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I was analyzing the possibility of varying an exterior magnetic field, that's aligned in direction with the magnetic field produced by the displacement currents(both parallel in the same plane) shown in the figure above.

And when I used this equation to model the case:

enter image description here

It seems possible to discharge the capacitor faster/slower depending on the orientation of the exterior magnetic field, and it's rate of change, the exterior magnetic field would affect the displacement current's magnetic field, wouldn't that help in discharging/charging it? In addition, increasing those rates or decreasing it?

Discharging faster by increasing the displacement currents, and the conduction currents flowing in the loop, if the field was in the opposite direction(out of the page) it would discharge slower since now it's opposing/reducing the displacement currents.

NOTE: The magnetic field will be confined mostly in the separation gap of the two plates, to reduce it's effects on the overall loop(that includes the load/resistor).

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    \$\begingroup\$ How can the external magnetic field affect the displacement current when its not going through it? \$\endgroup\$ – Bruce Abbott Aug 12 '18 at 22:43
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    \$\begingroup\$ Could you clarify what you mean by: "through it?" \$\endgroup\$ – e.d.m Aug 12 '18 at 22:48
  • \$\begingroup\$ @BruceAbbott The exterior magnetic field affects the displacement current's magnetic field, leading to an overall effect to the displacement currents? \$\endgroup\$ – e.d.m Aug 12 '18 at 22:52
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    \$\begingroup\$ If the fields don't coincide there won't be any interaction, so in order for the external field to affect the capacitor, the capacitor must also produce an external magnetic field. But your diagram only shows the capacitor producing an internal field. \$\endgroup\$ – Bruce Abbott Aug 13 '18 at 0:24
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Bd is around the axis of the capacitor. An externally caused field Bext doesn't boost Bd if it isn't around the same axis. I think that you do not want to add a piece of metal to make a direct current between the plates. That would cause just the right Bext and surely it accelerates the discharging.

Another possible way to make more field around the axis of the capacitor is to somehow increase diminishing rate of the electric field between the plates. That's Maxwell's famous virtual displacement current (which in insulators can also be a real molecular de-polarization current). But that is just accelerating the discharging somehow. You must accelerate discharging to get a field which is hoped to accelerate the discharging. How ingenious!

You can accelerate discharging by adding an opposite voltage in series with the resistor.That voltage can be induced by a changing magnetic field. The localization of that field is not critical, it's enough it goes through the discharging current loop. If you want to restrict it between the plates, it's ok and there's no need for the changing magnetic field go just around the axis of the capacitor.

There exists a trick to insert external changing magnetic field which really is around the axis of the capacitor: Insert a circularly magnetized permanent magnet between the plates. That will accelerate discharging or make it slower depending on the polarity of the magnet, because during the insertion movement the field changes in the discharging loop. The effect exists only when the magnet moves.

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  • \$\begingroup\$ Thanks for the response, just a point of clarification. There are two B's here, one that is produced by Id(defined in the diagram as Bd) and the other, is an exterior source which is supplying Bext. The exterior magnetic field would vary as the capacitor is discharging, so I_d is decaying and the exterior magnetic field is increasing with time, from vector analysis of Bd + Bext it seems that it would discharge faster? \$\endgroup\$ – e.d.m Aug 12 '18 at 23:48
  • \$\begingroup\$ @e.d.m Ok I must revise the answer. I kill it temporarily soon. \$\endgroup\$ – user287001 Aug 13 '18 at 0:06
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    \$\begingroup\$ I think that even assuming it worked, manipulating the current in this way would also manipulate the voltage, meaning that given the voltage rating of the cap cannot be exceeded, you would be just as well off directly applying a greater voltage to produce the higher current as you would by applying a smaller voltage and manipulating the current with a magnetic field. \$\endgroup\$ – K H Aug 13 '18 at 0:07
  • \$\begingroup\$ @e.d.m a revised answer is ready \$\endgroup\$ – user287001 Aug 13 '18 at 12:41

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