Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If you have seen my previous questions then this may make better sense. If you put an insanely powerful superconducter on the outside layer of a ROV that is underwater, would it form an outer layer around it of near nothingness, if the water is only H2O? What I mean is, since water is a polar molecule and in the scenario I am using no other molecule is in the water, shouldn't it make a smooth layer of nothing around it? And if this is feasible, would it keep neutral buoyancy?

share|cite|improve this question
Please define "insanely powerful superconductor". Do you mean a coil? Is there a current flowing through this thing? – adavid Oct 25 '11 at 19:29
A superconductor is not a magnet. I think you need to get some basics sorted out, but I can't recommend a good read right now, sorry. – adavid Oct 25 '11 at 21:38
How should i go aroung creating a uncostly superconductor so that it can repel the water from the ROV? – Bored915 Oct 25 '11 at 21:45
If you find one, let us know :-) – adavid Oct 25 '11 at 21:54
@Rudy Gruse - What makes you think that a superconductor will repel water? – Richard Terrett Oct 26 '11 at 3:09
up vote 1 down vote accepted

OK - if you want to repel water from the surface of your device, you may have better luck finding a hydrophobic coating for your device, a la: ORNL scientists discover superhydrophobic coating

Superconductors themselves are perfect diamagnets, which is why they are able to expel magnetic fields from the bulk, with different levels of magnetic field support for different superconductor classes and applied fields. Type II superconductors are capable of producing strong magnetic fields in a mixed state (see MRI, and the superconducting magnets at RHIC/LHC), but bear in mind that these operate at extremely low temperatures using liquid helium which is expensive. The critical fields for high temperature superconductors like cuprates are lower, which is why applications requiring strong magnetic fields are still using the helium based versions. Unfortunately, there is no low cost, high temperature superconductor, and in no way is this due to lack of effort by thousands of people.

For the actual question about repelling the water, if you can get a superconductor that produces a strong enough magnetic field and you can get the field orientation issues sorted, then sure, you could use a superconductor to repel water, but this is due to the diamagnetism of the water, and not the polar character of the water. There are ton of youtube videos demonstrating this phenomena.

Further, I don't know that I understand what you mean by "keep neutral buoyancy", but since buoyancy is driven by the balance of an objects density and the density of the suspending liquid, I suspect that the normal rules of buoyancy would still apply, but with some modification to what one would consider to be the boundary of the suspended object extended out the the finite spatial extent of the magnetic field.

Finally, an event where your superconducting magnet suddenly became non-superconducting (see LHC magnet quench of 2008) might blow your device up, but I'm not sure what would be the effect of dumping compressed cryogens into the ocean at any depth. Maybe it would become a floating block of ice? Just something to consider.

share|cite|improve this answer
well thank you it is very helpful. do you know how much that powder will cost? i wasn't going to put that ROV into deep water that has enough water pressure to crush it but yah thanks now i think i can start making it cause now i know all the resources i need. – Bored915 Oct 26 '11 at 18:03

Water is weakly diamagnetic so is aligned by a strong magnetic field.
But salt water is a conductor and so if the field in your coil was changing you would form a current of water around the hull. Basically just half a transformer and the left hand rule.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.