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Generator using a superconductor « on: 2022-11-03, 16:45:03 »

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I have started this new bench because I think there is a route to an OU generator, but unfortunately it needs a superconductor so beyond the reach of most of us here.  I do think this is a serious contender so I will gradually talk my way to the design here.  It comes from my work on the Marinov generator where the electrons in a slip ring are driven through a non-curl magnetic vector potential field.  This will be a solid-state version, so it has no moving parts.  I won't post any more today as I have to take my dear wife out to dinner now.

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Re: Generator using a superconductor « Reply #1 on: 2022-11-04, 02:10:30 »

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Yeah, I know superconductors are a massive pain, and that physical measurements become difficult due to the liquid nitrogen and convection currents caused by boiling.

YBCO is not completely beyond the realm of the amateur, but certainly near so.  A few successful attempts have been made in the recent past:
https://www.youtube.com/watch?v=sLFaa6RPJIU
https://www.youtube.com/watch?v=-JLWeYQFA2I


Is it a static/DC condition or varying?   As long as it is a varying condition, then perhaps a large core of copper/silver would have low enough conductivity to demonstrate some part of the intended effect?   Or is Meissner effect explicitly required?
How big/small do you think a system would need to be?


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Re: Generator using a superconductor « Reply #2 on: 2022-11-04, 09:02:13 »
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Quote from: Smudge on 2022-11-03, 16:45:03
...
It comes from my work on the Marinov generator where the electrons in a slip ring are driven through a non-curl magnetic vector potential field.  This will be a solid-state version, so it has no moving parts.  I won't post any more today as I have to take my dear wife out to dinner now.

Smudge

The questions around Marinov in your previous thread, related to the vector potential, were a great moment and I still think about it.
The fact that a spatial gradient of A can create an electric field is really worth testing. Is your new idea related to this topic? I found the right (and simple) idea to get a spatial gradient of A from a current gradient along the conductor in a circuit (which may sound amazing). I haven't talked about it yet because the spatial gradient of A can't make a loop, so I don't see how in practice to get a current in a circuit from it.



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Re: Generator using a superconductor « Reply #3 on: 2022-11-04, 16:29:26 »

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I have had a problem understanding what the product of charge q with vector potential A (bold denotes a vector), (q*A), really means.  It is regarded as a form of hidden momentum in the charge q and any change in momentum results in a force F.  Hence F=-q*dA/dt, then if we recognize F as due to an effective induced E field we get E=-dA/dt.  That is fine when A is changing with time.  But when you have a static but non-uniform A field, when the charge q moves through it at some velocity it sees a time changing A field so should endure a force due to that non-uniformity.  For non-uniformity such that the charge sees a magnetic field we have well known laws dealing with forces, but we can also have non-uniformity that is free of magnetic fields and that is the interest here.  If you take the simple case of an A field that points along (say) the x axis everywhere along that axis but its magnitude changes then a charge moving linearly along that axis sees that change and you can deduce that Fx=-q*(dAx/dx)*v where v is the velocity along the x axis.  That is OK but if we want to make use of that the charge cannot continue along that axis indefinitely.  We need charge flowing in closed circuits and that means a change of velocity direction as charge moves around the circuit.  You can no longer just use the component of A along the velocity direction (the tangential component) and use it to evaluate the change in A seen by the charge as it gives the wrong answer.  This becomes clear if you consider charge moving in a circular orbit within a uniform A field, the tangential component follows a sine wave that suggests a sinusoidal force, but in fact the hidden momentum remains constant, there is no change hence no force.  That realization that the moving charge can change direction within an A field yet endure zero force takes on a serious meaning if we have a hairpin circuit where the conductor is folded back on itself.  Over the tiny dimensions of the fold the A field is virtually uniform even if it changes value away from the fold.  Thus, we can have charge moving within a conductor that sees a 180-degree change in A field direction but without seeing any force.   That is a form of regauging.  If the charge moving towards the fold is seeing an increase in A that is a positive dA/dt yielding a negative force.  If at the point of maximum A it reaches the fold, it then finds itself seeing a negative A and as it continues is reverse journey that negative A reduces in value.  That reverse journey is still seeing a positive dA/dt yielding a negative force.   I am not aware that this rectifying aspect of circuits within an A field has been noted before, and it leads to closed circuits enduring an induced voltage, something that current science {and F6  ;)} says is impossible.

The interesting aspect of the induced voltage is its magnitude being proportional to the velocity of the charge.  For a given size of conductor the velocity is related to the current and the volume density of the charge carriers, hence the induction appears as a resistance value that can be either positive or negative depending on the direction of the current.  Of course it is the induced negative resistance that is of interest here.  For copper conductors within a practical embodiment the negative resistance is likely to be a tiny fraction of its normal positive resistance, but with superconductors that have a much smaller volume density of carriers (Cooper pairs) the velocities are much greater and there is zero positive resistance.  Thus, it is possible to conceive of a system where current builds up from circuit noise to some high value within a multi-turn closed "coil" of hairpins, the coil is opened and connected to a load so that it discharges its energy, then it is closed again and the cycle repeats.  Luckily I have experience of super-regeneration so I can work all this out.

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Re: Generator using a superconductor « Reply #4 on: 2022-11-04, 18:05:52 »
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@Smudge

I have read your text but I don't have time to study it now. This is a small partial answer. I agree that the charge must have a non-zero velocity in the spatial area of the gradient of A to be able to see the electric field, I was well aware of this but I had omitted to specify it in my previous answer.

As for the rest, your text is dense and full of ideas, I will have to read it again tomorrow more carefully before I can talk about it. It all looks very promising ;)


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Re: Generator using a superconductor « Reply #5 on: 2022-11-05, 10:57:41 »

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Here I explain how I evaluate the force on moving electrons.  Luckily I have a finite element program that gives me the A field values (more on that program later).  The image below shows a non-curl (zero magnetic field) A field where the arrows depict the magnitude (arrow length) and direction of the field vectors.  I show a current path through the field.  Taking two adjacent vectors the change in field magnitude δA between those points is A2-A1.  Note that not only are the vector amplitudes changing but also the vector directions.  The fem program enables the two vector directions relative to the current path, θ1 and θ2, to be evaluated.  Taking the average value of these two angles θAV gives the direction of the change in hidden momentum, hence allows the magnitude of the tangential force on the moving electrons to be evaluated as proportional to
(A2-A1)*cos(θAV).
This simple procedure carried out in a spreadsheet allows me to calculate this hidden force for any shape of the current path.  I say hidden force because to my knowledge this has not been discovered before.  But it is only hidden in the sense that for practical currents flowing in copper conductors the effect of the force is obscured by the finite resistivity of the copper.  And the surprising feature is that for some unusual shaped closed paths (like the hairpins mentioned in my previous post) the induced voltage is non-zero.  In my opinion this is a significant finding that goes against current teachings on closed paths, and could open the door to many new inventions.

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Re: Generator using a superconductor « Reply #6 on: 2022-11-05, 15:09:53 »
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Quote from: Smudge on 2022-11-04, 16:29:26
...
If you take the simple case of an A field that points along (say) the x axis everywhere along that axis but its magnitude changes then a charge moving linearly along that axis sees that change and you can deduce that Fx=-q*(dAx/dx)*v where v is the velocity along the x axis.  That is OK but if we want to make use of that the charge cannot continue along that axis indefinitely.  We need charge flowing in closed circuits and that means a change of velocity direction as charge moves around the circuit.

I agree.

Quote
...
Thus, we can have charge moving within a conductor that sees a 180-degree change in A field direction but without seeing any force.

I agree that the 180° turn is not a problem.

Quote
If the charge moving towards the fold is seeing an increase in A that is a positive dA/dt yielding a negative force.  If at the point of maximum A it reaches the fold, it then finds itself seeing a negative A and as it continues is reverse journey that negative A reduces in value.  That reverse journey is still seeing a positive dA/dt yielding a negative force.

I disagree.

Quote
I am not aware that this rectifying aspect of circuits within an A field has been noted before, and it leads to closed circuits enduring an induced voltage, something that current science {and F6  ;)} says is impossible.
...         

I agree with F6 and current science ;). I have bad news.

We arrived at F = -q.v.∂A/∂x. Ok. This means that in a reference frame from which the observer sees a charge q moving at speed v along the x axis in a gradient ∂A/∂x, he also sees a force F exerted on the charge such that (1) F = -q.v.∂A/∂x.
∂A/∂x is constant as seen from the observer, always of the same sign, whatever the motion of the charge. So the force in the forward direction is always the opposite of the force in the return direction if the forward and return directions are in the same gradient and v changes sign.

The problem is that in your reasoning, you switch to the point of view of the charge by saying that it sees an inverse ∂A/∂x on the return, which is correct. But if you switch to the point of view of the charge, then relative to its reference frame, v=0 and therefore F=0. You can't mix the two points of view by taking the force and velocity in one, but the gradient in the other. ∂A/∂x is not with respect to the charge but with respect to the observer in the frame of reference having the x axis and seeing the charge at speed v. 

In the frame of reference of the charge, v=0, then what does it see? As you say, the charge sees an electric field E = -∂A/∂t because its motion translates into time variation the spatial gradient of A seen from the observer. The charge is thus subjected to a force (2) F=q.E with E = -∂A/∂t. This is all it sees from its frame of reference, it is this field E which seen from the observer is equal to -v.∂A/∂x. As for the charge, ∂A/∂t changes sign on the way back compared to the way forward since it follows the gradient of A in the opposite direction, so E changes sign, and so does F. The point of view of the charge joins that of the observer, in complete coherence, current science wins again.
The sign of v is implicit in ∂A/∂t. You cannot again use v from formula (1) valid only in the observer's frame of reference, to maintain the same sign of the force seen from the charge through the product with ∂A/∂x. For the charge, only ∂A/∂t counts. The force on the return is therefore the opposite of the forward, and over one revolution, its work is zero.


In fact we face a more general problem. The vector potential is to current what the scalar potential is to voltage. But it is a real potential too, and when a force derives from a potential, the work does not depend on the path followed. This is not a question of physics in a particular case, but of math in the general case. As in a circuit the starting point is the same as the ending point, so at the same potential after one turn, the difference in vector potential is zero, so if the potential does not vary in time, the work of the force over one turn is zero. And when A varies in time, this is what gives rise to the current in a classical induced circuit.

So we will have to find a more twisted way to get around the "impossibility".



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Re: Generator using a superconductor « Reply #7 on: 2022-11-05, 15:43:12 »

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Quote from: F6FLT on 2022-11-05, 15:09:53
We arrived at F = -q.v.∂A/∂x. Ok. This means that in a reference frame from which the observer sees a charge q moving at speed v along the x axis in a gradient ∂A/∂x, he also sees a force F exerted on the charge such that (1) F = -q.v.∂A/∂x.
∂A/∂x is constant as seen from the observer, always of the same sign, whatever the motion of the charge. So the force in the forward direction is always the opposite of the force in the return direction if the forward and return directions are in the same gradient and v changes sign.

We are agreed on that.  So taking movement along the x axis, in the forward direction the force is (say) directed as +x, and in the reverse direction the force becomes -x.  If we are agreed on that then all your reasoning that follows is wrong.  The +x force is attempting to accelerate the electron along the forward path, and the later -x force is attempting to accelerate the electron along its reverse path.

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Re: Generator using a superconductor « Reply #8 on: 2022-11-05, 17:20:20 »
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Quote from: Smudge on 2022-11-05, 15:43:12
We are agreed on that.  So taking movement along the x axis, in the forward direction the force is (say) directed as +x, and in the reverse direction the force becomes -x.  If we are agreed on that then all your reasoning that follows is wrong.  The +x force is attempting to accelerate the electron along the forward path, and the later -x force is attempting to accelerate the electron along its reverse path.

Smudge

If the observer sees the force in the direction of the increasing x, it is true both on the forward and the return part. And vice versa.

When I say that from the observer's point of view, "the force in the forward direction is always the opposite of the force in the return direction", I mean that the effect on the current in the forward direction is the opposite of what it is in the return direction, sorry for not having been clear. And seen from the charge, there it is obvious with the dA/dt.

So a current loop where the electrons would rotate at a linear speed v will have a half part where the force will try to oppose the movement of the electrons while it will favor it on the other side. The total work is zero.
I hope it is clearer with the diagram.
So we would not agree.




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Re: Generator using a superconductor « Reply #9 on: 2022-11-05, 17:49:39 »

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I totally disagree!  Your image does not show the direction of the vector A field.  You have treated it like a scalar that has no direction.  If it were a scalar I would agree with your reasoning (except the math is wrong).  When you take account of the fact that on one journey the A vector is parallel to the velocity while on the return journey it is anti-parallel then you find I am right.

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Re: Generator using a superconductor « Reply #10 on: 2022-11-06, 11:11:53 »

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@F6
If I may take you to task on something you wrote earlier
Quote
But if you switch to the point of view of the charge, then relative to its reference frame, v=0 and therefore F=0.
That is wrong!  From our perspective it is like looking down (a stationary satellite view) onto a field of A vector arrows.  We see a point charge moving across that field and use our math to deduce what force there is on the charge as it moves from arrow to arrow.  If we widen our field of view, we see one edge of the field where we see the source of the arrows, and that is a permanent magnet in the form of a magnetized ring-core where the magnetic field is contained within the ring.  That ring is stationary as are we, so our reference frame is the same as the ring.  From our perspective the charge is enduring a changing A field as it moves because the arrows are of different sizes and directions.  We see a static arrow field and a moving charge.

Now from the reference frame of the charge there is no movement.  Let’s imagine we give our charge a smart phone camera on a pole so that it can see an image of itself in the field of arrows.  It sees itself as stationary, but it also sees the field of arrows moving past itself.  From its perspective without the camera it sees a time changing A, and the camera allows it to see why the A is changing.  Give the camera a wider field of view and it sees the ring core moving.  For both reference frames the force on the charge is the same, the only difference being that which is moving, the charge or the ring-core source of the field.

When you get this perspective into your head (and it is difficult to do, it has taken me years), you come to that Eureka moment where voltage induction into a closed loop is possible.

This problem with reference frames is not new and exists in some of the gedanken experiments of Einstein.  Take the famous one of the carousel within a circular room where you are standing on the outer edge of the carousel and enduring centrifugal force.  You are looking outward and all you can see are the circular walls of the room moving past you.  How can you tell whether the force you are enduring is because you are on a rotating platform while the walls are stationary, or you are on a stationary platform and the walls are rotating?  In the one case you are moving through the space of a stationary Universe, while in the other case the Universe and its space are rotating around you.  This has led to the perspective that the two are equivalent in producing the force you endure.  That is wrong!  Rotation per se cannot produce an inertia force, only translation through space does this.  If you look at your translation distances in the x-y plane they take on the form of x= Rcos(ωt) and y=Rsin(ωt).  To move the reference frame to one where you are stationary the Universe and its space are moving the Universe must have translation x=- Rcos(ωt) and y=-Rsin(ωt).  That is the Universe taking on the movement like that of that in an orbital sander, it is an orbital movement not a rotation.  In the wrong interpretation where the Universe is rotating, the outer edges of the Universe are continually translating in a circular motion at enormous (>>c) velocity, and that does not accord to the actual relative motion between you and the Universe.

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Re: Generator using a superconductor « Reply #11 on: 2022-11-06, 12:24:27 »
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@Smudge

The triangle with A just indicates on which side the greatest field strength is (at position x=0).
A is obviously supposed to be oriented along x, and identical, at the same level, on the upper and lower branches of the circuit, the whole circuit is bathed in the same field.

I have made the effort of a diagram, and all I see from you is "that is wrong" rather than a request for clarification on possible obscure points or your own presentation of an explanatory diagram of your device.

If "it is wrong" that the velocity of an object in its own frame of reference is zero, then I don't really see how we can continue to have a serious discussion.

So I will now wait for a real experiment proving what you say, done by others than me, and I predict that it will not happen for the reason I said above, the force deriving from a potential cannot provide work on a closed circuit, a matter of maths.
The underlying principle related to dA/dx may have a future, but it is with my own ideas that I will attempt it.



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Re: Generator using a superconductor « Reply #12 on: 2022-11-06, 14:37:17 »

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Quote from: F6FLT on 2022-11-06, 12:24:27

I have made the effort of a diagram, and all I see from you is "that is wrong" rather than a request for clarification on possible obscure points or your own presentation of an explanatory diagram of your device.
I have tried to explain why I think it is wrong.

Quote
If "it is wrong" that the velocity of an object in its own frame of reference is zero, then I don't really see how we can continue to have a serious discussion.

I have not said that is wrong.  I have merely pointed out that in that frame of reference the field (and its source) is moving, so the object still endures a changing field.

Quote
So I will now wait for a real experiment proving what you say, done by others than me, and I predict that it will not happen for the reason I said above, the force deriving from a potential cannot provide work on a closed circuit, a matter of maths.

Perhaps you will study my notes added to your image and tell me why I am wrong.  I have replaced your triangle with a set of arrows depicting the direction and magnitude of the A field.

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Re: Generator using a superconductor « Reply #13 on: 2022-11-07, 09:07:58 »
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In the upper segment, we agree that dA/dt<0.

The electric field is E=-dA/dt (the arrows are missing but we are talking about a vector) and dA/dt<0 so E>0, directed opposite to A gradient, so directed towards the positive x.
Now F=q.E so F is directed like E, towards the positive x, as I had indicated.
F is in the direction of v at the top, and opposite to v at the bottom. We do not gain anything on a round trip.

« Last Edit: 2022-11-07, 11:00:26 by F6FLT »


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Re: Generator using a superconductor « Reply #14 on: 2022-11-07, 13:42:39 »

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Quote from: F6FLT on 2022-11-07, 09:07:58
In the upper segment, we agree that dA/dt<0.

The electric field is E=-dA/dt (the arrows are missing but we are talking about a vector)
The arrows are not missing.  I show the arrows where you had your triangle.  I also show the arrows in  the text boxes.
Quote
and dA/dt<0 so E>0, directed opposite to A gradient, so directed towards the positive x.
Now F=q.E so F is directed like E, towards the positive x, as I had indicated.
I must disagree.  The direction of A is towards negative x.  Thus the direction of dA/dt is towards positive x since dA/dt is negatve.  Then the direction of -dA/dt is towards negative x, and that is opposite to your perception.

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Re: Generator using a superconductor « Reply #15 on: 2022-11-07, 18:24:35 »
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Quote from: Smudge on 2022-11-07, 13:42:39
The arrows are not missing.  I show the arrows where you had your triangle.  I also show the arrows in  the text boxes.

I just meant the arrows above E or v in my text here, not the ones in the picture.

Quote
I must disagree.  The direction of A is towards negative x.  Thus the direction of dA/dt is towards positive x since dA/dt is negatve.  Then the direction of -dA/dt is towards negative x, and that is opposite to your perception.

Smudge

Not clear to me yet. I'll take some more time.

« Last Edit: 2022-11-08, 10:49:20 by F6FLT »


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Re: Generator using a superconductor « Reply #16 on: 2022-11-08, 10:57:45 »
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@Smudge

Well, I think I have arrived at a coherent result between the observer's view, the charge's view, and your view since the forces now seem to be in agreement with your previous diagram.

However we can see that the force vector is always opposite to the velocity vector. Attempting to cross a spatial gradient of the vector potential would therefore result in a braking force.
Am I missing something again?



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Re: Generator using a superconductor « Reply #17 on: 2022-11-08, 15:42:56 »

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That braking force applies to a positive charge, in the case of a negative charge it becomes an accelerating force.  In either case it represents an effective E field around the loop whose closed-line integral is non-zero.  I can't emphasis that last phrase enough.  We have discovered a holy grail that goes against everything that physicists like you and I have been brainwashed into thinking is impossible.  I am so pleased we are on the same page at last.  Of course this needs evidence and that may be forthcoming, I am working my way to establishing the magnitude of the effective negative resistance for a scheme using copper wire and I hope to show that a suitable experiment can be designed that will allow the wire resistance to be measured while the special coil is not within the A field from some magnetized ring cores to be compared to its resistance while within the A field.  We should see a small reduction in resistance.

For the drift velocities within copper it is only a small effect, but for a superconductor that changes enormously.  A closed loop superconductor having some induced negative R effect should see the current build up following a positive exponential law e+|R|t/L.  That is what happens in super-regenerative receivers where it builds up from thermal noise if there is no signal.  We should be able to switch the coil onto a positive resistance to gain the energy built up in the coil, then repeat the process at some controlled rate.

Of course the question must be asked as to where the energy comes from.  You will find it no surprise that I will claim that the sudden change of electron velocity direction at the hairpin fold results in an E field that influences the electron dipoles within the magnetized ring cores, as I explained in my Marinov generator bench, except there the velocity change was at the slip-ring brush tips.

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Re: Generator using a superconductor « Reply #18 on: 2022-11-08, 16:20:10 »
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Quote from: Smudge on 2022-11-08, 15:42:56
That braking force applies to a positive charge, in the case of a negative charge it becomes an accelerating force.
...

This would be a strong violation of CP invariance. This is the Nobel Prize for sure!  :)


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Re: Generator using a superconductor « Reply #19 on: 2022-11-08, 16:32:08 »

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In the first image below I show the A field from two magnetized ring cores placed one above the other.  The FEM program uses an axisymmetric solution to give the field in two dimensions within a cylindrical r θ z coordinate system, displaying just the r z plane so showing only one half of the usual field pattern.  I show a closed hairpin loop in orange colour around which the fem program gives charts of the tangential component At, the normal component An and it also gives the magnitude |A|.  Of major interest is the tangential At.  The red arrows show the route taken around the loop that decides the direction of At.  The starting point is denoted by the red A then other points around the loop follow in alphabetical progression.

The second image shows At against distance in mm.  I have added a red A at the starting point then put the other points there in sequence.  For a fixed velocity around the loop the distance axis can become a time axis so there we have a plot of A against time.  It can be seen that along both straight-line sections dAt/dt is negative thus any induced E is supported, not cancelled.  However you can’t use only At to evaluate E.  In my post #5 I show how I evaluate E using a spreadsheet and I have done this for the hairpin loop here.  The fifth image shows the normal component An and the third image the magnitude |A|.  The fourth image is the result of the spreadsheet calculations where the E field is predominantly positive around the loop.

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Re: Generator using a superconductor « Reply #20 on: 2022-11-08, 17:21:16 »
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1) I don't see why you assume that the effect would be small. Going from temporal variation to spatial variation is only a change of reference frame. So a classical AC current induced in a circuit by a temporal variation of A seen by the charge, can also be seen as a spatial gradient around the circuit by the observer, this spatial variation rotating at the speed of the charges of the current. The speed of the charges of an ordinary current would thus be sufficient to show a very significant effect.

2) To have a gradient of A, it is not necessary to play on the speed of the charges.

A is proportional to the current I, and parallel to it. A straight conductor along x, with a spatially non-constant current I of the form k*x, would be sufficient to generate a variable A of the form k'*x along x too.

I think I have the device for this, I'm preparing a diagram, but with the disadvantage that A would not only have a spatial gradient, but would also have to be time-varying, and thus be generated by an AC current.
This may complicate the experiment, but we should see a modulation of the DC current in our loop, induced by the spatial grad.A also time modulated, and the AC component would be amplified in proportion to the DC current while the AC current generating A would remain constant. The signature of the effect would thus be very clear and without superconductor.


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Re: Generator using a superconductor « Reply #21 on: 2022-11-08, 19:25:56 »

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Quote from: F6FLT on 2022-11-08, 17:21:16
1) I don't see why you assume that the effect would be small. Going from temporal variation to spatial variation is only a change of reference frame. So a classical AC current induced in a circuit by a temporal variation of A seen by the charge, can also be seen as a spatial gradient around the circuit by the observer, this spatial variation rotating at the speed of the charges of the current. The speed of the charges of an ordinary current would thus be sufficient to show a very significant effect.
Not when you take account of the actual values of the A field from practical ring cores at their saturation.  Note that the closed loop induced V is proportional to the velocity hence proportional to the current.  That proportionality means it appears as a resistance and I know from my previous work on the Marinov generator that the value will be below the positive value of a copper loop.

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A is proportional to the current I, and parallel to it.
That is for current in the conductor.  It is not the A applied to the conductor, that comes from the flux in the ring cores.
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A straight conductor along x, with a spatially non-constant current I of the form k*x, would be sufficient to generate a variable A of the form k'*x along x too.
Interesting concept but the A magnitude cannot be anything near what you get from magnetized cores.

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I think I have the device for this, I'm preparing a diagram, but with the disadvantage that A would not only have a spatial gradient, but would also have to be time-varying, and thus be generated by an AC current.
This may complicate the experiment, but we should see a modulation of the DC current in our loop, induced by the spatial grad.A also time modulated, and the AC component would be amplified in proportion to the DC current while the AC current generating A would remain constant. The signature of the effect would thus be very clear and without superconductor.
I look forward to seeing your diagram.
Smudge
   
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Re: Generator using a superconductor « Reply #22 on: 2022-11-08, 19:57:57 »
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Quote from: Smudge on 2022-11-08, 19:25:56
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That is for current in the conductor.  It is not the A applied to the conductor,
...

Yes, it is the current that is used to generate the A field, not the current in our test circuit. ∇.I => ∇.A along I.

In a quasistationary regime, the current is the same everywhere in a circuit. But we can make it not, and so will be A, not constant along I.



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Re: Generator using a superconductor « Reply #23 on: 2022-11-09, 17:00:39 »

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Further to the results shown in my reply #19, if that hairpin loop is made of copper wire of 1mm square cross section its resistance would be 8.57x10-4 ohms.  The induced negative resistance is 6.84x10-8 ohms.  Thus the induced effect is very small but could be measurable using a precision ohmmeter.

The interesting feature of this induced effect is not induced voltage, it is genuinely a change of resistance.  We are familiar with some features that change the resistivity of a material.  Heat is one such, so this is very much like changing the temperature of the copper.  If the two ring cores have alternating magnetization instead of permanent magnetization the effect would be like varying the temperature, but without the time delay caused by the thermal capacity of the copper.  To see a voltage a current would have to be applied, then we should see a small induced alternating voltage.  This could be a better way of discovering the effect is real.  Instead of a single turn, a multi-turn coil could be wound onto a short length of plastic pipe to form an elongated toroidal winding.  With 100 turns and 1 amp current the voltage would be 6.84 microvolts, and that is easily measurable.  I will look into practical sized ring cores and come up with the details of an experiment.
   
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Re: Generator using a superconductor « Reply #24 on: 2022-11-09, 17:15:27 »
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Quote from: Smudge on 2022-11-08, 19:25:56
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I look forward to seeing your diagram.
Smudge

So as not to clutter your thread with another process for generating the A-gradient, I opened a thread on the generation of a gradient of A from a gradient of I: New generator from a spatial gradient of the vector potential and current.
The diagram is there.


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