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Author Topic: Homopolar Generator, a new one?  (Read 252 times)
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Posts: 235
Hi All,

I would have liked to experiment a lot more on this homopolar generator, to do things cleaner and better measured, but I'm too bad at mechanics. I hope it's a new model, and not known for a century as the last "good idea" I had!  :-[  :)

Two years ago, but 3 or 4 years after my first experiments where I saw this effect, I discussed the subject on a physics usenet forum (in French here), but nothing conclusive came out.
After at least 6 years, I think it's time to entrust this to better experimenters and analysts than me. The experiment is very easily reproducible and the voltage can be measured beyond any doubt (several hundreds mV).

I summarized my experiment  in the pdf here (I prefer to leave it there rather than post it to be able to rework it and have the updated version at this link).
I invalidated the hypothesis of a "Faraday disk" type of operation, and included an operating hypothesis of which I am not sure, a force on electrons with spin oriented, in a magnetic field gradient.


It's all in the pdf but for those who want an overview first, here's the principle:

Along a ferromagnetic conductive axle rotating in front of an axially magnetized coaxial cylindrical magnet, a potential difference is measured between 2 sliding contacts, one close to the magnet and the other further away. The direction of the PD depends on the direction of rotation and the polarity of the magnet. The order of magnitude of the PD is about half that obtained by a Faraday disc the size of the magnet, driven by the same motor.

Here is the diagram:


Here is my experiment where you can see the poverty of my mechanical skills:


Thank you for your feedback


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"Chance favours only the prepared mind."  Louis Pasteur
   

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How does it behave when the spinning rod is replaced with a Copper, Aluminum or Brass rod ?
   
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How does it behave when the spinning rod is replaced with a Copper, Aluminum or Brass rod ?

I just tested with Al, it's mentioned in the pdf. No significant current could be measured (< 100µV). I think there should be a current, but too weak because the magnetic field lines are not "attracted" by the rod. So not only the magnetic field is much weaker, but the field gradient is even weaker.


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"Chance favours only the prepared mind."  Louis Pasteur
   
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You can create a homopolar motor or generator using a cylinder instead of a disc.  If magnetic flux lines pass through the cylinder wall (i.e. have a radial component) then rotation of the cylinder will induce a potential along the cylinder.  Your ferromagnetic rod will have radial flux lines along the surface as the flux from the magnets leaks out sideways.  Then your results are simply explained by the v X B induction.  Just a thought.
Smudge
P.S.  If you have the ability to measure DC microvolts you wouldn't like to try the experiment without the rotation would you?  If you measure anything then it could be that magical dragging of spin polarized electrons through a magnetic field gradient.
   
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You can create a homopolar motor or generator using a cylinder instead of a disc.  If magnetic flux lines pass through the cylinder wall (i.e. have a radial component) then rotation of the cylinder will induce a potential along the cylinder.  Your ferromagnetic rod will have radial flux lines along the surface as the flux from the magnets leaks out sideways.  Then your results are simply explained by the v X B induction.  Just a thought.
Smudge

I don't think so in our case, because how would you explain that "if the sliding contact near the magnet is not on the periphery of the axis, but on the end of the axis, no difference in the order of magnitude of the voltage is observed"?


Quote
P.S.  If you have the ability to measure DC microvolts you wouldn't like to try the experiment without the rotation would you?  If you measure anything then it could be that magical dragging of spin polarized electrons through a magnetic field gradient.

I have a HP 3468A that can measure µV. The problem here is that the sliding contacts are noisy, as well as the motor that is powered by the mains and vibrates. Under 1 mV, the voltage is not stable so that my measurement with an aluminium axle was not significant.

Without rotation as you suggest, the test is feasible. I'll try tomorrow. But if the electrons are dragged through the gradient, they must also escape the gradient to be able to loop the circuit and form a current, otherwise the forces will balance them at some point, and therefore no current and no voltage will be observed. How do you think we could get around the difficulty?


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"Chance favours only the prepared mind."  Louis Pasteur
   

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Thanks for sharing your find F6FLT :)    I'll have to try replicating over here when I get a chance.

I wonder if the effect is more related to the surface velocity or to the RPM?  If it is Lorentz-like induction, then a larger diameter cylinder should have a higher gradient due to the higher surface velocity..
   
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Thanks for sharing your find F6FLT :)    I'll have to try replicating over here when I get a chance.

I wonder if the effect is more related to the surface velocity or to the RPM?  If it is Lorentz-like induction, then a larger diameter cylinder should have a higher gradient due to the higher surface velocity..

Hi Reiyuki,

For me, it doesn't depend on the diameter, as the sliding contacts can be made at the center of the ends instead of the cylinder surface.
With the configuration shown in your picture, and for the same magnet size, the gradient in the cylinder will be surely less than with a thin axis (but should work).

With a much stronger neodimium magnet, I had less voltage because the diameter was half that of the ferrite magnet. The gradient is here the most important thing.



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"Chance favours only the prepared mind."  Louis Pasteur
   
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Posts: 1069
Without rotation as you suggest, the test is feasible. I'll try tomorrow. But if the electrons are dragged through the gradient, they must also escape the gradient to be able to loop the circuit and form a current, otherwise the forces will balance them at some point, and therefore no current and no voltage will be observed. How do you think we could get around the difficulty?
If the external circuit is not ferromagnetic, say copper (which it likely is) then there is no spin polarization there and the gradient has no effect on that external circuit.  In going from Fe to Cu the spin polarization decays over a very small distance like micro-meters.  However that means there are two dissimilar metal contact points and although the (thermocouple) Seebeck effects cancel there is such a thing as the magnetic Seebeck effect where the Seebeck voltage is affected by a magnetic field.  Maybe that magnetic Seebeck effect exactly cancels the voltage gained through the Fe.
Smudge
   
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If the external circuit is not ferromagnetic, say copper (which it likely is) then there is no spin polarization there and the gradient has no effect on that external circuit.

I've tried.  I let the HP 3468A heat for a long time to stabilize it. I fixed an enamelled copper wire near one end of the ferromagnetic rod and sent it back along the rod to the other end, where I connected the voltmeter which displays 10 µV +/-1, without a magnet. It is difficult to say where this gap comes from, it depends on the contacts, but it was stable.
 
When I approach the magnet, the voltage fluctuates by several tens of µA (induction), then it stabilizes at the same value as without the magnet. Reverse polarity of the magnet does not change anything.
 
Update 2019-02-26
Contrary to my first idea, the force on a magnetic dipole should be proportional to B, not H, even for an electron inside a ferromagnetic material. So I looked for another explanation for the null result.
When an electron moves through the ferromagnetic material due to the field gradient and reaches the copper wire through which it can escape, there is another field gradient in the opposite direction, due to the fact that copper does not concentrate the magnetic field unlike a ferromagnetic material. Thus, the field outside is much less intense than the field inside, reversing the direction of the gradient. The electron gets stuck somewhere in between.

The idea of replacing copper wire with ferromagnetic wire would not change anything. The gradient along one half-circuit will always be the opposite of the gradient of the other half-circuit. The electrons will move in both half-circuits to the area of the highest gradient until the Coulomb force balances them.

We are therefore obliged to rotate the axis to obtain the imbalance causing the current.




« Last Edit: 2019-02-26, 10:46:22 by F6FLT »


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"Chance favours only the prepared mind."  Louis Pasteur
   
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