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Author Topic: Displacement Current - Does it Exist?  (Read 126341 times)

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tExB=qr
Coincidentally, Spherics did state that the RE particles:

As they stream out they are not deflected by anything, affect everything they touch, and reduce affect according to inverse square of the distance from source.
   

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Coincidentally, Spherics did state that the RE particles:

As they stream out they are not deflected by anything, affect everything they touch, and reduce affect according to inverse square of the distance from source.

Inverse square, or inverse cubed?


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Inverse square, or inverse cubed?

He distinctly said "inverse square".
   
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I just came across an interesting paper that explains why displacement currents are a myth.
Let's be clear, displacement currents are a simple way of dealing with problems, and can be used as such, but they are not a physical reality.

In http://www.asps.it/miller.pdf Miller starts from Jefimenko's work and explains to us that Maxwell's equations are non-local since they instantly connect distant objects, like B and E, or D and ρ. If they can do this, it is not that B instantly creates E somewhere else (or vice versa), it is that these objects have common causes.

What are the causes?
1. E is a function entirely of electric charge density and current density.
2. H is a function entirely of current density

So when we write e.g. ∇xE = ∂B/∂t, it is not that a varying magnetic field creates an electric field.
The only causes of the effects are the charges, and the movements of the charges. So a displacement current, unlike a charge current, cannot create a magnetic field or electromagnetic waves. This justifies justifies my replies #180 and #184.
Cherry on the cake, this is perfectly in line with relativity.


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I just came across an interesting paper that explains why displacement currents are a myth.
Let's be clear, displacement currents are a simple way of dealing with problems, and can be used as such, but they are not a physical reality.

In http://www.asps.it/miller.pdf Miller starts from Jefimenko's work and explains to us that Maxwell's equations are non-local since they instantly connect distant objects, like B and E, or D and ρ. If they can do this, it is not that B instantly creates E somewhere else (or vice versa), it is that these objects have common causes.

So when we write e.g. ∇xE = ∂B/∂t, it is not that a varying magnetic field creates an electric field.
The only causes of the effects are the charges, and the movements of the charges. So a displacement current, unlike a charge current, cannot create a magnetic field or electromagnetic waves. This justifies justifies my replies #180 and #184.
Cherry on the cake, this is perfectly in line with relativity.

Seems to have a world of possibilities, as a displacement current lacking an associated magnetic field would not be subject to normal ohmic losses (R=V/I). ;D


Even more provocative since it doesn't require fundamental formulas+principles to be rewritten, just re-interpreted.


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Seems to have a world of possibilities, as a displacement current lacking an associated magnetic field would not be subject to normal ohmic losses (R=V/I). ;D
...

You are going too fast. The absence of a magnetic field created by a displacement current does not mean the absence of a magnetic field "as if it were created by the displacement current".
The movement of charges along the capacitor plates creates a magnetic field of non-trivial topology.


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You are going too fast. The absence of a magnetic field created by a displacement current does not mean the absence of a magnetic field "as if it were created by the displacement current".
The movement of charges along the capacitor plates creates a magnetic field of non-trivial topology.
Quote
a displacement current, unlike a charge current, cannot create a magnetic field or electromagnetic waves.

That isn't to say there are no losses over distance.  Dielectric conductance loss still applies as well as loss/scattering from impedance mismatch.
Theory provides a good foundation for experimenters to know what to look for.


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That isn't to say there are no losses over distance.  Dielectric conductance loss still applies as well as loss/scattering from impedance mismatch.
Theory provides a good foundation for experimenters to know what to look for.

C.C
You juxtapose things without any logic between them.


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C.C
You juxtapose things without any logic between them.

Just deductive/dimensional analysis.

If
Quote from: F6FLT
"a displacement current cannot create a magnetic field or electromagnetic waves."
Then I=0 for a given displacement action. :o

Then, if
V=IR
R=V/I

then, the DC resistance for a displacement action would be 0 or undefined.

(which doesn't mean lossless however because that is only one field component)


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Just deductive/dimensional analysis.

IfThen I=0 for a given displacement action. :o

Then, if
V=IR
R=V/I

then, the DC resistance for a displacement action would be 0 or undefined.

(which doesn't mean lossless however because that is only one field component)

Always a nonsense. A displacement current is never in DC.
And what you say would also be true between the plates of a capacitor, but there can be losses even when the resistance is infinite.



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Always a nonsense. A displacement current is never in DC.
And what you say would also be true between the plates of a capacitor, but there can be losses even when the resistance is infinite.

Of course the magnetic loss tangent is not the only factor.  I stated as such in both posts as to other loss factors that would affect and limit real performance. 
It makes sense too, as otherwise we would have probably detected superconductivity in every capacitor over a century ago. C.C


But back to the subject, "a displacement current, unlike a charge current, cannot create a magnetic field or electromagnetic waves" clearly has some interesting ramifications.
One could envision a circuit with conduction current mostly acting in one direction with displacement current mostly acting in the other direction.  Would not such activity appear in-practice as a magnetic monopole?


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Why I said there is no displacement current around a coil:

It is clear from the attached diagram that in the space between the plates of a capacitor, the dipoles orient themselves by rotation in the electric field.
Around a coil, the same thing, the dipoles also rotate around their electric centre and orient themselves according to the EMF loop of the electromagnetic induction.

But only in the first case, the dipoles oppose their field to the one imposed by the plates, which creates this displacement that we note D=Ɛ.E.
In the second case, as the dipoles are in a loop, their electric field cancels out over one turn. They cannot oppose the induction EMF.

This explains why if you place a coil in water, which has a high permittivity (80), you will not see any effect on the primary current.

In the case of induction, can the rotation of the dipoles on themselves but without displacement be detected or even used? I think this is a subject that should be worked on.




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The displacement occurs when the charges in one plate of the capacitor "displace" the charges in the other plate of the capacitor.

During this act of displacing charges (charging and discharging), a magnetic field should exist.  Many sources state that this magnetic field is between the plates.

See figure 17.2 here:
https://phys.libretexts.org/Bookshelves/University_Physics/Radically_Modern_Introductory_Physics_Text_II_(Raymond)/17%3A_Capacitors_Inductors_and_Resistors/17.01%3A_The_Capacitor_and_Amperes_Law
   
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...
During this act of displacing charges (charging and discharging), a magnetic field should exist.  Many sources state that this magnetic field is between the plates.
...

It's likely. It remains to be seen what it is created by, most probably by the movement of charges in the plates, even if the magnetic field is around the area between plates.
Nothing but moving charges can create a magnetic field.
Moreover, your link explains it:
"The quantity  ϵ0.dΦE/dt  was called the displacement current by Maxwell since it has the dimensions of current and is numerically equal to the current entering the capacitor. However, it isn’t really a current — it is just an electric flux that changes with time!"


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You can look at displacement current as the separation of opposite charges.
A means to continually apply a displacement might resemble a battery that never runs out of charge.
   
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The displacement occurs when the charges in one plate of the capacitor "displace" the charges in the other plate of the capacitor.

During this act of displacing charges (charging and discharging), a magnetic field should exist.  Many sources state that this magnetic field is between the plates.

See figure 17.2 here:
https://phys.libretexts.org/Bookshelves/University_Physics/Radically_Modern_Introductory_Physics_Text_II_(Raymond)/17%3A_Capacitors_Inductors_and_Resistors/17.01%3A_The_Capacitor_and_Amperes_Law

Hello, been following this thread with interest, some good opinions.

@ Grumpy, looked @ that link, and his description of the source of the magnetic field doesn't make sense to me. How is the A vector orthogonal to the changing E Field?  Isn't the E field from a changing current source is -dA/dt? It's not a cross product?  Whats interesting about that is that if you look at the displacement current in parallel plate capacitor, with equal sized plates, with a uniform but changing E field across them with respect to the A vector, then you should come to the conclusion that there is no magnetic field between the plates.  Visually, without any math.
   
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Hello, been following this thread with interest, some good opinions.

@ Grumpy, looked @ that link, and his description of the source of the magnetic field doesn't make sense to me. How is the A vector orthogonal to the changing E Field?  Isn't the E field from a changing current source is -dA/dt? It's not a cross product?  Whats interesting about that is that if you look at the displacement current in parallel plate capacitor, with equal sized plates, with a uniform but changing E field across them with respect to the A vector, then you should come to the conclusion that there is no magnetic field between the plates.  Visually, without any math.

See the right view of the following image:



It is clear that there is a radial current from the connection point in the center of the plates, current that gives rise to a vector potential, radial as well.
As B=∇×A, there is also a magnetic field which is circular. As its influence extends beyond the plates, it is also found between the plates, where those of the two plates add up.



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Why I said there is no displacement current around a coil:

It is clear from the attached diagram that in the space between the plates of a capacitor, the dipoles orient themselves by rotation in the electric field.
Around a coil, the same thing, the dipoles also rotate around their electric centre and orient themselves according to the EMF loop of the electromagnetic induction.

What happens in vacuum where there are no dipoles?

Quote
But only in the first case, the dipoles oppose their field to the one imposed by the plates, which creates this displacement that we note D=Ɛ.E.

I think you read too much into the word displacement.  Yes D is known as displacement but it is also a flux density that occurs in vacuum where there are no charges to displace.  And what do you mean by the dipoles oppose when their presence increases the outward force on the charges residing on the plates and increases the flux density?
 
Quote
In the second case, as the dipoles are in a loop, their electric field cancels out over one turn. They cannot oppose the induction EMF.

What do you mean by cancellation?  Their presence increases the flux density around the loop.  The E field remains the same, the induction EMF remains the same.

Quote
This explains why if you place a coil in water, which has a high permittivity (80), you will not see any effect on the primary current.

Really?  Can you give us details of this experiment?  Or is this just your opinion?  You use the words primary current so presumably this is not just a coil but a transformer.

Quote
In the case of induction, can the rotation of the dipoles on themselves but without displacement be detected or even used? I think this is a subject that should be worked on.

That is an interesting remark.  It suggests that a DC charged parallel plate capacitor with a dielectric between the plates could have the effect of the dielectric modulated by a crossed magnetic field that is alternating in magnitude, thus inducing circular E field attempting to rotate the dipoles.  Then we should see an alternating current in the DC supply to the capacitor.

Smudge
   

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See the right view of the following image:



It is clear that there is a radial current from the connection point in the center of the plates, current that gives rise to a vector potential, radial as well.
As B=∇×A, there is also a magnetic field which is circular. As its influence extends beyond the plates, it is also found between the plates, where those of the two plates add up.

If the conducting wire had conical end pieces bringing the current to the whole plate surface and not just to the center, then the radial currents would not exist.  Whether the so-called displacement current "flowing" across the gap creates a magnetic field is not considered in that paper.

Smudge
   
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If the conducting wire had conical end pieces bringing the current to the whole plate surface and not just to the center, then the radial currents would not exist.  Whether the so-called displacement current "flowing" across the gap creates a magnetic field is not considered in that paper.

Smudge

In cases where there is no radial spreading of electrons, the longitudinal current does create a circular field around it.
Near the gap between plates, which must remain small to be in the approximation of the plane capacitor, it will be difficult to say if it is not this field that we detect, rather than a hypothetical magnetic field resulting from a current in space which is not a current of electric charges.

Having said that, I can see where you're going with this, perhaps this displacement current would be a current in the aether, wouldn't it?


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In cases where there is no radial spreading of electrons, the longitudinal current does create a circular field around it.
Near the gap between plates, which must remain small to be in the approximation of the plane capacitor, it will be difficult to say if it is not this field that we detect, rather than a hypothetical magnetic field resulting from a current in space which is not a current of electric charges.

Having said that, I can see where you're going with this, perhaps this displacement current would be a current in the aether, wouldn't it?

Yes, that is one way to look at it.  I think the formula for displacement current density JD = ∂D/∂t applies everywhere and gave rise to Maxwell's fourth equation Curl B = μ0(J + ∂D/∂t).  That tells you displacement current produces a magnetic field that has curl.

Smudge
   
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Yes, that is one way to look at it.  I think the formula for displacement current density JD = ∂D/∂t applies everywhere and gave rise to Maxwell's fourth equation Curl B = μ0(J + ∂D/∂t).  That tells you displacement current produces a magnetic field that has curl.

Smudge

This equation doesn't prove anything. Maxwell's equations don't indicate causality. It's like B=∇xA, it doesn't prove that B would generate A or vice versa, it only proves that they have a common cause. And we know what that is: charges and currents.
In all conductor configurations, conservation laws cause charges and currents to arrange themselves in such a way that their mutual influences are equivalent to what the displacement current would produce. An ether will therefore be difficult to prove experimentally, and until we have this proof, there is no point in assuming it since everything is explainable without it.



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This equation doesn't prove anything. Maxwell's equations don't indicate causality. It's like B=∇xA, it doesn't prove that B would generate A or vice versa, it only proves that they have a common cause. And we know what that is: charges and currents.
In trigonometry we have functions like sine, cosine, tangent.  We also have the inverse functions written sin-1and so on.  You mention B=curl A. There is also an inverse function A=curl-1B which means the vector A whose curl is B.  It means they both coexist.  So a B field coexists with a displacement current.

Smudge
   
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...
So a B field coexists with a displacement current.
...

The problem is not specific to electromagnetism but to physics in general. The concepts we choose for the models do not correspond to a precise physical reality.
Nobody denies the displacement current. Nobody denies the electronic current. Nobody denies the magnetic field...  But these concepts are redundant, the proof is the equalities of the equations. If B=∇xA, then what is physical reality? B ? A ? Neither?

These notions are not tangible realities as if we have identified objects in the universe with intrinsic properties, when all this is just a representation of what we measure.
The question "do displacement currents exist" is nonsense. Our mathematical concepts exist, but only as mathematical concepts, not as physical objects. Each mathematical concept does not correspond directly to a physical object, only to a representation that we have of the effects. An electron is a particle that corresponds to a physical reality that we can discern but are unable to describe precisely. A displacement current is even worse, it's just C/s, what reality can we attach to it? Anything at all.


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@F6,

I agree that B, A and displacement current are all just math entities.  But to get an idea of whether displacement current really is present around a magnetic core of changing flux (as the math tells us it is) you can carry out a gedanken experiment.  You know that if you take a single turn connected to a parallel plate capacitor current will flow, and that includes the displacement current through the capacitor.  You could connect a whole chain of these capacitors in series around the loop and current will still flow.  Then gradually increase the number of capacitors where each time the actual conductor length gets smaller, we still get current.  Taken to the limit we simply have a ring of dielectric material around the core.  So a good practical experiment would be to have two wound ring cores that are coupled via a dielectric ring, i.e. the three rings form links in a chain.  Then observe the coupling between the two windings.  We know that if the center ring were made of conductive material there would be coupling because of the conduction currents in that ring.  If coupling still existed with the dielectric ring then that is absolute proof of displacement current.

Smudge
   
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