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Author Topic: Does Dielectric Displacement Current generate a magnetic field?  (Read 18573 times)

Dave45

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #15 on: May 30, 2016, 02:50:12 PM »
Transformer cores are constructed to eliminate eddy currents but what if you built one to accommodate and collect eddy currents.
Work with the system
www.olympus-ims.com/en/ndt-tutorials/eca-tutorial/what-is-eca/basic

Dave45

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #16 on: May 30, 2016, 03:15:13 PM »
Eddy currents cause bemf in a dc pulsed coil

Dave45

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #17 on: May 30, 2016, 03:27:31 PM »
Wind a coil on a welding rod then position welding rods around the circumference of the coil then wind a coil on top of the rods then more rods then another coil, repeat, repeat.

What happens, the eddy currents circulating in the coils magnify the magnetic field every layer, work with the structure of the field.

SolarLab

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #18 on: May 30, 2016, 06:28:30 PM »
Hi there
Here is my take on the issue.
In order to try to address your question, the question should be properly defined.
1, If your question is based on the Maxwell displacement current then “Maxwell displacement current” either in vacuum or in dielectrics, neither generate magnetic field nor are sensitive to external magnetic fields. In other words current in dielectrics “The polarization current” does not act with potential forces on other currents and “external magnetic field” does not react with kinetic forces to the action of other currents.
2, Displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.
To conclude the above I quote from
“Harry McLaughlin” https://www.quora.com/profile/Harry-McLaughlin
a. A "changing" electric field CANNOT create a magnetic field
b. A "changing" magnetic field CANNOT create an electric field “end of quote”
3, The so-called "displacement current" term(1/4π)  ∂E/∂t  is not some current density generating magnetic field, as Maxwell supposed. This term gives information about the conduction currents which have been interrupted in the neighborhood of the reference point.
4, For what it worth, we cannot measure magnetic field produced by displacement currents but we can measure exactly the field of the interrupted conduction currents. Even if the details are not so obvious and require a skill to understand but what the equations imply is that the electric and magnetic fields depend only on the source charges. It is our orientation relative to the source charges and their motions that give rise to the details of the fields we measure.

From Maxwell basic equations the only sensible is the existence of ε0 and µ0
∂E/∂x=-Z0 ϵ0  ∂E/∂t
∂H/∂x=-μ0/Z0 ∂H/∂t

And they express that the E field causes the E Filed and The H field causes the H field
WAW what a discovery!!!!

The Equations only express that E and H fields are co-existent, co-substantial, and co-eternal like any two perpendicular sides of a brick neither the length affects the width or the other way around.

Hope it helps
jj

F.Y.I.

IMHO you are definitely on the "right track!" 

However consider/explain this: A CRT TV tube uses a "gun" structure to create a fast (HV) electron beam that moves towards a phosphorous coated "screen" to form the picture (glowing pixels); and this beam is steered using magnetic fields (the yoke coils)?

Also consider: An RF "Slotted Line" technique (an older but common RF Test and Measurement Instrument) - e.g. two parallel wires spaced close together; a generator at one end and a termination at the far end. If the line is impedance matched at the far end, with respect to the generator sine wave frequency, there is no reflection (narrow band). If the far end termination is a short; current is at maximum (minus line losses) and voltage is at minimum. If the far end termination is open; voltage is maximum and current is minimum. In all cases, except a matched termination, a "Standing Wave" (VSWR for example) will appear along the line, max and min are sub-multiples of the test frequency. It gets a bit more complicated when a "pulse" is used since the pulse contains multiple sine waves (see Fourier analysis).

A "piece-wise linear model" of this parallel wire yields: the conductors are modeled by placing many inductors along the line(s) and between these lines the model contains many capacitors connecting the two lines (the impedance formed between the lines is a complex number consisting of Resistance, Inductance and Capacitance with respect to Frequency and the physical characteristics of the lines and the adjacent environment. It becomes difficult at best to "guess analyze" the characteristics and performance (even when using a Smith Chart and simple coaxial or twin lead lines).

Basics: initially consider how a capacitor, an inductor and a resistance will respond to both voltage (potential difference) and current (movement of charge - and associated magnetic field) with respect to time (t~0 through many cycles or pulses).

Now consider one or more helical coils of conductive wire (a slow wave structure); fed by either or both a sine wave and a high voltage pulse; while also considering the signals velocity of propagation through the system; electron velocity modulation (kinetic energy); the systems impedance changes along the line and terminations; reflections (standing waves); and so forth... Add to this task the fact that  for nearly a hundred years we've used incorrect/incomplete Maxwell's equations which do not consider a Longitudinal wave (a.k.a Scalar) electromagnetic component.

From what I know, Displacement Current is the "gotcha" of Maxwell's equations. It's the "thing" that has no logical/mathematical solution via present day theory (or at least trying to get a good engineering answer from the guru's has yet proved elusive). Someone here mentioned "Tetryonic" theory - it may or may not be a good postulation, time will bare it out, but it is well worth a detailed review/study in my opine.

jj good stuff - [beware of Harry however] - check out Dr./Prof. Constantine Meyl...

FIN

SolarLab

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #19 on: May 30, 2016, 06:55:05 PM »
F.Y.I.

Sorry, forgot an interesting reference and answer the first sentence:

http://agni.phys.iit.edu/~vpa/electromagnetic.html

Quote from within the above:

"Why not use magnets (instead of electric fields) to accelerate the beam? The force that the electric charge (beam) feels due to a magnetic field is always perpendicular to path the charge follows. Since the magnetic force always acts at right angles to the motion of a charge, it can only turn the charge, it cannot do work on the charge. So, electric fields, which can be oriented to act parallel to the motion of the particles, are used to accelerate particles. Although a particle accelerator complex often has many magnets, these are used not to increase the beam energy, but to control the direction of motion of the particles, pointing or focusing the beam.

Explaining this with a simple analogy, imagine the chair in which you are sitting. You want to roll across the room. The chair is pushing you with an upward force, while you are naturally pushing against it with a downward force. This upward force is perpendicular to the direction you want to go. In order to go across the room, you need another force that will push you parallel to the floor, i.e. perpendicular to the normal force of the chair. You still want the have the force of the chair as well, to keep you off the floor/carpet itself (because dragging on the carpet really burns). Now replace yourself with a small particle, replace the chair with a magnetic field, and replace the person nice enough to push you across the room with an electric field. Now you have a simplified acceleration situation, except in the acceleration process, the beam is propelled to travel because of an electric potential difference, or voltage difference, due to an electric field."

FIN

SolarLab

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #20 on: May 31, 2016, 06:38:29 AM »
F.Y.I.

Or even more simply:

Electric charges or Potential Differences can be "exchanged or transferred" either by direct contact (e.g. wire conductor) or by {close} proximity (electrostatic influence).

Current is generally associated with electron "flow" through a conductor and occurs when an electric charge is exchanged between potential charge differences (e.g. a wire circuit with resistance and a battery or oscillatory generator). Within the conductor, electrons "flow" [current] while external to the conductor, magnetic fields "flow."

Electrostatic influence potential difference (charge) transfer does not use a conductor to facilitate "electron flow" therefore no current appears and thus there is no magnetic field. This assumes dry air or a pure dielectric. Arc discharges are beyond the scope here.

The conundrum with Maxwell's equations occurs when a capacitor is considered since there is no direct connection (conductor) to facilitate current "flow" so, it appears, to preserve the "logic" a fudge factor was created. This has come to be known by many names including "Displacement Current". So the question might become; "Does a capacitor have a magnetic field?"

However, if you consider the missing "scalar" part of Maxwell's equations or refer to the original complete equations initially created by Maxwell, there is no anomaly. But this would create another problem (?); that being the likely existence of the illusive {forbidden} aether and the existence of light as both a particle and a wave! Have faith however, there are work-arounds - not too elegant, but, break the problem into both Electromagnetics and Particle Physics!

Also see: http://agni.phys.iit.edu/~vpa/information.html

FIN

Reiyuki

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #21 on: May 31, 2016, 06:27:08 PM »
Thanks for all that SolarLab, there's a lot to fully process there.  In general, it seems that exploiting the well defined but often neglected concepts is going to be the way forward, to getting a mathematically (or at least conceptually) modeled device out of all this.


I've had plenty of luck getting voltage from HV displacement, but getting current attached to it is a difficult matter.  Usually I get orders of 100's of volts output but in mere microamps.

Using the concepts you've presented, how do you think that issue could be solved?  Separating source-from-discharge like Bearden's 'degenerate semiconductor'?  Or 'boosting' running currents with high voltage like TK devices seem to do?


thanks,
Rei



F.Y.I.

Or even more simply:

Electric charges or Potential Differences can be "exchanged or transferred" either by direct contact (e.g. wire conductor) or by {close} proximity (electrostatic influence).

Current is generally associated with electron "flow" through a conductor and occurs when an electric charge is exchanged between potential charge differences (e.g. a wire circuit with resistance and a battery or oscillatory generator). Within the conductor, electrons "flow" [current] while external to the conductor, magnetic fields "flow."

Electrostatic influence potential difference (charge) transfer does not use a conductor to facilitate "electron flow" therefore no current appears and thus there is no magnetic field. This assumes dry air or a pure dielectric. Arc discharges are beyond the scope here.

The conundrum with Maxwell's equations occurs when a capacitor is considered since there is no direct connection (conductor) to facilitate current "flow" so, it appears, to preserve the "logic" a fudge factor was created. This has come to be known by many names including "Displacement Current". So the question might become; "Does a capacitor have a magnetic field?"

However, if you consider the missing "scalar" part of Maxwell's equations or refer to the original complete equations initially created by Maxwell, there is no anomaly. But this would create another problem (?); that being the likely existence of the illusive {forbidden} aether and the existence of light as both a particle and a wave! Have faith however, there are work-arounds - not too elegant, but, break the problem into both Electromagnetics and Particle Physics!

Also see: http://agni.phys.iit.edu/~vpa/information.html

FIN

SolarLab

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #22 on: June 01, 2016, 06:13:21 PM »
F.Y.I.

The following video's illustrate only one of several approaches to be investigated; others include Velocity of Modulation as seen in Traveling Wave Tubes; electrical current related magnetic field interactions; earth ground effects with respect to potential difference; Breakaway Electrons; and so forth.

Large credits to those who have provided these valuable insights.

Playlist [5 total] - High Voltage Step Down Transformer:
https://www.youtube.com/playlist?list=PLIF_6L5V8Qm79grxbveQTBfKLxwl5RVgv

Individual Videos (some repeats of above):
https://www.youtube.com/watch?v=Zilvl9tS0Og
https://www.youtube.com/watch?v=9LtPeCBHtoY
https://www.youtube.com/watch?v=k06S-01HBqQ
https://www.youtube.com/watch?v=g6ipm6r7h-Y
https://www.youtube.com/watch?v=NNZFDaEYiDM
https://www.youtube.com/watch?v=aZcZkyd9wco


Consider - a foil (open or split) can be electrically similar to a 'close wound' coil (an "area" of conductive material). Wire insulation is dielectric (similar to PVC pipe). Note also, the system "capacitor - energy storage" mechanism.

Consider - Wimshurst RPM at KHz or MHz (electronic operation at radio frequencies {RF}). Bovin Kacher, TT, etc. replacing the Wimshurst HV generator.

Consider - timing is {most likely} critical (injection; SWR; pulse, load shed/sync, etc.).

One general objective is to "model" the system employing modern CAD/CAM/CAE followed by experimental engineering design verification. As Chris B. (SA) once quoted "Anyone can make anything work once; but it takes a good designer to make it reproducible and work well over time and temperature!"

FIN

SolarLab

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #23 on: June 04, 2016, 04:36:31 PM »
F.Y.I.

Note: this is a modified re-post of the above - an account problem (ongoing) makes posting/editing almost impossible (???).

High Voltage Step Down Transformers is one of several approaches being investigated; others include Velocity Modulation (VM) as seen in Traveling Wave Tubes (TWT); electrical current related magnetic field interactions (typically the most pursued theory but little success has been demonstrated to date, by most investigators); earth ground effects with respect to potential difference (new theories are emerging relating to formation/cause of Lightning - "Breakaway Electrons;" {Atmospherics/Geophysics}); and so forth.

Credit to those who have provided valuable insights.

Playlist [5 total] - High Voltage Step Down Transformer:
https://www.youtube.com/playlist?list=PLIF_6L5V8Qm79grxbveQTBfKLxwl5RVgv
Individual Videos (some repeats of above):
https://www.youtube.com/watch?v=qwVOp-HPIVE
https://www.youtube.com/watch?v=Zilvl9tS0Og
https://www.youtube.com/watch?v=9LtPeCBHtoY
https://www.youtube.com/watch?v=k06S-01HBqQ
https://www.youtube.com/watch?v=g6ipm6r7h-Y
https://www.youtube.com/watch?v=NNZFDaEYiDM
https://www.youtube.com/watch?v=aZcZkyd9wco

Consider - a foil (open or split) can be electrically similar to a "close wound" helical coil (an "area" of conductive material). Wire insulation is dielectric (similar to PVC pipe).

Note the system's "capacitor energy storage/transfer" mechanism- rapid electrostatic charge [KHz, MHz]; electrostatic influence of a second (or more) output coil [very rapid - without current/magnetic slowdown]; then "on demand" withdrawal of energy from the output coil [current/magnetic] to load [controlled by impedance of load]. Extremely rapid Source charging - feeding a relatively slow (current limited by magnetics) discharge.

Resonance assists the mechanism since "R" is infinite (theoretically) and the source/load {coils - parasitic capacitance, intrinsic inductance and lumped resistance} will likely Oscillate as well {if the feed HV feed pulse, etc., are "tuned" to the system} and may include a Standing or Traveling Wave. OHM's Law applied in quick fashion! Just a postulate, but hopefully we will soon know if it holds water, so to speak. Inductive or Capacitive coupling of the system will also have significant implications. Investigate using Slotted Line techniques with (1) series L,shunt C and (2) series C, shunt L [piece-wise linear transmission line model].

Consider - Wimshurst RPM at KHz or MHz (electronic operation at radio frequencies {RF}). Bovin Kacher, TT, etc. replacing the Wimshurst HV generator.

Consider - timing is likely critical (injection; Standing Wave - Oscillation; pulse feed, load shed/sync, etc.).

One general objective is to "model" the system employing modern CAD/CAM/CAE followed by experimental engineering design verification. As Chris B. (SA) once quoted "Anyone can make anything work once; but it takes a good designer to make it reproducible and work well over time and temperature!"

Electrostatic stuff:
https://www.youtube.com/watch?v=XofdRjwuAu8&list=PLw28_n7AgcmAAWJzDI_6E6C_gtHoh9u2J

Lots could be said about these "ou account" problems, and such, but enough for now.
Good luck to all in sorting through this fascinating mystery!

FIN

SolarLab

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Re: Does Dielectric Displacement Current generate a magnetic field?
« Reply #24 on: September 14, 2018, 03:29:25 AM »
F.Y.I.

The Derivation of the Formula for Displacement Current

https://www.youtube.com/watch?v=MjOly3h7brs

The magnetic circulation around any path depends on the electric current linked through the path. The only way to be sure that the current goes through the path is to imagine a membrane bounded by the path. And the same current must pass through the membrane , no matter what shape the membrane takes.

But that is no longer true if the current flows into a capacitor. Then even though current is flowing , there may be none through the membrane. If so, what is the magnetic circulation equal to?
 Seeking an answer to that question, Maxwell took a page from Faraday's book. Changing magnetic flux creates electric circulation, and then as usual, he looked at things the other way around.

Could changing electric flux, he wondered, create magnetic circulation. The answer promised to solve the capacitor problem. As current flows into the capacitor, charge builds up, which creates an increasing electric field between the plates. The electric flux through the membrane can be deduced by Gauss's Law by imagining a closed surface . All the flux goes through the dome shaped membrane and its equal to the charge on the capacitor plate over epsilon naught.

The rate of change of electric flux can be found by differentiating both sides of the equation. It is given simply by the current flowing in the wire. In other words, epsilon nought times the rate of change of electric flux through the dome shaped membrane is the same as the electric current through the flat membrane.

This was Maxwell's crucial discovery :the precise manner in which changing electric flux can generate a magnetic field as if it were a kind of electric current.

In fact, Maxwell himself called this Mathematical term "the displacement current." In other words, the magnetic circulation around the closed path is given, not only by the electric charge for it but also by the rate of change of electric flux through it.

This is how James Clark Maxwell completed the laws of electricity and magnetism.

FIN