superconductivity - superconductor levitating in earth's magnetic field?

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?

Answer

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 B_t and B_b, 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.