I have been thinking about an inductor made out of superconductors, would it work? How would it affect the total inductance compared to a normal inductor? Would it be possible to calculate such a device with a formula? (I'm not an engineer yet, I'm a student so be considerate of that)

  • \$\begingroup\$ If the wire is what is super conducting, you remove the resistive loss, however capacitance and inductance still exists to the same formulas \$\endgroup\$ – Reroute May 5 '20 at 11:15
  • \$\begingroup\$ ... and inductance is unchanged. But removing the resistance can be a biggie. \$\endgroup\$ – user_1818839 May 5 '20 at 11:17
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    \$\begingroup\$ You're a student, so before asking I would expect you to have studied the basics. Like what makes an inductor an inductor? What is the fundamental difference between a conductor and a superconductor? Anything can be calculated with a formula provided you have the formula or can derive it and you have accurate enough input parameters for the formula. Formulas just a tool to describe things it is not like there's a formula or there isn't. \$\endgroup\$ – Bimpelrekkie May 5 '20 at 11:18
  • \$\begingroup\$ One of the biggest inductors I've come across was the superconducting magnet to an MRI machine. That had timescales of minutes, when you tried to change the current through it by putting a voltage across the terminals. \$\endgroup\$ – Neil_UK Nov 14 '20 at 19:53

I worked with a rather large superconducting inductor in my early career, the magnet for an MRI scanner. It had a roughly 1m bore, was about 3m outside diameter, and about 4m long, though admittedly most of that was insulation and liquid N2 and He.

A critical part of using it was to 'ramp the current', when you got to see at first hand and at the sort of timescales only usually associated with capacitors, how inductance behaves.

The magnet was normally short-circuited internally, with a superconducting switch, which allowed the current to circulate indefinitely. A probe was put in, to make contacts on either side of the switch, and an external power supply then set to match the current running in the magnet. Then the exciting bit, the superconducting switch was heated to turn it resistive. Once the magnet current was running through the power supply, we could apply a voltage, to slew the magnet current, as dI/dt = V/L. If I recall correctly (it was a long time ago), roughly 75 V applied gave us a slew rate of less than 1 Amp per second for that particular magnet. Estimate the inductance from that. This was a 'low field' 0.15T magnet, present ones are more than 10 times the field strength.

The only difference between superconducting wires and normal resistive wires to make an inductor is whether they have any loss.


how would it affect the total inductance compared to a normal inductor?

Any formula or rule for inductance does not involve the conductivity of the material that might make the wire good for passing a current with low loss. Inductance is more like a measure of area and "good" areas (those that yield higher values of inductance) are more rounded. "Bad" areas are more like long elongated loops having a high ratio of length to breadth. When you consider a the wider 3D model inductance becomes more like a measure of volume.

EM waves

Consider an electromagnetic wave - it is composed of H field and E field (the physics guys might argue that it's something else). The H-field is measured in amps per metre and the E field is measured in volts per metre and if we dig down it's more like volt-metres per metre-squared (which becomes volts per metre) and similarly for inductors.

Then there is the ether (that thing that EM waves travel through from the sun or stars to our eyes). That has two parts namely \$\mu_0\$ and \$\epsilon_0\$ and they are measured in henries per metre and farads per metre.

They define the speed of light as in: -

$$c = \dfrac{1}{\sqrt{\mu_0\epsilon_0}}$$

Last time I checked the ether didn't use any form of current conducting material.

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    \$\begingroup\$ The ether is not a thing. \$\endgroup\$ – user110971 May 5 '20 at 12:06
  • \$\begingroup\$ Is the permeability of free space a thing? Is dark energy a thing? Were you being rhetoric @user110971? Do we really understand stuff enough to make a limited guess at what it is? \$\endgroup\$ – Andy aka May 5 '20 at 12:10
  • \$\begingroup\$ What I meant is that the ether theories, i.e. a medium necessary for the propagation of EM waves, have all been discredited. The permeability of free space is a relativistic effect of the electrostatic force. So EM waves in vacuum do not travel in something. \$\endgroup\$ – user110971 May 5 '20 at 12:17
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    \$\begingroup\$ Would you rather me call it free-space? \$\endgroup\$ – Andy aka May 5 '20 at 12:37
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    \$\begingroup\$ Units was a bad choice. \$\endgroup\$ – Andy aka May 5 '20 at 13:41

It's already done. I guess you have seen a levitating magnet experiment video where a magnet floats in the air above a piece of superconductor. The superconductor has virtually zero resistance so the current induced by the magnet can stay stable a long time. The eddy current flows in a 1 turn shorted inductor.

Some experiments surely are done to get strong electromagnets which do not generate heat, but the cooling needed for superconductivity still costs more than achieved energy savings, so the effect is not in common use, it's used only where it's really needed. There's material dependent upper limit how strong fields are possible because too strong magnetic field removes the superconductivity, say the physicists.

The lack of resistance only removes one loss mechanism but there's no effect to the inductance. Or actually there is some effect. The skin effect becomes stronger, so there's no inductance caused by the interior of the conductor.


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