal in both directions and co-rotating.


This is the complicated part and also my question - what happens if we take a non axis symmetric magnet and rigidly co-rotate it with the disc in such a way that the disc surface area has a field in just one direction. Lets say for example a half ring magnet, it is attached to the disc and co-rotates with it.The flux is routed back outside the disc area. There is azimuthal mean flux through the disc, but the flux is co-rotating. Do I get an EMF in this case or no?




Thanks.

 

[anuttarasammyak]

In order to understand the configuration correctly, could you provide a sketch of your experiment?

 

[FusionJim]

In order to understand the configuration correctly, could you provide a sketch of your experiment?

Just imagine the usual toroidal ring magnet but cut in half. this half ring is rigidly attached to the disc and rotates with it. Assume the flux doesn't return back through the disc plane but instead loops around and comes back through the stationary loop part.

The red half in the drawing represents the magnet, blue arrows are flux direction and dark red line is stationary conductor and brushes that close the disc circuit.

I know from experience that if a non symmetric magnet rigidly co-rotates with the disc no EMF is generates , but in that case the magnet flux was looping back through each side of it, so the flux was cutting the disc twice, once fourth and once back. therefore the total flux the disc integrated was zero. But how it is when the flux only cuts once and doesn't come back through the disc?
Assuming a non symmetric magnet that co-rotates rigidly!

 

[anuttarasammyak]

Thanks for the explanation for confirmation I also draw a sketch.

A is a contact point of the disk. The path OA matters. When A is on white half disk edge, there is no B through line OA, emf V=0. When contact A is on red haldf disk edge, the pass OA could be on the B spots thus v X B 's should contribute to emf V.

 

[FusionJim]

Thanks for the explanation for confirmation I also draw a sketch.View attachment 365008
A is a contact point of the disk. The path OA matters. When A is on white half disk edge, there is no B through line OA, emf V=0. When contact A is on red haldf disk edge, the pass OA could be on the B spots thus v X B 's should contribute to emf V.

No , the magnet is not a toroid like that, it s a permanent ring magnet but cut in half, so closer to this.

 

[anuttarasammyak]

Thanks for the teaching and clarification. Now I guess:

1. In case that the metal disk is all white, i.e. no magnets at all, emf V=0

2. In case that the metal disk is all red, i.e. magnets all around, we will observe emf V $\neq$ 0.

3. In case that the metal disk is made of half red, half white and insulator band between them, V=0 for contact point A is on white and V $neq$ 0 when contact point A is on red. Here I have assumed no B leakage on white half.

4. In case that the metal disk is made of half red, half white with direct contact, "short circuit“ of current takes place within the disk, however, the case 3 stands here at least in quality though not in quantity. Here I have assumed no B leakage on white half.

How do you see it and what is your own guess ?

I found a open access paper, "Resolving the paradox of unipolar induction: new experimental evidence on the influence of the test circuit" saying
https://www.nature.com/articles/s41598-022-21155-x .
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To recall the phenomenon and associated paradox, an illustration is given in Fig. 1 where a magnet is situated next to a conducting disc, both of which are free to spin around their cylindrical axis. Usually, the following three cases are considered for discussion:

(i)The disc is rotating and the magnet remains stationary. A voltage can be measured across the radius of the disc.

(ii)The disc remains stationary and the magnet rotates. No voltage can be measured across the disc.

(iii)Both disc and magnet rotate at the same speed and direction, and again a voltage is observed across the disc.
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I hope you are interested in it too.