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Author Topic: Confirming 90deg coupling to ferromagnetic wire.  (Read 1847 times)
Group: Experimentalist
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Why think that the output signal would be a consequence of a significant but doubtful variation of the magnetic wire or strip permeability with the current, when the simplest explanation is the capacitive coupling between the coil and the strip?
Was the experiment done with a shielded strip?






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If any coupling is 90º, perpendicular, between any two conducting materials then the coupling is, IMO, capacitive as F6FLT has stated.

Regards

Mike 8)


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Why think that the output signal would be a consequence of a significant but doubtful variation of the magnetic wire or strip permeability with the current, when the simplest explanation is the capacitive coupling between the coil and the strip?
Was the experiment done with a shielded strip?

If any coupling is 90º, perpendicular, between any two conducting materials then the coupling is, IMO, capacitive as F6FLT has stated.

Capacitive coupling is a good question, and thankfully it is just as easy to test as the original experiment.

The first-post experiment was re-created, except with an additional layer of aluminum foil between the nickel strip 'core' and the copper magnet wire helix (galvanically isolated of course).
There was no detectable change in response when the shield was left floating or grounded to any other component in the circuit.  Physical earth ground was also used to confirm this.
 ^-^ O0
« Last Edit: 2021-05-06, 02:08:55 by Reiyuki »


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Hi Reiyuki,

This was the right thing to do, but there is always a possible loophole when shielding, because the shielding is also capacitively coupled to the circuit.
So you have to connect each end of the shield to the ground with the shortest possible connections, as on the attached diagram which is the one you provided and which I modified. Is this what you did? Only after this test, if the measured levels are always of the same order of magnitude, we can be sure of a magnetic effect.




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"Open your mind, but not like a trash bin"
   

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This was the right thing to do, but there is always a possible loophole when shielding, because the shielding is also capacitively coupled to the circuit.
So you have to connect each end of the shield to the ground with the shortest possible connections, as on the attached diagram which is the one you provided and which I modified. Is this what you did? Only after this test, if the measured levels are always of the same order of magnitude, we can be sure of a magnetic effect.

There is a slightly diminished output when both shielded ends are shorted, but that is more likely caused by magnetic induction of the foil rather than electrostatic.  A ground-loop.
To an extent you can often 'feel' electrostatic vs magnetic effects on a bench circuit, as small things like lead positioning or moving your hands usually has a noticeable impact on output.

Another way to confirm predominantly magnetic action is with a partial ferrite core placed near the ferromagnetic wire, which results in a significant output as well (see attached). :)


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The decreased output is probably due to the fact that the shield is a shorted loop, magnetically coupled to the line, so it acts as a secondary winding of a shorted transformer, weakening the signal in the primary, i.e. the line.
To know if there is a real effect linked to the fact that the line is ferromagnetic, it is to replace it by a non-magnetic Cu or Al line, as suggested by Poynt99, and see if the difference is significant.

Classical electromagnetism tells us that there will be very little difference, because the ferromagnetic material here is not a core in the center of the winding but along the conductor, which it is.
If the difference is significant, then there is indeed more to investigate.


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One possibility for the induction being considered here is the conduction electrons being spin-polarized and being dragged along the Fe wire by a magnetic field gradient.  I have crudely modified the image posted in reply #12 to illustrate this.  Here the coil is not uniformly wound on the wire but the turns are bunched up towards one end to create non-uniform field.  The FE wire terminates just inside the coil where it is connected to Cu wire where the electrons lose their spin alignment and don't get a reverse force where the gradient reverses direction as the wire leaves the coil.  For a gradient field that changes magnitude by 1 Tesla along its length the expected voltage is only 5uV, so a pretty small affect, but maybe worth exploring.  Unlike inductive or capacitive effects the voltage is not a function of frequency and can exist even at DC.

Smudge
   
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One possibility for the induction being considered here is the conduction electrons being spin-polarized and being dragged along the Fe wire by a magnetic field gradient.
...

If this works in AC, it should also work in DC.
The wire could then be placed above a cylindrical permanent magnet, collinear with the axis of the magnet, so that we would have the field gradient due to a greater distance to the magnet for one end of the wire than for the other.

We could easily measure a current of 5 µV because it is now a direct current.

But why would electrons of opposite spin go in the same direction?


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If this works in AC, it should also work in DC.

The wire could then be placed above a cylindrical permanent magnet, collinear with the axis of the magnet, so that we would have the field gradient due to a greater distance to the magnet for one end of the wire than for the other.
Agreed, and I did say DC.
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We could easily measure a current of 5 µV because it is now a direct current.
I would agree for a current of 5µA, but this is a voltage not a current.  As such it is pretty useless as a power source unless we have circuits using superconductors.  And with DC, detecting 5µV is difficult on oscilloscopes unless you have additional amplification.  Itsu has a 1,000 times DC amplifier built by Graham Gunderson for me, then passed to him via Grumage, and that would enable anyone to see that 5µV DC.

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But why would electrons of opposite spin go in the same direction?
I did say spin polarized electrons and in the world of spintronics that means spins aligned with a magnetic field, so it depends on the degree of spin-polarization.  At a field that reaches 1 Tesla the spin-polarization would be high, meaning a greater proportion of positive spin than negative spin.  My 5µV was base on 100% polarization, and I should have said that.

Smudge
   
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Hi Smudge

I mistakenly used the word "current" instead of "voltage" I was thinking of. I have a HP3468A multi-meter, and the 5µV is at the limit of what it can do, but still visible. On the oscilloscope, it's unlikely that you'll be able to read anything correct.

In fact I think it unlikely that we would have a DC current, which would be a proof of perpetual motion without us knowing where the energy comes from.
I think that at the end of the iron wire where the field is strongest, we have a gradient opposite to the one established along the wire.
This inverse gradient is over a very short distance, at the end of the wire, at the point where it meets the copper wire. At the end of the iron wire, in fact, the field lines spread out again since the wire is no longer there to channel them.
The opposite effect of the two reverse gradients will prevent any current.


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Hi Smudge

I mistakenly used the word "current" instead of "voltage" I was thinking of. I have a HP3468A multi-meter, and the 5µV is at the limit of what it can do, but still visible. On the oscilloscope, it's unlikely that you'll be able to read anything correct.

In fact I think it unlikely that we would have a DC current, which would be a proof of perpetual motion without us knowing where the energy comes from.

If you treat the electron dipole as a changing current loop then the changing magnetic field as the electron moves through the field gradient induces a voltage into that current loop.  That voltage of of a polarity whereby the source of the loop-current delivers power.  If you treat the electron as a spinning spherical charge then the induced induced E field tries to slow down the spin, which of course it cannot do.  Whatever keeps the electron permanently spinning could be the energy source.

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I think that at the end of the iron wire where the field is strongest, we have a gradient opposite to the one established along the wire.
This inverse gradient is over a very short distance, at the end of the wire, at the point where it meets the copper wire. At the end of the iron wire, in fact, the field lines spread out again since the wire is no longer there to channel them.
The opposite effect of the two reverse gradients will prevent any current.
I disagree.  Within the copper the electrons are no longer spin-polarized (they lose their spin-polarization within a few Angstroms of the Fe-Cu interface) hence are not influenced by the presence of a field gradient.  Of course the Fe-Cu interface must occur at the magnetic field maximum, which is within the Fe.  That means drilling a hole in the Fe rod and ensuring that the Fe-Cu connection is at the bottom of the hole.

I think a more likely reason for zero induced voltage is the Magnetic Seebeck Effect, which I presume means the Seebeck coefficient changes value if a magnetic field is present.  Because of the Seebeck Effect the temperature must be the same at each end of the Fe rod.  Maybe the Magnetic Seebeck effect exactly cancels the induced voltage because the magnetic field isn't the same at each end of the rod.  Only experiment will tell us this.

Smudge

P.S.  The Marinov generator is a more likely candidate for "where does the energy come from?".  In my experiments I achieved 3 millivolts at 1000 RPM, and there is the possibility that E field radiation from electrons as they are accelerated from the brush onto the moving slip-ring similarly applies a load to the spinning electrons within the magnet. 
   
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...If you treat the electron as a spinning spherical charge then the induced induced E field tries to slow down the spin, which of course it cannot do. 

I agree. This is a consequence of the quantization of the magnetic flux.

Quote
Whatever keeps the electron permanently spinning could be the energy source.

No energy is needed to maintain motion, either linear or angular, only to overcome losses or to do work. No energy is required for electron spin.

Quote

I disagree.  Within the copper the electrons are no longer spin-polarized (they lose their spin-polarization within a few Angstroms of the Fe-Cu interface) hence are not influenced by the presence of a field gradient...

I agree, but that was not my point. I am saying that there is a reverse gradient at the end of the iron wire  (probably in copper too a bit, but that's not my point).
The field lines follow the wire. But when they reach the end, they meet the copper which cannot channel them anymore. The field lines will not continue to exit in the continuity of the wire, because the break in permeability is felt throughout the field line loop.  Indeed, the field lines use the "shortest magnetic path", i.e. the field at the end of the iron wire rearranges its topology according to the environment (this is what we see in the representations of rectangular permanent magnets: the field lines do not come out at the end but escape before :
 

The field lines then start to emerge with a larger and larger angle as we approach the end of the wire. So we will have a gradient in a small section before the end of the iron wire, and as the flux is conservative, this gradient will be identical to the other one along the wire and will compensate it exactly. I don't believe in the Seebeck effect here.



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"Open your mind, but not like a trash bin"
   
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