sci.physics.electromag

Hi designers of electronic stuff

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Here is a related newsgroup where I posted about why iron is magnetic.

Reply to
Alan Folmsbee
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Reply to
Long Hair

That group looks like 100% lunatics. There are a lot of physics lunatics.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

I was repelled by this. Not sure if it was by my north or south pole.

Wait maybe I am unipolar. Nah, I just checked. I am bipolar.

Reply to
Beakman

Are you suggesting that the guy who legally changed his name to match his nym, Archimedes Plutonium, is a lunatic?

I once asked there, why some textbooks show EM waves with 2 sine waves at right angles in phase and some texts show them 90 degrees out of phase, but there didn't seem to be a concensus. In phase seems to imply that each peak is a photon, while out of phase seems to imply that energy is constant along the path and is changing form. Or maybe there is no meaningful distinction.

Reply to
Tom Del Rosso

"Tom Del Rosso" wrote in message news:p5666o$fhd$ snipped-for-privacy@dont-email.me...

If you're wondering, the correct description is both (phase shifted and polarized at right angles).

The circuit equivalent is an LC tank having current peak at voltage zero crossing, and vice versa.

One must read textbooks carefully, as they too are prone to errors.

Tim

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Seven Transistor Labs, LLC 
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Reply to
Tim Williams

But not as many as Usenet. ;)

In a wave propagating in a lossless medium, E and H are exactly in phase.

Cheers

Phil Hobbs (In the customs lineup at the Ft Lauderdale cruise terminal)

Reply to
pcdhobbs

Of course they are always depicted at right angles, as I know they should be. The books differ in the phase.

So energy is constant along the path and changing form.

But Phil says otherwise. Still a tie breaker.

Well, Phil's book is one of the most expensive. :)

Reply to
Tom Del Rosso

Does lossless medium mean vacuum?

Reply to
Tom Del Rosso

They certainly are. But AoE was unique in not having any errors at all (the final 2nd edition I mean). But mostly yeah, they're awful.

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Reply to
Cursitor Doom

Yeah, that ^^ :^)

Tim

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Seven Transistor Labs, LLC 
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Reply to
Tim Williams

This has not really clarified the matter. So is it 90 out of phase in a lossy medium? Then is follows the tank circuit analogy, so if Q goes up as the medium becomes lossless then they move into phase? That doesn't make sense either. And what is lossless anyway?

Reply to
Tom Del Rosso

Well, real materials have loss, but many dielectrics are pretty close to lo ssless. For instance, optical fibre's loss tangent is on the order of 1 um / 2pi*10 km or 1E-10. Much of that loss is due to Rayleigh scatter, i.e. t he second order effect from the uniform medium being made up of atoms.

You can see that the two fields are in phase by looking at the curl equatio ns. A plane wave moving towards the +x direction in a lossless medium goes as

exp(i(kx - omega t))

so when you apply the curl equation you get a factor i from both space and time derivatives, i.e. both see a pi/2 phase shift.

Thus since curl E equals some real number times dB/dT, so the two are in ph ase, E is likewise in phase with B. For anybody who's worried about the energy density not being constant with time, you can look at the nulls of |E| and|B| as beats (interference fringe s) between left- and right-circular polarizations.

Cheers

Phil Hobbs

Reply to
pcdhobbs

AFAIK it doesn't make the slightest bit of difference to the phase angle what the hell the medium is.

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Reply to
Cursitor Doom

n a

No, only under total internal reflection--the evanescent field has an imagi nary-exponential time dependence and a real-exponential space dependence, s o the curl equation introduces a pi/2 phase shift.

In mildly lossy dielectrics the phase shift is small--microradians to picor adians--rising to pi/4 in a so-called "normal conductor".

Cheers

Phil Hobbs

Reply to
pcdhobbs

The phase shift between E and B is the arctan of the amplitude loss per radian. If you stick a weak real-exponential decay with x onto the plane wave, the curl equation tells you what the phase shift is--B/E is no longer exactly a real number.

Cheers

Phil Hobbs

Reply to
pcdhobbs

Earlier you said E and H are in phase in a lossless medium. So in a lossy medium are B and H still in phase but E is shifted?

Reply to
Tom Del Rosso

Depends on the loss mechanism. If mu = 1, as it generally does in optics, then epsilon is complex, so E and H aren't perfectly in phase. Amusingly, in the IR, Cu, Ag, and Au have epsilons that are very close to a _negative_ real number.

Cheers

Phil Hobbs

Reply to
pcdhobbs

It's just a phase you're going through.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

Yeah, rather handy, you can derive skin effect that way:

First of all, a capacitive material has real e_r; it follows that a conductive material has imaginary e_r, so that its capacitive impedance is real (resistivity). To maintain analyticity, |e_r| therefore has to be proportional to frequency, so that the resistivity is constant. Similarly, the propagation velocity is lower, proportional to sqrt(|e_r|), and a wave attenuates strongly as it propagates through such a medium. The -3dB cutoff is... the skin depth. :-)

This leads to some interesting observations. In a wire that's relatively small (a few skin depths in diameter), the wave won't be fully attenuated inside the core. Interference occurs, and peaks and nulls arise at certain points. You can actually have greater current density inside a wire than at the surface, for certain combinations (IIRC?). For a cylindrical wire, the solutions are -- of course -- Bessel functions, as often arises in cylindrically symmetric problems. So the nulls are zeroes of those functions.

The drop-off and analyticity dance occurs another layer deeper, where sigma itself (giving rise to imaginary e_r) drops off at some point. Roughly speaking, this is the cutoff frequency of the material, determined by quantum energy levels and the periodic atomic structure. Term is plasma frequency. The lowest PF among pure metals occurs in copper, gold and cesium, having a cutoff in the blue range (reflection below, absorption above). Most metals, this falls somewhere in the UV.

Again, very roughly; if you want to know more, Phil has actual working knowledge in this regime. :)

Tim

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Seven Transistor Labs, LLC 
Electrical Engineering Consultation and Contract Design 
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Reply to
Tim Williams

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