Lets make a simple Gedankenspiel with only a wire made from LK-99 and a bunch of electrons sitting on the left end of the wire.

After a moment, the electrons will be spread out evenly through the wire, right?

So after a moment, the energy that was stored in the system (When all electrons were on the left) is lost.

If no heat is emitted, where did the energy go?

Current in a conductor does not involve the potential energy of charges (electrons) being concentrated. This should be clear by the fact that currents cause magnetic fields, but do not give the conductive wire a charge gradient (or an electric dipole moment).

That's not how electric current works. It's more like a wave in an ocean. A wave may travel from Hawaii to California but no actual water molecule moves that far (or even very much at all). Electric current is a wave of excitation of electrons, which is what moves at the speed of light. The actual drift of individual electrons in the current is much, much slower (and IIRC in a superconductor is zero).

Veritasium has a cool video on this:

https://www.youtube.com/watch?v=bHIhgxav9LY&pp=ygUVdmVyaXRhc...

Interesting video, So it is sad that they did such a terrible job explaining why the the answer is 1/c.

It makes it sound like the bulb is getting full power at 1/c, when it wont. As I understand from further reading, the power will build up over time, and that time depends highly on the shape of your circuit.

Why would the electrons spread evenly?

They move over to the other side and back and forth resulting in an oscillator.

That causes all your energy being radiated into the surrounding via radio waves.

That is how your microwave oven works but instead of a superconductor it uses a magnetron with free flowing electrons.

Nowhere, it's still there - a loop of superconducting wire would also be a perfect energy storage because the electrons would just keep going around without any loss(it would be an extremely small capacity storage though, superconductors lose their superconductivity at high enough amperage which is basically what introducting more energy into the system would be).

I suspect that a circular trajectory implies acceleration and when a charged particle accelerates, it radiates away energy.

Superconducting circuits can still have AC loss. You’re basically describing discharging a capacitor through a superconductor. If there’s no loss anywhere, it will be be an LC oscillator. If there is loss, the energy will dissipate.