Can superconducting magnets fly (or repel the earth's core)?

If a superconducting magnet and appropriate power supply had just enough $I\cdot s$ (current $\cdot$ length) so that when it was perpendicular to the earth's magnetic field, the force of the interaction was just enough to excede the force exerted on the object from gravity. And it was rotating so the angular momentum of the vehicle was just high enough so it wouldn't flip over, would the vehicle fly?

Assuming the vehicle is a 1000 kg (and the earth's magnetic field is $0.3$ gauss) I calculated that with $6.54\cdot10^8$ meter amperes you just about reverse the force on the vehicle.

Now assuming a $100$ meter diameter, that leaves $6.54 \cdot 10^6$ A, which is less then the current in a railgun, but still a lot.

The problem is that the force normal is no longer so normal. It will want to flip the vehicle so the magnet is the other way. Now we would need to spin the vehicle fast enough, so that it has rotated 180 degrees faster then it would take for the force of the magnet to flip the vehicle 180 degrees. How would you go about calculating this part?

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haha, that is kind of an awesome idea. I'm curious to hear what the limitations with it would be. – Alan Rominger Aug 27 '11 at 21:38
@Zassounotsukushi: I think the biggest limitation is that it only would work at one location in the north of Canada. :-) – Alexander Nov 15 '11 at 23:50
Superconductors don't flip over when above a magnet. They don't repel magnetic fields, they expel magnetic fields. Your vehicle shouldn't need any rotation. Also, they have limits to the amount of current they can carry before they stop being superconductors. I suspect the mass required to carry a given current will always overcome the lift of that current. – endolith Nov 17 '11 at 21:33
@endolith he is talking about a superconducting magnet, not a superconductor. – P3trus Jan 25 '12 at 17:32
@endolith The question is independant of type of magnet. It's just needed to create a magnetic field.Btw instead of a simple dipole field one could use a more advanced field profile, to avoid the necessity of the rotation. – P3trus Jan 25 '12 at 18:27

The entire craft wouldn't need to be rotating, only a hollow torus containing sufficient mass (of a heavy superfluid?), rotating at a sufficient speed. It would also make an incredibly good flywheel, assuming you could easily put energy in and take it out again. (super-ferrofluid?)

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The bigger problem would be keeping it cool and not going over the magnetic threshold of the material. You really wouldn't need to worry about the spinning since your using the flux trapping property of the super conductors. Essentially once you produced the field needed you could trap it in a conducting loop, and as long as it was cooled properly you would have almost a perfect field for decades.

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I think the basic problem here is that a magnetic gradient is required for levitation, whereas the Earth's magnetic field is very close to uniform on the scale of any human-scale crafts. Try using the right-hand rule to figure out the direction of the magnetic force from a uniform magnetic field at two points on opposite sides of a circular current loop (i.e. each point is at 180 degrees from the other), by considering the velocity vector for a moving charge at both points and taking the cross product with the magnetic field vector (which will be in the same direction at both points for a uniform field). You'll see that no matter how you orient the external uniform field, the magnetic force at the two points will be equal and opposite because of the different velocity vectors for charge passing through both points, and thus there is no net magnetic force on the loop as a whole. Analyzing a non-circular current loop would be a little more complicated, but this page confirms that "the total magnetic force on a magnet in a uniform magnetic field is exactly zero, and the forces we normally associate with magnets repelling or attracting are proportional to the rate of change of the field strength with position."

Here is a page that gives some specific equations for magnetic levitation--it's focused on diamagnetic levitation, but the page mentions that the specific form of magnetism enters into the equation in the variable $\chi$, the magnetic susceptibility, which is about $10^{-5}$ for a diamagnetic material but equal to -1 for a superconductor. Apparently the field gradient $\nabla B$ must have a value greater than $2 \mu_0 \rho g / \chi$ for levitation to occur, where $\mu_0$ is the vacuum permeability, $\rho$ is the density of the object to be levitated, and g is the gravitational acceleration (9.8 meters/second^2 at sea level).

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protected by Qmechanic♦Aug 20 '15 at 11:28

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