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Author Topic: Marinov Generator  (Read 24377 times)
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... If you mistakenly put (v×∇).A when you meant to put the convective term (v.∇)A that I have banged on about then I still don't see why you expect the Hall sensor might detect the A field.

You're right. It's (v.∇)A that I wanted to talk about. By dint of jumping from the vector product of Lorentz force to the vector potential, we end up mixing the formulas.  :-[
However, the idea remains a good one.

Unlike a disc where the speed of electrons is obtained by its rotation, in a Hall effect sensor it is an input current that makes the speed.
But as in a Faraday disc where a transverse voltage (radial) is measured, it is this transverse voltage that the component outputs.
The transverse voltage is due to the Lorentz force according to the classical explanation, and to the force related to the spatial gradient of A in the explanation by the vector potential.

This spatial gradient of A must therefore produce the same effect as the Lorentz force where it exists and B does not, and consequently influence the Hall effect sensor outside of a magnetic field, which is to be verified.



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You're right. It's (v.∇)A that I wanted to talk about. By dint of jumping from the vector product of Lorentz force to the vector potential, we end up mixing the formulas.  :-[
However, the idea remains a good one.

Unlike a disc where the speed of electrons is obtained by its rotation, in a Hall effect sensor it is an input current that makes the speed.
But as in a Faraday disc where a transverse voltage (radial) is measured, it is this transverse voltage that the component outputs.
The transverse voltage is due to the Lorentz force according to the classical explanation, and to the force related to the spatial gradient of A in the explanation by the vector potential.

This spatial gradient of A must therefore produce the same effect as the Lorentz force where it exists and B does not, and consequently influence the Hall effect sensor outside of a magnetic field, which is to be verified.
OK but I must point out that a primary feature of (v.del)A is its longitudinal components that are absent in vXB,  And the longitudinal induction will not create the Hall effect.  Still worth doing though.
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OK but I must point out that a primary feature of (v.del)A is its longitudinal components that are absent in vXB,  And the longitudinal induction will not create the Hall effect.  Still worth doing though.
Smudge

So what is for you the equivalent of vXB for A?

For me: there is no longitudinal induction in the Faraday disk, only a transverse one due to the relativistic effect of length contraction, because there is no length contraction when the observed length is colinear to the speed vector. The equivalent of vXB for A comes from the Lorentz transformation of the electromagnetic 4-vector which combines A and the scalar potential.
It is named Aµ. Aµ=(φ/c,Ax,Ay,Az). Each coordinate of Aµ are function of the space-time position (ct,x,y,z) we are looking at. Aµ is transformed into A'µ as follows:

|φ/c|    |γ -γβ  0 0 | | 0 |
|A'x | = |-γβ  γ 0 0 | |Ax|
|A'y |    | 0  0  1  0 | |Ay|
|A'z |    | 0  0  0  1 | |Az|

At the beginning, we have no scalar potential in the referential at rest: Aµ=(0,Ax,Ay,Az). Suppose that A is only along x, so Ay=Az=0. You obtain φ/c=-γβAx and A'x=γAx. We are interested here in φ/c which is the scalar potential that appears in the referential of the moving charge from the relativistic effect.
When the velocity v is different at different places in space, as is the case along the radius of a rotating Faraday disc, then φ/c=-γβAx changes with the speed which is included in β=v/c and increases from the center of the disc to the rim. Therefore we obtain a potential difference along the radius, which corresponds to the electric field E=vXB that you see with the Lorentz force.

About the longitudinal induction:
in absence of a scalar potential, a longitudinal induction (say along x) means a longitudinal potential difference viewed by the moving observer (the charge) 𝝯φ/c = γAx1-γAx20, or γAx(ct1,x1,0,0)-γAx(ct2,x2,0,0)0, i.e. that Ax has not the same value at both positions, which implies either a time dependent variation of A (classical induction), or a longitudinal spatial gradient of A along x, or a not constant speed of the observer, the charge (this last case is to be verified with GR, as the observer's referential is no more inertial).

« Last Edit: 2019-06-14, 11:44:27 by F6FLT »


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So what is for you the equivalent of vXB for A?
I am not well versed in using the determinant form of vectors to it will take me some time to convert your 4 vector into something meaningful for me.  My version is in equation A2 of my paper in Reply #51 of this thread.
Quote
About the longitudinal induction:
in absence of a scalar potential, a longitudinal induction (say along x) means a longitudinal potential difference viewed by the moving observer (the charge) γAx1-γAx20, or γAx(ct1,x1,0,0)-γAx(ct2,x2,0,0)0, i.e. that Ax has not the same value at both positions, which implies either a time dependent variation of A (classical induction), or a longitudinal gradient of A along x, or a not constant speed of the observer (the charge).
I don't disagree with that and the non curl A fields can have just that longitudinal gradient of A along x.  Those longitudinal gradients appear in the convective derivative (v.del)A.  But they don't appear in the (vXB) term.  My take on this is that the curl function creates a field where you can't get a longitudinal effect.  So they don't appear in the Faraday disc.  But in a non-curl A field they do exist.  And it is up to us to find that effect.
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I am not well versed in using the determinant form of vectors to it will take me some time to convert your 4 vector into something meaningful for me.  My version is in equation A2 of my paper in Reply #51 of this thread.I don't disagree with that and the non curl A fields can have just that longitudinal gradient of A along x.  Those longitudinal gradients appear in the convective derivative (v.del)A.  But they don't appear in the (vXB) term.  My take on this is that the curl function creates a field where you can't get a longitudinal effect.  So they don't appear in the Faraday disc.  But in a non-curl A field they do exist.  And it is up to us to find that effect.
Smudge

The advantage of using relativity in electromagnetism is that it is complete, because we are really in a 4D universe, and contrary to what we would think, it also makes it easier to imagine qualitative effects.

By using 4-vectors, no more approximations or oversights are made.
For example, a movement relative to the magnetic vector potential will show you a scalar (electric) potential. But also you will no longer see the same vector potential.
However, the electrical potential, as well as the magnetic vector potential, change in a covariant way, hence the interest of treating them together in a formalism that supports it, rather than separately.

For example, we know that the field is derived from a potential. In a 4D space, a vector, like the potential vector, has 4 coordinates.
If we want to move to the fields, we have 16 gradients, one for each difference between each of the 4 coordinates of the 4-vector potential at one point in space-time and each of the 4 coordinates of the 4-vector potential at another point in space-time.
So we get a 4x4 matrix, and this one really gives us _all_ the electromagnetic information at one point in space-time, it's the electromagnetic tensor.

It is noted Fµν=∂µAν - ∂νAµ, according to Einstein's convention. And this gives us the fields we are familiar with:
Fµν=
| 0      -Ex/c  -Ey/c  -Ez/c |
| Ex/c   0      -Bz      By    |
| Ey/c   Bz     0       -Bx    |
| Ez/c  -By     Bx      0      |

If we see so many contradictory opinions in electromagnetism, it is because traditional formalism is insufficient.
Special relativity is not new to me, but its use in electromagnetism, yes. I am trying to familiarize myself, and I regret that electromagnetism was not taught to me on this basis, it is really elegant, complete and uncompromising.



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For anyone interested in here is my latest paper looking into motional induction from the magnetic vector potential A field.  By taking account of both the change of vector amplitude and the change of vector orientation with distance it is possible to obtain classical E = (v × B) motional induction directly from the time-rate-of-change of the magnetic vector potential A. When this procedure is applied to a non-curl A field where B is zero it predicts both a longitudinal {equations 8 or 11} and a transverse {equations 10 or 12} induced force on moving charge.
Enjoy!

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Hi Smudge

I find your paper fascinating, can I pass a question by you as what you think would be the outcome in the following:-

A coax cable which has the center core of iron with a coating of copper, a dielectric insulator and then a copper screen.
The coax is bent to form a loop or multiple loops (coil) where the ends are magnetically and electrically looped together.
So we now have a ferromagnetic core and an inductor/capacitor with an insulating dielectric between the inner and outer copper conductors.
Around the now toroidal looking loops we wind a secondary coil or coils (our output) at 90º.
To the two copper conductors of the coax, we supply an alternating HF current which is at the natural resonant frequency of the capacitive/inductive coax coil/ toroid.

I am thinking of the magnetic fields interreacting with the capacitive charge (a possible magnetic or electron accelerator which would induce charge into the secondary coil).

Thanks in advance

Regards

Mike 8)
ADDED
PS the whole has a ferromagnetic core, the inner conductor
« Last Edit: 2019-07-28, 17:27:15 by Centraflow »


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Hi Smudge

I find your paper fascinating, can I pass a question by you as what you think would be the outcome in the following:-

A coax cable which has the center core of iron with a coating of copper, a dielectric insulator and then a copper screen.
The coax is bent to form a loop or multiple loops (coil) where the ends are magnetically and electrically looped together.
Not sure what you mean here.  Are the inner and outer ends connected so we have a shorted turn of it iron wire that is surrounded by a shorted turn of the coax outer?  Please clarify.

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Not sure what you mean here.  Are the inner and outer ends connected so we have a shorted turn of it iron wire that is surrounded by a shorted turn of the coax outer?  Please clarify.

Smudge

Sorry I was not clear.

Take a piece of coax which has the center core made of iron but coated with copper (that is then a ferromagnetic core and a conductor. Make multiple turns and connect the center core ends together, then the screen to screen ends together so as both inner and outer wires are independent. We would now have both a capacitor and a coil which has a magnetic core.

Hope that explains a bit better :)

Regards

Mike 8)


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OK Mike,
I get it now.  The coax outer will be a shorted turn to the secondary coil wound over it.  Thus the field from the secondary induces eddy currents around the coax and those eddies will cancel out the field thus acting as a magnetic shield preventing the inner core from being magnetized.  Only at very low frequency will that inner core work as an inductor to get any excitation.  And then it can't be resonant.
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OK Mike,
I get it now.  The coax outer will be a shorted turn to the secondary coil wound over it.  Thus the field from the secondary induces eddy currents around the coax and those eddies will cancel out the field thus acting as a magnetic shield preventing the inner core from being magnetized.  Only at very low frequency will that inner core work as an inductor to get any excitation.  And then it can't be resonant.
Smudge

Hi Smudge,

ok, both the inner (copper coating on the iron) is a shorted turn and the outer is not shorted turn so they are like the two plates of a capacitor, in my case 1.22nF, and 0.415mH inductance on the outer screen if not shorted, the inner of course is a shorted turn. The resonant frequency is 223.7kHz.

Around this wound at 90º is the secondary, question is will the secondary see the primary? will the iron center core have eddy currents? what will happen to the capacitive charge in relation to the magnetic field of the core?

Can you see what I am getting at?

Regards

Mike 8)


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"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

As a general rule, the most successful person in life is the person that has the best information.
   
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Can you see what I am getting at?

Regards

Mike 8)
Not really.  Can you show a sketch or a picture of the set up please?  Just the single loop of coax would do to show me where the input is fed.

Smudge

 
   

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Smudge

see your PM's

Regards

Mike 8)


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"All truth passes through three stages. First, it is ridiculed, second it is violently opposed, and third, it is accepted as self-evident."
Arthur Schopenhauer, Philosopher, 1788-1860

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Here are a few images showing the A field external to a core.  If the flux in the core is varying with time, for any closed circuit that encloses the core we of course get an E field driving current around that circuit.  For a closed circuit that does not enclose the core there are still E fields, but they do not drive current around the circuit because they cancel out wrt to voltage induction.  But they can polarize the circuit, produce surface charges at different parts of the circuit.  Could this be used to advantage?

For constant flux the A field doesn't vary with time, but electrons moving through that non-linear A field can "feel" a time changing A and therefore an E field.  That E field at any point is proportional to the electron velocity.  In any closed circuit, if electrons travel at constant velocity the sum of any induced voltage is zero, there is no motion-induced voltage to maintain any current.  But there are some parts of the loop where the motion-induced E field supports the current and other parts where it opposes the current.  Again this will create surface charge on the conductor.

In the Marinov generator the motion induced voltage across the slip-ring, coming from relatively high slip-ring velocity, is not negated by the motion-induced voltage from the rest of the closed circuit, because there the velocity is only trivial drift velocity.  So the generator can drive current through a load.  And since it doesn't work in reverse as an electric motor it should be inherently OU.

Smudge

   
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