I Counterexample to the Poynting theorem

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A proposed counterexample to the Poynting theorem involves a magnetized ferrite ring with capacitor plates, suggesting a scenario where the Poynting vector indicates energy flow without free current. However, participants argue that the Poynting vector is zero in this configuration, as the electric field in the conductor is also zero, contradicting the claim of a violation. The discussion highlights that the scenario may not represent a true counterexample, as circulating energy flow is possible without propagating waves. Additionally, it is emphasized that any configuration must satisfy Maxwell's equations and cannot simultaneously violate the Poynting theorem. The overall consensus leans towards the need for clarification or correction in the original analysis.
goran d
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The counter-example is as follows: We have a rectangular toroid ferrite(ring ferrite), magnetized in a closed loop around the ring. We put capacitor plates on top and bottom surfaces, with suitable direction. Now the Poynting vector points inwards or outwards. We look at a cylindrical surface passing through the middle of the thickness of the ring. There is surface integral of the Poynting vector! But there is no free current, and electromagnetic energy is constant. Which seems to violate the Poynting theorem. Any exmplanation?
 
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Very clever ! Just a guess=I'm no expert on the topic, is that the Poynting vector applies to electromagnetic wave propagation, and here there is no propagating wave.
 
Feynman talks about just this case. See Figure 27-6 in the attached link. What you have is a circulating energy flow. I think you have no way to prove that electromagnetic energy is not circulating around the ferrite ring, so it is not really a counterexample.
 
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goran d said:
Which seems to violate the Poynting theorem.

That means either that your statement of the problem is wrong, or the analysis is wrong, or that you may misunderstand Poynting's Theorum.

Either post a link to your source, or show diagrams and your math coming to that conclusion. Otherwise, this thread may be deleted.
 
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Here is a diagram of what I mean by this description.
 

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anorlunda said:
or that you may misunderstand Poynting's Theorum.

It can't be that! It's a paradox, I tell you! A paradox!

As drawn, the Poynting vector is zero. The E field in a conductor is zero and the B field, as drawn, does not leave the iron.

I fear the next step will be the Yeahbuts. I stipulate that had he asked a different question he would get a different answer. But for what he actually drew, S is zero.
 
Vanadium 50 said:
It can't be that! It's a paradox, I tell you! A paradox!

As drawn, the Poynting vector is zero. The E field in a conductor is zero and the B field, as drawn, does not leave the iron.

I fear the next step will be the Yeahbuts. I stipulate that had he asked a different question he would get a different answer. But for what he actually drew, S is zero.

Ferrites are typically non-conductive, or at least very resistive. So I don't think you can conclude that E=0.
 
The case of a toroidal wound solenoid, without any core, carrying a DC current would follow the diagram of the OP, and would illustrate what the OP was trying to show.
 
Charles Link said:
he case of a toroidal wound solenoid, without any core, carrying a DC current

Our very first Yeahbut!
 
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goran d said:
Here is a diagram of what I mean by this description.
It appears that you have ignored what @phyzguy posted, and have not read his link... That's not conducive to a good discussion...
phyzguy said:
Feynman talks about just this case. See Figure 27-6 in the attached link. What you have is a circulating energy flow. I think you have no way to prove that electromagnetic energy is not circulating around the ferrite ring, so it is not really a counterexample.
 
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It is not possible for a source/field configuration to both satisfy Maxwell’s equations and violate Poynting’s theorem. However, I must admit that in this case it isn’t clear to me which is the problem.

@goran d can you show that your proposed configuration satisfies Maxwell’s equations?
 
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Dale said:
It is not possible for a source/field configuration to both satisfy Maxwell’s equations and violate Poynting’s theorem. However, I must admit that in this case it isn’t clear to me which is the problem.

@goran d can you show that your proposed configuration satisfies Maxwell’s equations?
In the "link" supplied by @phyzguy in post 3, Feynman basically writes off the dilemma posed by the OP as a non-issue.
 

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