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Faraday shields and radiation and misinterpretations
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE WHILE THE OTHER FIELD VECTORS CANCEL OUT ( I believe that this was the object intended in the cross field antenna) As with a applied varying current leaves a xmitter to create radiation, so must the receiver obtain a time varying current. Maxwells equations show equations with the electric field, the magnetic field and a time varying current. When you have a electrical field or vector of a static particle at rest outside the boundary opposing the static vector on the inside of the boundary you have nothing left EXCEPT a time varying current in the closed circuit. For informative descriptions of how radiation occurs view the QRZ forum of ( antenna construction and design ) threads (3) on the double helix antenna ( see you there) Somebody some where should re write the above such that a definition is left for those who follow and remove the garbage which is now in place |
Faraday shields and radiation and misinterpretations
On Nov 30, 8:10*pm, Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways *THE CHARGE WHEN ACCELERATED *CREATES A TIME VARYING CURRENT ALONE WHILE THE OTHER FIELD VECTORS CANCEL OUT ( I believe that this was the object intended in *the cross field antenna) As with a applied varying current leaves a xmitter to create radiation, so must the receiver obtain a time varying current. Maxwells equations show equations with the electric field, the magnetic field and a time varying current. When you have a electrical field or vector of a static particle at rest outside the boundary opposing the static vector on the inside of the boundary you have nothing left EXCEPT a time varying current in the closed circuit. For informative descriptions of how radiation occurs view the QRZ forum of *( antenna construction and design ) threads (3) on the double helix antenna ( see you there) Somebody some where should re write the above such that a definition is left for those who follow and remove the garbage which is now in place hey there groups archivist, if there is such a thing... remove this post in accordance with his own request that such garbage be removed. |
Faraday shields and radiation and misinterpretations
Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways snip crap, but left plenty WOW!!! Look everybody he's totally making up physics again! Art - get back on your meds, you're nuts again. tom K0TAR |
Faraday shields and radiation and misinterpretations
Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE WHILE THE OTHER FIELD VECTORS CANCEL OUT ( I believe that this was the object intended in the cross field antenna) As with a applied varying current leaves a xmitter to create radiation, so must the receiver obtain a time varying current. Maxwells equations show equations with the electric field, the magnetic field and a time varying current. When you have a electrical field or vector of a static particle at rest outside the boundary opposing the static vector on the inside of the boundary you have nothing left EXCEPT a time varying current in the closed circuit. For informative descriptions of how radiation occurs view the QRZ forum of ( antenna construction and design ) threads (3) on the double helix antenna ( see you there) Somebody some where should re write the above such that a definition is left for those who follow and remove the garbage which is now in place TROLL! |
Faraday shields and radiation and misinterpretations
orfus wrote:
Art Unwin wrote: I have been reading the groups archives on shield antennas and Faraday shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE WHILE THE OTHER FIELD VECTORS CANCEL OUT ( I believe that this was the object intended in the cross field antenna) As with a applied varying current leaves a xmitter to create radiation, so must the receiver obtain a time varying current. Maxwells equations show equations with the electric field, the magnetic field and a time varying current. When you have a electrical field or vector of a static particle at rest outside the boundary opposing the static vector on the inside of the boundary you have nothing left EXCEPT a time varying current in the closed circuit. For informative descriptions of how radiation occurs view the QRZ forum of ( antenna construction and design ) threads (3) on the double helix antenna ( see you there) Somebody some where should re write the above such that a definition is left for those who follow and remove the garbage which is now in place TROLL! Nope. Local loony. You, however, are a troll until proven otherwise. tom K0TAR |
Faraday shields and radiation and misinterpretations
tom wrote in
. net: orfus wrote: Art Unwin wrote: I have been reading the groups archives on shield antennas and Faraday shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE WHILE THE OTHER FIELD VECTORS CANCEL OUT ( I believe that this was the object intended in the cross field antenna) As with a applied varying current leaves a xmitter to create radiation, so must the receiver obtain a time varying current. Maxwells equations show equations with the electric field, the magnetic field and a time varying current. When you have a electrical field or vector of a static particle at rest outside the boundary opposing the static vector on the inside of the boundary you have nothing left EXCEPT a time varying current in the closed circuit. For informative descriptions of how radiation occurs view the QRZ forum of ( antenna construction and design ) threads (3) on the double helix antenna ( see you there) Somebody some where should re write the above such that a definition is left for those who follow and remove the garbage which is now in place TROLL! Nope. Local loony. You, however, are a troll until proven otherwise. tom K0TAR Ok, at the risk of stirring muddy water, I'm curious now, I'm new to this group, and the subject as there clearly seems to be more to it than I knew. I also don't know of those archives mentioned so I haven't seen the context. So in simple terms (hopefully) what is the truth of it? As far as I knew, a photon at RF with energy but no mass will produce a current that changes over time in a metal that it hits, though I imagine that as metal has resistance there must also be a voltage too. I've also heard of the 'skin effect' that means that at high RF frequencies, current flow tends to stay on the surface, so clearly the picture isn't as simple as DC and Ohm's law. I also know that when photons in optical fibres meet boundaries between layers they don't reflect simply on one side, within one region of specific refractive index, there's apparently some more complex information exchange that amounts to the photon crossing the border before returning. Which makes me suspect that equally exotic action happens when RF photons hit metal sheilds. So what IS correct? And even if there is more to it, does the aggregate of many photons, and the wave analysis of their behaviour, reduce to a simple model that makes the OP correct? I'm asking this because calls of 'troll' and 'loony' aren't working for me. |
Faraday shields and radiation and misinterpretations
On Nov 30, 9:05*pm, Lostgallifreyan wrote:
tom wrote e.net: orfus wrote: Art Unwin wrote: I have been reading the groups archives on shield antennas and Faraday shields and the different auguments regarding how shielding or the Faraday shield works. Frankly it is a total mess and should be removed so that hams are not mislead. Shielding is very simple. A particle with a electromagnetic field strikes the outside of the shield. The magnetic field of same passes thru the shield some might say it is coupled to the inside of the shield. The magnetic vector component is out of phase with the electrical field so it will be just a static particle at rest on the inside but no inline with the electrical field vector which is now a staic particle at rest on the outside We now have a arbitrary boundary as discused by Gauss For equilibrium all vectors impinging on the boundary must be aligned such that they cancel. To accomplish this the inner vector or charge MUST move sideways *THE CHARGE WHEN ACCELERATED *CREATES A TIME VARYING CURRENT ALONE WHILE THE OTHER FIELD VECTORS CANCEL OUT ( I believe that this was the object intended in *the cross field antenna) As with a applied varying current leaves a xmitter to create radiation, so must the receiver obtain a time varying current. Maxwells equations show equations with the electric field, the magnetic field and a time varying current. When you have a electrical field or vector of a static particle at rest outside the boundary opposing the static vector on the inside of the boundary you have nothing left EXCEPT a time varying current in the closed circuit. For informative descriptions of how radiation occurs view the QRZ forum of *( antenna construction and design ) threads (3) on the double helix antenna ( see you there) Somebody some where should re write the above such that a definition is left for those who follow and remove the garbage which is now in place TROLL! Nope. Local loony. You, however, are a troll until proven otherwise. tom K0TAR Ok, at the risk of stirring muddy water, I'm curious now, I'm new to this group, and the subject as there clearly seems to be more to it than I knew. I also don't know of those archives mentioned so I haven't seen the context.. So in simple terms (hopefully) what is the truth of it? As far as I knew, a photon at RF with energy but no mass will produce a current that changes over time in a metal that it hits, though I imagine that as metal has resistance there must also be a voltage too. I've also heard of the 'skin effect' that means that at high RF frequencies, current flow tends to stay on the surface, so clearly the picture isn't as simple as DC and Ohm's law. I also know that when photons in optical fibres meet boundaries between layers they don't reflect simply on one side, within one region of specific refractive index, there's apparently some more complex information exchange that amounts to the photon crossing the border before returning. Which makes me suspect that equally exotic action happens when RF photons hit metal sheilds. So what IS correct? And even if there is more to it, does the aggregate of many photons, and the wave analysis of their behaviour, reduce to a simple model that makes the OP correct? I'm asking this because calls of 'troll' and 'loony' aren't working for me. It's fairly straightforward, actually, if you believe in Faraday's law of magnetic induction. That law says that for any closed loop (through air, through a conductor, through anything), there is an electromotive force (a voltage source, if you will) whose magnitude is proportional to the rate of change of magnetic flux enclosed by the loop. As there is no voltage drop along a perfect conductor, if your closed loop follows the path of a perfect conductor, there is no voltage drop around that loop, and therefore the rate of change of the total magnetic flux enclosed by that loop must be zero. If the perfect conductor is a closed box, then you can draw loops anywhere through that conductor and you will never see a changing magnetic field enclosed by that loop. Thus, the inside of the box and the outside are magnetically independent; things happening on one side (magnetically) are not sensed on the other side. You can understand how this works if you realize that a changing magnetic field outside the box that would penetrate the box if it weren't there will induce currents in the conducting box (or even just in a closed loop of wire). Those currents will (in a perfect conductor) be exactly the right magnitude to cause a magnetic field that cancels the external one everywhere inside the closed box (or the net flux enclosed by a loop of wire). An example: if you short the secondary of a mains transformer, the primary will draw lots of current at its rated voltage: it's very difficult for the primary to change the magnetic flux in the core. Does the electric field shielding from a perfect conductor need any explanation? Of course, an imperfect conductor will be an imperfect magnetic shield. But a perfect conductor won't let any change of field through, no matter how slow (no matter how low an EMF it generates), so a perfect conductor works as a shield all the way down to DC. A box made with an imperfect conductor is essentially a perfect shield if the box's wall thickness is at least many skin-depths thick at the frequency of interest. That's a quick beginning. You can find lots more about this in E&M texts. There's even useful stuff about it on the web. ;-) Cheers, Tom |
Faraday shields and radiation and misinterpretations
K7ITM wrote:
I'm asking this because calls of 'troll' and 'loony' aren't working for me. It's fairly straightforward, actually, if you believe in Faraday's law of magnetic induction. That law says that for any closed loop (through air, through a conductor, through anything), there is an electromotive force (a voltage source, if you will) whose magnitude is proportional to the rate of change of magnetic flux enclosed by the loop. As there is no voltage drop along a perfect conductor, if your closed loop follows the path of a perfect conductor, there is no voltage drop around that loop, and therefore the rate of change of the total magnetic flux enclosed by that loop must be zero. If the perfect conductor is a closed box, then you can draw loops anywhere through that conductor and you will never see a changing magnetic field enclosed by that loop. Thus, the inside of the box and the outside are magnetically independent; things happening on one side (magnetically) are not sensed on the other side. You can understand how this works if you realize that a changing magnetic field outside the box that would penetrate the box if it weren't there will induce currents in the conducting box (or even just in a closed loop of wire). Those currents will (in a perfect conductor) be exactly the right magnitude to cause a magnetic field that cancels the external one everywhere inside the closed box (or the net flux enclosed by a loop of wire). An example: if you short the secondary of a mains transformer, the primary will draw lots of current at its rated voltage: it's very difficult for the primary to change the magnetic flux in the core. Does the electric field shielding from a perfect conductor need any explanation? Of course, an imperfect conductor will be an imperfect magnetic shield. But a perfect conductor won't let any change of field through, no matter how slow (no matter how low an EMF it generates), so a perfect conductor works as a shield all the way down to DC. A box made with an imperfect conductor is essentially a perfect shield if the box's wall thickness is at least many skin-depths thick at the frequency of interest. That's a quick beginning. You can find lots more about this in E&M texts. There's even useful stuff about it on the web. ;-) Here is a link to a generalized proof of the skin effect: http://www.ifwtech.co.uk/g3sek/misc/skin.htm This is exactly equivalent to Tom's explanation above. The detailed proof is quite mathematical but it is solidly based in classical physics - Faraday's Law and Ampere's theorem (both of which are embodied in Maxwell's equations). This derivation produces the well-known equations for current density as a function of depth, conductivity and permeability. The special feature of this particular proof is that it's much more general than the ones you see in better-known textbooks - and therefore much more powerful. It shows that if RF current is flowing in/on *any* conducting surface, for *any* reason, then the skin effect will be present. The possible reasons why RF current may be flowing can be divided into two main groups: * "Circuit conditions" - the conductor is part of a circuit that makes RF current flow. * "Electromagnetic induction" - the conductor is intercepting an incident electromagnetic wave which induces a current. In either case, an RF current flows... and wherever that happens, there you will also find the skin effect. -- 73 from Ian GM3SEK http://www.ifwtech.co.uk/g3sek |
Faraday shields and radiation and misinterpretations
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Faraday shields and radiation and misinterpretations
Ian White GM3SEK wrote in
: http://www.ifwtech.co.uk/g3sek/misc/skin.htm This is exactly equivalent to Tom's explanation above. The detailed proof is quite mathematical but it is solidly based in classical physics - Faraday's Law and Ampere's theorem (both of which are embodied in Maxwell's equations). This derivation produces the well-known equations for current density as a function of depth, conductivity and permeability. Thanks, that linking of laws and theories will help me (and the confirmation that classical physics will be enough to describe it, as I hoped). The OP mentioned Maxwell too... so did he make some error I have yet to grasp? Other than taking pot shots at an establishment, that is... :) The special feature of this particular proof is that it's much more general than the ones you see in better-known textbooks - and therefore much more powerful. It shows that if RF current is flowing in/on *any* conducting surface, for *any* reason, then the skin effect will be present. That appeals to me. I think the more something can be seen to apply generally, the more it helps. Proportion can't be gauged with a model that denies it. |
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