Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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

I've seen superconductors levitating on magnets. But is it possible for superconductors to levitate on Earth from Earth's magnetic field?

share|improve this question

marked as duplicate by Qmechanic, Georg, dmckee Nov 18 '11 at 15:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1  
Another duplicate that is arguably better written than the original. Certainly the answer here is superior. Any thoughts from on merging them or leaving them separate? –  dmckee Nov 18 '11 at 15:07

1 Answer 1

up vote 5 down vote accepted

The lift generated by magnetic field B on a superconductor of area S is:

\begin{equation} F = \frac{B^2S}{2\mu_0} \end{equation}

disregarding lateral forces and assuming superconducting cylinder (or similar shape) with area S at the top and bottom and height h, we need three forces to remain in the equilibrium: magnetic pressure on top, bottom and gravity force:

\begin{equation} F_{b} - F_{t} = F_{g} \end{equation}

denoting density of the superconductor as ρ, Earth' gravity as g and magnetic field at the top and bottom of the object as Bt and Bb, we have

\begin{equation} \frac{1}{2\mu_0}(B_{b}^2-B_{t}^2)=\rho gh \end{equation}

assuming the vertical rate of change of magnetic field is nearly constant and denoting the average magnetic field as B, we have

\begin{equation} -B\frac{dB}{dz}=\mu_{0}\rho g \end{equation}

Compare with diamagnetic levitation (superconductor's magnetic susceptibility is -1).

Now, Earth magnetic field is between 25 to 65 μT. For the derivative I have found this survey from British Columbia with upper point on the scale being 2.161 nT/m. Assuming this to be the maximum for vertical derivative we get the required density of 1.1394e-08 kg/m3. For comparison air density at the sea level at 15C is around 1.275 kg/m3, so required density is 8 orders of magnitude smaller.

Even assuming a very high vertical derivative where B goes from its maximum 65 μT to 0 on 1 m of height results in density required of 0.00034272 kg/m3.

share|improve this answer
    
so it's possible for levitation if the superconductor has maximal area and minimal density? What about metallic hydrogen? I hear that it is the lightest metal in the universe. Though I can't find any numbers.... –  mugetsu Nov 17 '11 at 21:33
    
Well, you certainly couldn't keep on decreasing the thickness beyond London penetration depth. –  Adam Zalcman Nov 17 '11 at 21:51
    
thats like in the nm right? Which would be enough in terms of thickness for levitation. –  mugetsu Nov 17 '11 at 22:12
    
since a sheet so thin would break easily, it should theoretically work in the same way if instead I had superconductor powder. I'm assuming that each particle is like a very small thin square sheet, so given the right density, these particles can levitate under gravity, even when in close proximity to each other. Is this assumption correct? –  mugetsu Nov 17 '11 at 22:22
    
just saw your update about the magnetic field obstacle. so in effect, given the perfect material, levitation cannot be achieved practically? quite a shame.... –  mugetsu Nov 17 '11 at 22:32

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