superconductor levitating in earth's magnetic field?

Question

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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 1

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.