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In a super conducting multi-filament wire, if a particular filament within the wire has a break for instance from a nanometer size occlusion, will induction partially or completely make up for the discontinuity and allow full conduction in that filiment or is that filament out of the circuit for conduction?

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  • $\begingroup$ You mean displacement current filling the gap? $\endgroup$ – Buzz Nov 29 '17 at 0:47
  • $\begingroup$ I don't have a clear idea about what to think actually does happen.. It seems reasonable that there would be some type of coupling that could make up for the break since the filaments are so close together. On the other hand it seems that there should be some type of degradation. If you have 40 or 50 filaments on one wire drawn to such fine dimension it seems reasonable that not all of them would make the connection. I wouldn't think that current would flow across the actual gap. I would expect something more like induction between the filiments equalizing the current across the wire. $\endgroup$ – DMac Nov 29 '17 at 7:19
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The filament with the occlusion will conduct by induction, but only when the current is increasing or decreasing. That means that during steady state operation as in a MRI magnet when the power supply is disconnected from the circuit, an occlusion will de-rate the current capacity of the wire by each filament that doesn't connect 100% completely from end to end. Each filament must be perfect to have maximum current carrying capacity.

Once a superconductive magnet or storage ring or other such device is in steady state conduction, the filament is out of the circuit. The defective filament will conduct only under the following three conditions.

  1. during current rise of the circuit
  2. during current fall of the circuit
  3. during quench which will of course produce current rise and fall in various sections of the SC wire depending on the winding of the wire.
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