So with your flying car example CraigD, say it was 10cubic metres and 10kg total weight (about the overall density of air). Then how much current would be needed to lift it and could the flux density be managed within limits of current superconductors?

The short answer to the last of this question appears to be “no”. Here’re some details:

From standard references, such as

this wikipedia section, and the fundamental definition of force, we have:

[math]M g = F = B I L[/math],

where: [math]M[/math] is mass of vehicle, 10 kg in this example;

[math]g[/math] is acceleration of gravity, about 9.8 m/s/s at Earth’s surface;

[math]F[/math] is force, 98 N in this example;

[math]B[/math] is magnetic flux density, about 0.00003 and 0.00006

T at Earth’s surface;

[math]I[/math] is current;

[math]L[/math] is length of the conductor. The given 10 m^3 can be interpreted as many possible lengths, so I’ll just use a reasonable “big ship” size of 100 m.

So [math]I = \frac{M g}{B L}[/math], which calculates to between roughly 16000 and 33000

A.

Rather than a working vehicle, let’s imagine a basic engineering experiment: Somewhere not close to Earth’s magnetic poles, we lay a length of superconductive wire (chilled to its operating temperature) on an insulating surface in an east-west direction, attaching each end to a powerful positive and negative terminal. We then pass a current thought it sufficient to lift it from the surface.

American Superconductors sells superconducting wire with a cross-section area of about .000002 m^2 and a maximum current capability of about 160 A. It’s made mostly of bismuth and silver, so should have a density of about 10000 kg/m^3.

Notice that [math]M = k L[/math],

where [math]k[/math] is the wire’s density times its cross-section area, .025 kg/m for the American Superconductor wire.

So we can rewrite the current equation as: [math]I = \frac{k g}{B}[/math],

and calculate a current required to levitate just the wire of about 3200 to 6500 A,

**about 20 to 40 times its capability**.

Note that, with no changes in the wire’s direction, it shouldn’t induce a magnetic field that intersects itself, so loss of superconductivity due to magnetic field should not be an issue (as it would be in a self-contained vehicle). The issue here is the material’s ability to move electrons. Although superconductors have zero resistance, they have a finite current-carrying capacity.

It’s worth noting that, while it doesn’t appear practical to use the Earth or other bodies’ magnetic fields to lift objects from their surfaces, such technology is very promising for maneuvering objects already in orbit. Know as “

tether propulsion”, some preliminary experiments have already flown, with others scheduled. These systems are “reversible” – electric energy, such as from a solar cell, can be used to increase the craft’s kinetic and potential energy, boosting it into a higher orbit, or they can be used to generate electricity while braking the craft into a lower (or different eccentricity) orbit.

A drawback to tether propulsion is that “popular destination” bodies like the Moon and Mars have effectively no magnetic field, so it can’t be used around these bodies. Jupiter and the other giant planets, on the other hand, have greater magnetospheres than Earth, and are, IMHO, prime locations to use electromagnetic tethers for both navigation and power generation. (I elaborate on this in a “big picture” manner in

“Sheer human fecundity”, and

”Relevance of space elevators in a 1,000,000 times more energy rich civilization”) in the

”Colonizing the Solar System” thread)

The Sun’s magnetic field (

the IMF), while about .0001 T ([math]10^{-4}\mbox{T}[/math], about twice the Earth’s) at its surface, decreases with distance (though not as sharply as simple theory predicts). At 1 AU, the distance of the Earth’s orbit, it’s about [math]10^{-9}\mbox{T}[/math], about 1/5000th that of Earth’s surface strength, so interplanetary maneuvering using electomagnetics appears impractical.