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This question concerns the article Right Angle Circuitry.

Having passed a grad E&M course, I feel like this should make perfect sense to me and I should be able to answer all the questions it raises. Sadly that is not yet the case.

What are the answers to all the unanswered questions here?

In particular:

  • Why does the circuit in Figure 8 not act as a transformer?
  • Is it possible to extract energy from a toroid with capacitor plates as shown in Figure 9? If not, why not?
  • Have structures like that in Figure 10 ever been used in practice as transmission lines?
  • Last question in the article: Is it possible to create "insulating electric circuitry which conducts only displacement current"?
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closed as too localized by Sklivvz, Waffle's Crazy Peanut, Manishearth Jan 3 '13 at 13:25

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what i believe, just forget about that A-field thing. Assume rest same as what you have learnt in ur grad school. Even google is silent about "A-field" google.com/… –  Vineet Menon Oct 5 '11 at 9:53
    
I don't understand. I'm already familiar with the magnetic vector potential field, denoted A. The A field is not mysterious. –  Keenan Pepper Oct 15 '11 at 18:35
    
ooh...its that!!! denotation...sry.. –  Vineet Menon Oct 16 '11 at 6:34
    
"a field of Vector Potential ... commonly called the "A-field." quoted from amasci.com/elect/mcoils.html –  raindrop Dec 28 '12 at 2:19
    
Please see our homework policy. We expect homework problems to have some effort put into them, and deal with conceptual issues.In addition, we prefer there to be one "question" per post. If you edit your question to explain (1) What you have tried, (2) the concept you have trouble with, and (3) your level of understanding, I'll be happy to reopen this. (Flag this message for ♦ attention with a custom message, or reply to me in the comments with @Manishearth to notify me) –  Manishearth Jan 3 '13 at 13:25

2 Answers 2

up vote 2 down vote accepted
  • The figure 8 object does not have a magnetic flux through it at any time, so it doesn't have a changing magnetic flux, so it doesn't have an EMF.
  • Yes, it is possible with AC current--- in fact it is routine! You take the EMF and feed it to a small capacitor. Circuits don't have to close to conduct AC current, or else capacitors wouldn't be very useful circuit elements.
  • The structure of figure 10 is nothing but a particular restriction of the E and B fields so that the electric fields mostly go in one material, and the magnetic fields in another. The transmission lines we have do the same thing, with the E and B in the space around the wire. Electricity is a property of the fields in the space around the materials, not of the materials themselves. The trick in the page is that they are using iron which channels magnetic fields inside it so effectively, so that the space-property becomes a material property.
  • There is nothing strange about conducting displacement current--- it happens every time you use a capacitor. If you take a normal electronic device with at least one capacitor on every path from power to ground (I am sure this is true in every complex enough device you own) then it will not conduct DC current, so it is "insulating" in this sense, but it conducts the AC currents just fine.
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I'll take a stab at your first two questions:

1.) The circuit does not act as a transformer because there is no linking current. The "A-field flux" is there, but no traveling charge. Putting in a ring of copper (as he does in other pics) allows the A-field to generate current which would link to the next core.

2.) Yes, but not in any practical sense as shown. The capacitor plates need to be very close together, as they would be in a commercial cap.

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Are you sure these two answers don't contradict each other? In 1.) you're saying that it doesn't work because there's no charge passing through the loop. In 2.) you're saying that it would work... even though there's no charge passing through the loop! What's the crucial difference? –  Keenan Pepper Jul 12 '11 at 18:21
    
If 2.) can work even though the only current passing through the loop is displacement current, then why can't 1.) work the same way? –  Keenan Pepper Jul 12 '11 at 18:22

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