I know that you can use induction to create a current in a superconducting loop, but this only works as long as the coil that induces the field has a current flowing through it. And obviously, this cannot work for the really big magnets that are used in MRI scanners or NMR machines.

How does it work?


1 Answer 1


Actually, induction works, although it is often used a bit differently than you described. You can place a warm superconductor loop into a normal coil. As you switch the coil on, there will be some current inside the superconductor, but since it is not cold yet, this current quickly dies down. Then you cool the superconductor below its critical temperature. When you now switch off the normal coil, the flux-change induces a current in the superconducting loop. This current is there to stay.

With the proper coil windings, you can get quite large currents, but you won't be able to get to field energies that are needed in NMR or MRI machines. These magnets are energized (sometimes called "charged") differently. The trick is to make part of the loop normal conducting during charging, and then "close the loop" when the coil is energized.

If you want to know details, this is the electrical schematic on how it is done:

Circuit for charging superconducting magnet

I used values from a typical 5 Tesla NMR magnet that I energized a few years ago.

I will leave out all those pesky details that make charging a real superconducting magnet a hassle. The procedure boils down to this:

  1. Start with the current supply and the voltage source $V_1$ in the off state. Cool the cryostat down until the superconducting (sc) wires are below their critical temperature. Most magnets use NbTi, which has a critical temperature of 9 Kelvin. You can reach this temperature by immersing the coils in liquid helium, which boils at 4 Kelvin.
  2. Switch on the voltage supply $V_1$. This sends a small current through $R_1$, creating a small amount of heat (10 mW). A part of this heat will warm up the adjacent sc wire so that it becomes normal conducting (nc), which means it will have a resistance of a few Ohm at that particular point. Let's call this part the "heated point".
  3. Dial the maximum current that the coil can take (30 A) into the current source and set its voltage limit to no more than $V_\text{charge} = 1\,\text{V}$ ; then switch it on. (Side note: The low voltage limit actually makes the current source a voltage source during ramp up.) The charging voltage is applied across the coil and the "heated point". It leads to two effects:
    • A constant current $$I_\text{heated point} = \frac{V_\text{charge}}{R_\text{heated point}} \approx 200\,\text{mA}$$ will flow through the "heated point" and continues to heat it. The external heater can be switched off at this point by setting $V_1$ to 0.
    • The current in the coil will ramp up at a rate of $$ \frac{dI_\text{coil}}{dt} = \frac{V_\text{charge}}{L} \approx 5\,\text{mA/s}\, $$ because the self-inductance of the coil "pushes" against a change of the current.
  4. After 100 minutes the current source will be at 30 A. Since the current is not changing any more, the only voltage that the current source needs to apply is whatever little amount it takes to push 30 A through the normal conducting (nc) part of the wires. The voltage drop across the coil, and across the heated point, is now 0. All current flows from the current source through the coil and then back to the current source.
  5. Because no current goes through the "heated point" anymore, this part of the wire is no longer heated. As its temperature drops to the temperature of the cryostat, it quickly (within seconds) becomes superconducting again.
  6. Ramp down (not too fast, maybe within a few minutes, just to be on the safe side) the current of the current source. The coil is now part of a superconducting loop, so its current will stay constant! The current that is not going through the current source anymore just goes through the (formerly heated) return path of the superconductor.

With the loop closed, the magnet is now in "persist" mode, so the current goes on forever. Well, not quite forever, there usually is a little bit of resistance that leads to "flux creep". For the magnet I am using this is an unusually low value of $10^{-11}$ Ohm, which makes the current decay with a time constant of a few 100 000 years. The designer of the magnet speculates that this resistance is caused by the weld that is needed to turn the wire into a loop (see the link above, near the end of the paper). Other reasons that are often given for this effect involve details that are specific to type-II superconductors.

  • $\begingroup$ Awesome explanation. Just to clarify, how was the approximate current of 200 mA calculated (i.e. is it an arbitrary $R_{heated \ point} $)? Also, what is the typical self-inductance, $L$, for a typical superconducting coil as you've described? What limits for $L$ are there in a typical superconducting coil to ramp up or down a current as quick as possible? $\endgroup$
    – Mathews24
    Aug 24, 2018 at 23:54
  • $\begingroup$ Martin, have you ever wondered that "because the self-inductance of the coil "pushes" against a change in current", a magnetic self-holding field is created from the aligned magnetic dipoles of the influenced subatomic particles and no current flows at all starting from your point 5? When the coil heats up later, all the aligned particles simply de-align again and a strong but short current flows that can cause the coil to melt because the energy is captured in the coil? $\endgroup$ Aug 1, 2021 at 13:34
  • $\begingroup$ We use superconducting materials streched into filaments. The filaments are certainly often interrupted in their length, but a superconducting current should still flow? Or are the magnetic dipoles of subatomic particles simply aligned as strongly as possible in the filaments? If an external magnetic field works against that of the coil, the chain-reaction de-alignment (quenching) of the alignment of the magnetic dipoles also occurs. $\endgroup$ Aug 1, 2021 at 13:34
  • $\begingroup$ This is a good explanation. You said the large BNR and MRI magnets are charged differently. How do they charge these magnets? $\endgroup$
    – Kent Davey
    Sep 21, 2021 at 22:52

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