PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2019-07-22, 04:31:55
News: If you have a suggestion or need for a new board title, please PM the Admins.

Pages: 1 2 [3] 4 5 6 7
Author Topic: Marinov Generator  (Read 22606 times)
Hero Member
*****

Posts: 591
Hi Smudge,

In the case of Marinov, the field lines of A must be difficult to define mathematically. Even with a magnetic field confined in a circular torus, the field lines of A must be something like ellipses, and if we have to make elliptical integrals, we are not out of the woods...

So to move forward step by step, and to verify that the potential vector approach is possible, I am trying to describe the Faraday disc in terms of potential vector, without using the magnetic field.
This is surely possible because the circular symmetry of A allows us easy integrals.

Indeed, if we can describe Faraday's disc without using the magnetic field, we obtain the proof of concept of a motor/generator with B=0 and A≠0, and the door is also open for many new innovative homopolar systems in addition to Marinov's.

Have you already done it for Faraday's disk, or can you try to do it? That's what I'm trying to do, but I'm not doing it yet.




---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
Hi F6,

I have looked into using the convective derivative of the vector potential A field to describe the Faraday disc induction but came to an abrupt halt part way through because it predicts half the voltage obtained from the usual v X B induction.  I am posting the unfinished document here.  I also post another unfinished document on the Marinov generator where the item of interest is the appendix which shows that using (v.del)A=delA(v.A)-v X B has some hidden subtleties that may explain the discrepancy.  You will see from my ramblings that I don't fully understand differentiation of vectors.
Smudge
   
Group: Moderator
Hero Member
*****

Posts: 1161
Hi Smudge,
In the case of Marinov, the field lines of A must be difficult to define mathematically.
They can be derived quite easily using finite element programs.  I use the free FEMM for simulations which admittedly is only a 2D model but can yield useful results.  Because of the math similarity governing H field lines around a conductor carrying current and A field lines around a core carrying flux you can use the former to predict the latter.  The A field lines in my paper "On Electrodynamic Formula" were obtained from FEMM that way, I simply used the H field plots around currents, where 1 amp represents 1 Weber and the resulting amps/meter are read as Weber/meter.   FEMM also does the integration for you along selected contours.

Smudge
   
Hero Member
*****

Posts: 591
Hi F6,

I have looked into using the convective derivative of the vector potential A field to describe the Faraday disc induction but came to an abrupt halt part way through because it predicts half the voltage obtained from the usual v X B induction.
...

Hi Smudge,

Thank you for the documents, excellent work.
For Faraday's disc, I get the same result and I don't see why we have this 1/2 factor in the solution with A.
We absolutely have to solve this. If we don't have the correct solution by A in the simple case of Faraday's disc, we have missed something big, and we risk failing in the more complex problems.

I wonder if the spatial gradient of A should be explicitly used. But I'm progressing very slowly and you seem to be well ahead of me, so go on with your idea.

Concerning the calculation of elliptical integrals, I meant to find the analytical solution. Of course we can do the numerical calculation by software.



---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
Having given more thought to the discrepancy when using the convective derivative of the A field to calculate voltage from the Faraday disc homopolar generator I have arrived at a new perspective.  See my new paper here for my definition of the E field.
Smudge
   
Hero Member
*****

Posts: 591
Having given more thought to the discrepancy when using the convective derivative of the A field to calculate voltage from the Faraday disc homopolar generator I have arrived at a new perspective.  See my new paper here for my definition of the E field.
Smudge

Hi Smudge,

I said, "I wonder if the spatial gradient of A should be explicitly used," and here's why.
In your work, you write "plus another term that takes into account the evolution of A as seen by an electron moving through a non-uniform A field". I think there is a weakness here because the analysis must be done in only one referential frame, usually the observer's.

When using E=-∂A/∂t, it is the time variation seen by the observer that sees the force F=q.E on the charge. A and E are related to the observer, not the charge. The charge does not see the same A because it is mobile, and worse, because it is accelerated, i.e. its reference frame is not even inertial.
The spatial gradient of A, if it is to be introduced, must be that seen by the observer, not the charge which can't see a spatial gradient because it is assumed to be point, it can only see a temporal gradient, the electric field. If a spatial gradient gave rise to an electric field, then a charge placed far from an infinite (very long) solenoid would be attracted to it, but this is not what we observe.

I think the problem turns more on how to integrate on a loop when the circuit of this loop has parts that move relative to each other. We cannot assume that the electrons will just follow the radius of the disc between the sliding contacts. The electrons are forced to move tangentially because of the rotation of the disc, so they will follow a non-radial path statistically longer than the radius. I admit I don't know how to translate that into the equations.


---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
Hi Smudge,

I said, "I wonder if the spatial gradient of A should be explicitly used," and here's why.
In your work, you write "plus another term that takes into account the evolution of A as seen by an electron moving through a non-uniform A field". I think there is a weakness here because the analysis must be done in only one referential frame, usually the observer's.

And if the observer is moving through a non-uniform A field he will see a temporal gradient as his local A field changes value.  Or are you suggesting that the moving observer will not see the same A field, his A field will be different?  I could agree with you if the source of the A field, which is itself some moving charge, is moving at low velocity and you invoke relativity, i.e your observer’s movement alters the relative motion between source and observer.  But our A fields come from spinning or orbiting electrons, and their velocities are so tremendous that our puny movement is immaterial.  That is why in the Faraday homopolar machine it matters not whether the magnet is fixed or moves with the disc.  Spinning the magnet at our puny RPM’s does nothing to the relative motion between source and observer.

Quote
If a spatial gradient gave rise to an electric field, then a charge placed far from an infinite (very long) solenoid would be attracted to it, but this is not what we observe.

Why would the charge be attracted to the solenoid?  The A field forms circles around the solenoid axis, and for the infinitely long solenoid that field is curl free, there is no B field there.  Yes the circular A field increases in magnitude as you move closer radially, but that seen as a temporal increase is a transverse force on the moving electron, and that is not an attractive force.

Quote
I think the problem turns more on how to integrate on a loop when the circuit of this loop has parts that move relative to each other. We cannot assume that the electrons will just follow the radius of the disc between the sliding contacts. The electrons are forced to move tangentially because of the rotation of the disc, so they will follow a non-radial path statistically longer than the radius. I admit I don't know how to translate that into the equations.

I think that spiralling action has led people to believe that the homopolar disc machines are OU.  My Marinov generator does not have that problem, and unlike other claims for OU I can show where the energy comes from.  I can show the possibility for the anomalous energy to be obtained from the PM’s spinning or orbiting electrons responsible for its magnetization, the machine tries to slow down those spins.

Smudge
   
Hero Member
*****

Posts: 591
And if the observer is moving through a non-uniform A field he will see a temporal gradient as his local A field changes value.  Or are you suggesting that the moving observer will not see the same A field, his A field will be different?    I could agree with you if the source of the A field, which is itself some moving charge, is moving at low velocity and you invoke relativity, i.e your observer’s movement alters the relative motion between source and observer.  But our A fields come from spinning or orbiting electrons, and their velocities are so tremendous that our puny movement is immaterial... 

The question of the origin of A is not relevant, because the potential vector is, like a field, a local property of space at the position of the charge. If the charge is one light year from the source of A, and the source of A is anihilated, the charge will continue to see A for one year, and if it starts moving in A, it will no longer see the same A. So it doesn't matter which source of A. In relativity A is a 4-vector to which the Lorentz transforms apply, and the change is quite simple by matrix calculation:

|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|

We even see, which surprised me at first, that a scalar potential φ can appear in the charge referential when it moves in a place where there is only the vector potential. And vice versa.

Quote
Why would the charge be attracted to the solenoid?  The A field forms circles around the solenoid axis, and for the infinitely long solenoid that field is curl free, there is no B field there.  Yes the circular A field increases in magnitude as you move closer radially, but that seen as a temporal increase is a transverse force on the moving electron, and that is not an attractive force.

Around the solenoid, the equipotentials of A are concentric circles that weaken proportionally with distance.
If the spatial gradient has an effect on the charge, it is because dA/dx results in an electric field. In our particular case, the differential can only be made on an axis transverse to A, i. e. x is radial. Unlike t, x is not a simple scalar, it has a direction. I don't see what makes you think that the E field related to dA/dx should be along A.

Quote
I think that spiralling action has led people to believe that the homopolar disc machines are OU.  My Marinov generator does not have that problem, and unlike other claims for OU I can show where the energy comes from.  I can show the possibility for the anomalous energy to be obtained from the PM’s spinning or orbiting electrons responsible for its magnetization, the machine tries to slow down those spins.

Smudge

I'm not that enthusiastic. I note that the effect behind Marinov's device is much weaker, at equal field strength, than Faraday's. So either we have an artifact due to a residual field, or the effect is real but not related to the first power of the spatial gradient.

However, the Marinov's idea raises a much more interesting question, to which a positive answer opens up much more prospects than the Marinov generator alone: can we create a homopolar generator with mobile conductors through a non-zero vector potential but a zero magnetic field B?
I am preparing a schematic of an experiment to prove it, it will be ready today, maybe even this morning.



---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
The question of the origin of A is not relevant, because the potential vector is, like a field, a local property of space at the position of the charge. If the charge is one light year from the source of A, and the source of A is anihilated, the charge will continue to see A for one year
I fail to see how this affects the argument.  We are dealing with systems where the distance between the source and the charge is so small that the propagation time is negligible.
Quote
and if it starts moving in A, it will no longer see the same A.
Explain why it will no longer see the same A.  You quote the Lorentz transform which invokes beta and gamma but since we are dealing with electron velocities that are small compared to c, beta tends to zero and gamma tends to unity.  So IMO we can forget the Lorentz transform, it has negligible effect and to all intents and purposes the moving electron sees the unmodified A.
Quote
Around the solenoid, the equipotentials of A are concentric circles that weaken proportionally with distance.
If the spatial gradient has an effect on the charge, it is because dA/dx results in an electric field. In our particular case, the differential can only be made on an axis transverse to A, i. e. x is radial. Unlike t, x is not a simple scalar, it has a direction. I don't see what makes you think that the E field related to dA/dx should be along A.
Inside the solenoid the equipotentials of A are also concentric circles but they strengthen proportionally with distance.  This also applies to the A field in the Faraday disc.  As you well know the force on any radially moving electron there is not radial, it is always transverse to the movement.
Quote
I don't see what makes you think that the E field related to dA/dx should be along A.
For E=-dA/dt the E is along A.  If you examine the components of (v.del)A the E is along A.
Quote
I note that the effect behind Marinov's device is much weaker, at equal field strength, than Faraday's.
Not true.  What equal field strength are you using?  The Marinov generator discussed in this thread has zero B field at the moving electrons.
Quote
However, the Marinov's idea raises a much more interesting question, to which a positive answer opens up much more prospects than the Marinov generator alone: can we create a homopolar generator with mobile conductors through a non-zero vector potential but a zero magnetic field B?
IMO the Marinov generator is just that.
Quote
I am preparing a schematic of an experiment to prove it, it will be ready today, maybe even this morning.
I look forward to seeing it.

Smudge
   
Hero Member
*****

Posts: 591
I fail to see how this affects the argument.  We are dealing with systems where the distance between the source and the charge is so small that the propagation time is negligible.

This argument was just related to your comment: "But our A fields come from spinning or orbiting electrons". A potential as well as a field is a tool to avoid to discuss the question of the source and of the propagation of an effect from a source.
If we deal with the potential vector, the question of the source is off topic.

Quote
Explain why it will no longer see the same A.  You quote the Lorentz transform which invokes beta and gamma but since we are dealing with electron velocities that are small compared to c, beta tends to zero and gamma tends to unity.  So IMO we can forget the Lorentz transform, it has negligible effect and to all intents and purposes the moving electron sees the unmodified A.

Electromagnetism effects are relativistic effects. It is a common mistake to think that the low electron velocity could be negligible while it is the only cause of the magnetic field. I found the best explanation is the book "Claustrophobic physics" from Paul Bickerstaff.
As the question comes up regularly, I have attached an extract from his chapter on currents, with the question of the force between 2 parallel wires carrying a current.

I quote him:
"So the magnetic field experienced by the test charge is correctly understood as arising from an electric force in another frame due to a line of net charge density γ.β².λ0 [...]. The relative strength of the magnetic and electric force is thus of the order of β² where β is determined by the drift velocity of the electrons. As remarked above, this is of the order of just a few mm/s and so
        β ≈ 10-11
for typical current bearing wires. The magnetic force they exert is therefore just 10-22 times that of an electric force due to just the positive (or negative) charges. That the magnetic force is observed at all (and is appreciable) is due to the enormous amount of charge carried by a number of electrons of the order of the Avogadro's number
".

In other words, the electrical force is really a huge force and the large number of charges makes visible the smallest relativistic effect. Without the natural balance of positive and negative charges in nature, this force would prevail everywhere.
The electrical effects come from the separation of charges.
Magnetic effects come from relative speeds between charges, the relativistic effects cancelling the apparent balance of the charges, in fact the equilibrium of their electric field.

Quote
Inside the solenoid the equipotentials of A are also concentric circles but they strengthen proportionally with distance.  This also applies to the A field in the Faraday disc.  As you well know the force on any radially moving electron there is not radial, it is always transverse to the movement.For E=-dA/dt the E is along A.  If you examine the components of (v.del)A the E is along A.

My interrogation was general. I wrote dA/dx by simplification but a spatial gradient is dA/dl where A and l are vectors. And we know there is no field gradient along A, so the question of the gradient direction arises, and this question doesn't depend on the speed of the charge that moves in.
A and dA/dt are independent of the load, they are evaluated by the observer.
So the spatial gradient of A should be as well. Let's not forget that the electric field is a property related to the place. If we assume that the spatial gradient gives rise to an electric field, this field cannot depend on the test charge that is placed in it. If charges at different speeds see different fields, this is understandable, it is related to relativity, but in any case the observer should only see a single electric field!

The inconsistency is due to the fact that I have already mentionned, that of calculating the spatial gradient in the referential of the charge instead of that of the observer, and mixing it with dA/dt or VxB which are relative to the observer's referential.
Can you give the spatial gradient of A, viewed in the referential frame of the observer ?

Quote
IMO the Marinov generator is just that.I look forward to seeing it.

Smudge

The Marinov generator is just that but perhaps there is simpler. In the Marinov generator there is a spatial A gradient, not a constant A. In the Faraday disk, A is constant, spatially and temporally. Just keep A and remove B in the Faraday disk, and you have a possible homopolar generator working on A, and not on a gradient of A, which is different from Marinov. I don't know yet the result. See my next post.



---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Hero Member
*****

Posts: 591
Here is the simplest idea of using A instead of B for a homopolar generator.

Take a infinite (long) cylinder magnet (like an infinite solenoid carrying a constant current). Make a hole in the disk and put the magnet through it. Place the sliding contacts between the inner and outer diameter, and rotate the disk, or the disk with the magnet as on the picture.

As for an infinite solenoid, there is no B field outside the magnet, but at the position of the disk, the vector potential is not null. If we have a voltage, it coudn't have the Lorentz force as explanation, unlike the Faraday disk. It could be only the result of the rotation in the vector potential.

But will we have this voltage?

If the Faraday disc can be explained with only the vector potential, then there are good reasons for this configuration to work, because unlike the magnetic field, there is continuity of A when one passes from the inside of the magnet to the outside.


(To avoid experimental biases as a magnetic field leakage, the long magnet could be replace by a toroid magnet where the B field would be confined).


---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   

Hero Member
*****

Posts: 3078
tExB=qr
See the attached document:

(Sorry if I posted this before, I have a poor memory.)
   
Group: Moderator
Hero Member
*****

Posts: 1161
See the attached document:

(Sorry if I posted this before, I have a poor memory.)
Grumpy,
You didn't post it but I did in the first post of this thread.  That's my paper. :)
Smudge
Edit.  Yes you did post it so that's three postings within this thread.
« Last Edit: 2019-05-20, 19:46:51 by Smudge »
   
Group: Moderator
Hero Member
*****

Posts: 1161
Here is the simplest idea of using A instead of B for a homopolar generator.

Take a infinite (long) cylinder magnet (like an infinite solenoid carrying a constant current). Make a hole in the disk and put the magnet through it. Place the sliding contacts between the inner and outer diameter, and rotate the disk, or the disk with the magnet as on the picture.

As for an infinite solenoid, there is no B field outside the magnet, but at the position of the disk, the vector potential is not null. If we have a voltage, it coudn't have the Lorentz force as explanation, unlike the Faraday disk. It could be only the result of the rotation in the vector potential.

But will we have this voltage?

If the Faraday disc can be explained with only the vector potential, then there are good reasons for this configuration to work, because unlike the magnetic field, there is continuity of A when one passes from the inside of the magnet to the outside.


(To avoid experimental biases as a magnetic field leakage, the long magnet could be replace by a toroid magnet where the B field would be confined).
If anyone does this experiment I would be interested in the result.  But even if it yields zero voltage there is still the possibility that a longitudinal force (along the velocity direction) could be present in an appropriately shaped A field.  The trick of using vector identities but with some parts removed (like the velocity derivatives) seems a poor way of describing the physics.  But if you are going to do this why not remove the transverse components and leave the longitudinal ones.  You need a different experiment to show up that longitudinal force, an experiment like the Marinov generator. :)

Smudge
   
Group: Moderator
Hero Member
*****

Posts: 1161
In the Faraday disk, A is constant, spatially and temporally.
It is not constant spatially.  If it were there would be zero B field.  But I guess you know this and didn't mean to write that.

Smudge
   
Hero Member
*****

Posts: 591
It is not constant spatially.  If it were there would be zero B field.  But I guess you know this and didn't mean to write that.

Smudge

I haven't been explicit enough. Of course A is not constant everywhere in space. I just want to say that unlike Marinov's device, the conductor movement is done in a constant A: each point of the disc rotates in a constant potential because A is circular and concentric to the axis of rotation.
Naturally each point is not in the same A, it depends on its position along the radius, but the mechanical rotational movement keeps the electron in the same A, so a spatial gradient which would be related to the rotation can't exist and be the cause for a current. In other words, the electron is not forced in a vector potential gradient.


---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
I haven't been explicit enough. Of course A is not constant everywhere in space. I just want to say that unlike Marinov's device, the conductor movement is done in a constant A: each point of the disc rotates in a constant potential because A is circular and concentric to the axis of rotation.
Naturally each point is not in the same A, it depends on its position along the radius, but the mechanical rotational movement keeps the electron in the same A, so a spatial gradient which would be related to the rotation can't exist and be the cause for a current. In other words, the electron is not forced in a vector potential gradient.
The same argument applies to the force between two infinitely long parallel current carrying conductors, Ampere's force law.  The electrons are moving along a constant A field line, yet they endure a force at right angles to that movement.  When you examine the individual components of the flux cutting v X B force as expressed by movement through the A field you find ones like Ex=vy*dAy/dx that brings in the spatial gradient at right angles to the movement.  So even though the electron is not moving through that spatial gradient its presence influences the force on the electron.  As you pointed out, the electron as a point particle can't "see" or "Know" that spatial gradient exists, it really is a matter of relativity and different frames of reference, but nevertheless we do have tried and tested vector-math formula to predict those forces and they do use that spatial gradient. 

Your suggested experiment has the same problem, the disc rotation moves the conduction electrons along constant A field lines.  If that really does yield an output voltage that can drive current through a load, the question must be asked "where does the energy come from".   IMO it comes from the electron spins or orbits responsible for the magnetic field in the core, as I have repeatedly tried to say in various ways.  Electrons passing from a brush contact onto a moving conductor must endure an acceleration, and therefore "radiate" an electric field.  It is the electric near-field that interests me and I think it possible that it has unusual characteristics in that a small closed circuit near the brush contact will obtain an induced unidirectional voltage impulse for each electron that jumps across.  For a continual stream of electrons that becomes a DC voltage.  Note that there is no dB/dt present to invoke the usual transformer induction rule for closed circuits.  Each magnetic dipole atomic electron circulation within the core gets this induced voltage attempting to slow down the spin or orbit, which it can't do.  So these spins are truly miniature quantum dynamos.  I attach an old paper written some years ago where I try to explain all this.   Also another old paper on possible experiments to discover longitudinal induction where I see a did a crude experiment using a square NdFeB magnet.

Smudge
   
Hero Member
*****

Posts: 591
The same argument applies to the force between two infinitely long parallel current carrying conductors, Ampere's force law.  The electrons are moving along a constant A field line, yet they endure a force at right angles to that movement.  When you examine the individual components of the flux cutting v X B force as expressed by movement through the A field you find ones like Ex=vy*dAy/dx that brings in the spatial gradient at right angles to the movement.  So even though the electron is not moving through that spatial gradient its presence influences the force on the electron.  As you pointed out, the electron as a point particle can't "see" or "Know" that spatial gradient exists, it really is a matter of relativity and different frames of reference, but nevertheless we do have tried and tested vector-math formula to predict those forces and they do use that spatial gradient. 

Apart from a mechanical force, there is only one force capable of moving an electron, the force F=q.E.
If we have a current, we have a force, so we have an electric field E seen by the electron.
In the case of the classical Faraday disc, this electric field E seen from the charge is VxB where v and B are seen by the observer.
In the case of the Faraday disc with an out-of-field conductor, as in my experimental proposal, VxB can no longer be used.
So I agree with you. When dealing with A, your idea that E is due to the transverse gradient of A, in this case as in the other, is a credible hypothesis to be tested, with real chances to be verified.

I won't be able to do this experiment because I'm too bad at mechanics. The long magnet (to be made from a series of smaller ones) will be difficult to assemble and rotate quickly enough. There is the problem of the magnet end leakage field which can create artifacts. And using a torus instead of the long magnet drastically complicates the rotation, because you need a fixed toroidal magnet, you lose the axis of rotation, so you need a transmission belt to rotate the disc. I don't feel like I can do it.

Quote
Your suggested experiment has the same problem, the disc rotation moves the conduction electrons along constant A field lines.  If that really does yield an output voltage that can drive current through a load, the question must be asked "where does the energy come from".   IMO it comes from the electron spins or orbits responsible for the magnetic field in the core, as I have repeatedly tried to say in various ways.  Electrons passing from a brush contact onto a moving conductor must endure an acceleration, and therefore "radiate" an electric field.  It is the electric near-field that interests me and I think it possible that it has unusual characteristics in that a small closed circuit near the brush contact will obtain an induced unidirectional voltage impulse for each electron that jumps across.  For a continual stream of electrons that becomes a DC voltage.  Note that there is no dB/dt present to invoke the usual transformer induction rule for closed circuits.  Each magnetic dipole atomic electron circulation within the core gets this induced voltage attempting to slow down the spin or orbit, which it can't do.  So these spins are truly miniature quantum dynamos.  I attach an old paper written some years ago where I try to explain all this.   Also another old paper on possible experiments to discover longitudinal induction where I see a did a crude experiment using a square NdFeB magnet.

Smudge

I don't agree. We are only dealing with a relativistic effect between two electron flows of different speeds. A force is exerted between the two parts of the circuit, and nothing acts on the source of the field or the potential. As for the normal Faraday disk, the energy comes only from the mechanical energy to rotate the disk. Even if the disc no longer rotates in a magnetic field, the current still generates the same magnetic field in both cases, so it will oppose rotation as in the ordinary Faraday disk. The magnetic dipoles including spins are just the vectors of the change of the mechanical energy into electrical energy.

We should think of simpler ways than mechanics to highlight all this, for example with a solid state device that would make electrical induction from the spatial gradient of A.


---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
I don't agree. We are only dealing with a relativistic effect between two electron flows of different speeds. A force is exerted between the two parts of the circuit, and nothing acts on the source of the field or the potential. As for the normal Faraday disk, the energy comes only from the mechanical energy to rotate the disk. Even if the disc no longer rotates in a magnetic field, the current still generates the same magnetic field in both cases, so it will oppose rotation as in the ordinary Faraday disk.

Then we will have to agree to disagree. You say a force is exerted between the two parts of the circuit, and that force is on the moving electrons.  That is the moving electrons in one part and the moving electrons in the other part.  It so happens that Nature in her perverse way has arranged things so that in our usual machines such as the Faraday disc when we extract electrical energy from one part the force acting on the electrons in the other part creates mechanical drag so we have to supply mechanical force and that is part of the energy balance.  However there are other machines where the force on the "driving" electrons is along the conductor even if it is circular (a coil) and we have to supply electrical energy to complete the energy balance.
Quote
The magnetic dipoles including spins are just the vectors of the change of the mechanical energy into electrical energy.
That is a very limited view of magnetic dipoles.  To me a magnetic dipole is a current loop and like any current loop it can get an induced voltage that either aids or opposes the current.  Hence whatever current generator is there driving current around that loop, it can either supply energy to whatever is creating that induction or receive energy from it.  I can argue that atomic dipoles are no different and that in many cases they already do just that, supply energy or receive energy, but that energy exchange is hidden from view because it doesn't appear in the usual energy audit, over complete cycles the exchange sums to zero, i.e they take back as much energy as they give out.  Note this is not any form of dipole rotation or flipping, this is not mechanical energy related.  The fact that this is happening under our very noses, and we ignore it, is a big mistake on our part.  We should be looking for ways to break that reciprocity, and I think that can happen when we have atomic dipoles close to brush contacts that are carrying DC.

Quote
We should think of simpler ways than mechanics to highlight all this, for example with a solid state device that would make electrical induction from the spatial gradient of A.
Well my suggestion for using a change of drift velocity at a junction between a normal conductor and a super-conductor is one possibility.  I think there are others, for instance the drift velocity along a conductor surface where the conductor is charged highly negative will be much greater than the normal drift velocity.  That is simply because there you have a region with electrons just above the rough surface (rough at atomic dimension scale) hence the collision rate with ions is much reduced.  It must be possible to use this greater electron speed at a junction with an uncharged conductor to get acceleration there, and use that instead of a junction of a brush against a moving conductor.

Smudge 
   

Hero Member
*****

Posts: 3078
tExB=qr
Can you use a simple experiment like this one?

https://pdfs.semanticscholar.org/88a9/37981ae9f9663c591f9aca5db108faf122f0.pdf

see page 203 of the linked paper:

Quote
The writer is not the first observer in this country to confirm the existence of nonzero torque in the Marinov motor. As reported by Kooistra (13),  Tom  Ligon  was  the  first  to  recognize  that  an  easy way to test the concept is to suspend a toroidal magnet “armature” shaped somewhat as in Fig. 1 of Wesley’s  report (1),  within  a  horizontal  copper  ring,  fixed  in  the  lab,  to  which  current  leads  are  attached. This eliminates the need for brushes, and emulates one very practical brushless motor design (which,  however,  requires  periodic  DC  polarity  reversals).  Ligon  observed  vigorous  turning  of  the armature up to almost 90 degrees when strong DC was put through the ring. At 90 degrees there is a torque null, and beyond that a counter-torque.
   
Hero Member
*****

Posts: 591
Then we will have to agree to disagree. You say a force is exerted between the two parts of the circuit, and that force is on the moving electrons.  That is the moving electrons in one part and the moving electrons in the other part...

That's what I'm saying, I confirm. You forgot VxB: electrical current and the mechanical force are transversal.
If you pass current through a Faraday disc, it turns into a motor, so the current in the circuit creates a torque.
When you use the disc as a generator, and draw current from a load, this current acts exactly the same way as in the Faraday disc as a motor, and the direction of the currents causes the torque to oppose rotation.
All this is completely reciprocal.

Quote
...To me a magnetic dipole is a current loop and like any current loop it can get an induced voltage that either aids or opposes the current.

Magnetic dipoles are current loops. But why do you thing that there would be induced voltage in these dipoles in the case of the Faraday disk? Why do you think that induced voltage would have an effect on the electron spins?...

The magnet may or may not rotate because there is no interaction of any "counter"-field with those of the magnet's dipoles. This is due to the fact that the magnetic force F=VxB doesn't work, the force being perpendicular to the displacement, so the reaction is not here.
Otherwise rotating a cylindrical magnet in front of another that is coaxial would cause a torque from one to the other. The experiment has been done many times, there is no effect. I'm afraid you're going to go into highly speculative speculation   :).

Replace the permanent magnet of a Faraday disk by a solenoid. I'm sure you will see no induced voltage.




---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   

Hero Member
*****

Posts: 3078
tExB=qr
Will the Marinov device work if the conductive ring encircles only one side of the toroidal magnet?
   
Hero Member
*****

Posts: 591
Will the Marinov device work if the conductive ring encircles only one side of the toroidal magnet?

It won't, unless the sliding contacts are between the inner and outer edge of the rotating ring, and the ring is sufficiently thick. This is what I proposed here, simply replace the long magnet by one of the two sections of the Marinov's magnetic circuit.


---------------------------
"Chance favours only the prepared mind."  Louis Pasteur
   
Group: Moderator
Hero Member
*****

Posts: 1161
That's what I'm saying, I confirm. You forgot VxB: electrical current and the mechanical force are transversal.
If you pass current through a Faraday disc, it turns into a motor, so the current in the circuit creates a torque.
When you use the disc as a generator, and draw current from a load, this current acts exactly the same way as in the Faraday disc as a motor, and the direction of the currents causes the torque to oppose rotation.
All this is completely reciprocal.
I didn't forget and I agree with everything you say here. 

Quote
Magnetic dipoles are current loops. But why do you thing that there would be induced voltage in these dipoles in the case of the Faraday disk?
I was not considering the Faraday disc, I was considering the Marinov generator and possibly your spinning annulus where the movement is not in a B field.  But you raise an interesting point, if electron acceleration across a brush contact can induce, should this not also apply to the Faraday disc?  The answer must be yes, so the next question should be is there any evidence of the Faraday disc exhibiting anomalous behaviour that could be attributed to this?  Well there are lots of claims for anomalous behaviour that are ridiculed so maybe it is possible.
Quote
Why do you think that induced voltage would have an effect on the electron spins?
If you replace the spins with current driven loops the current generator then plays its part in the energy audit either as an energy source or sink.

Quote
The magnet may or may not rotate because there is no interaction of any "counter"-field with those of the magnet's dipoles. This is due to the fact that the magnetic force F=VxB doesn't work, the force being perpendicular to the displacement, so the reaction is not here.
Otherwise rotating a cylindrical magnet in front of another that is coaxial would cause a torque from one to the other. The experiment has been done many times, there is no effect. I'm afraid you're going to go into highly speculative speculation   :).
You have misunderstood what I am trying to convey, it certainly is not a torque on the magnet.

Quote
Replace the permanent magnet of a Faraday disk by a solenoid. I'm sure you will see no induced voltage.
I am not so sure, I think there could be.  Has anyone ever tried it?  It would be a small voltage perhaps too small to register.  It would come from electrons flowing across the outer brush contact where their acceleration creates an unusual electric near-field.  Has anyone ever tried to measure that electric field?
Smudge
   

Hero Member
*****

Posts: 3078
tExB=qr
https://skemman.is/bitstream/1946/29521/3/KOK_MagneticVectorPotential.pdf

Marinov is on page 51.

--------------
EDIT:
the entire section 5.2.1 Vorticity, Acceleration and Induction of this paper is interesting and poses multiple possible induction methods related tot he vector potential.
   
Pages: 1 2 [3] 4 5 6 7
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2019-07-22, 04:31:55