# Induction cooker + superconductive pot [closed]

What will happen if I put a super-conductive object on an induction cooker, and turn the cooker on?

There will be no heating because there is no resistance (but current flows to cancel any penetrating magnetic field).

• Yeah, but will it levitate? Jump? Tremble? Commented Feb 23, 2013 at 7:11
• At first, it will start to levitate. But, it will soon tilt due to a misbalance and then accelerate this tilt...the 3D shape of the superconductor will then determine how the induced currents flow. Of the three options you gave, I'd say "tremble". Commented Feb 25, 2013 at 21:31
• I don't know the correct answer but I feel you're wrong here. Take a look: en.wikipedia.org/wiki/Magnetic_levitation#Superconductors for the example of stable levitation of real-live (imperfectly shaped) objects. Commented Mar 1, 2013 at 18:11
• This is not correct, superconductors have only a zero resistance for DC currents, not AC. There will be heating. The details depend on the material (i.e. type I or II, pinning, frequency, etc.). Commented May 20, 2014 at 11:28

This requires an experimental answer but in principle the following will happen:

The induction cooker provides a strong magnetic field in the kHz frequency range. In the case of a type I superconductor you will only have field penetration at the corners, this will cause heating and will most likely drive your material above the critical temperature within seconds.

In a type II superconductor with weak pinning the magnetic flux can enter and exit the material with relatively low losses. Depending on your cooling system you might be able to sustain the superconducting state. The upper critical field Hc2 in those materials is often quite high (up to 100T), so the cooker will not destroy the superconducting state.

In a type II superconductor with strong pinning a relatively large amount of energy is required to push the magnetic flux lines around, this will create large losses and the induction cooker will heat your material above the critical temperature.

This isn't a superconductor but here's a video of induction launching a disk. A superconducting disk in the same situation would probably go a bit higher, but I'm not sure how much.

• The difference between your experiment and mine is induction cooker creates alternating electromagnetic field. Commented Mar 1, 2013 at 18:07
• This experiment in the video link generates the field by discharging a large capacitor; this is a single huge pulse of current, which in turn causes the dramatic effect. This may not be relevant for the problem at hand.
– Dave
Commented Mar 7, 2013 at 13:12

Yes above answer is correct because induction cookers work on the Principal of current induced in a conductor when kept in a magnetic field. But if you connect your superconducting pot (if you have one because it will be too costly) or any other pot probably a force will act on it (F = ILxB) I=current flowing through your pot L=Length between to points were you have connected the wires to the pot B=Magnetic field due to induction cooker (Visit below given link for direction of foce magnetic field and current http://t2.gstatic.com/images?q=tbn:ANd9GcR5eKnu-3HIlrYNmVUbm-E2mg44OkzGthoDsjGXihEWepu9K3cq) Due to the force it will jump or press the induction cooker depending on the direction of current and magnetic field Well if you want to try it you can try it you can try it even wit a copper or any other good metal rod. If are going to try it ask me here how to do the connection and be safe. It works in theory I don't know it will work in real or not if it does please do record it and upload to you tube and post me a link here only.