What is the magnitude of the current? current n the wire that keeps it levitated just above the ground?

Question

Imagine a very long, uniform wire that has a linear mass density of 0.003 kg/m and that encircles the Earth at its equator. Assume the Earth's magnetic dipole movement is aligned with the Earth's rotational axis. The Earth's magnetic field is cylindrically symmetric (like an ideal bar magnet). The acceleration of gravity is 9.8 m/s2 and the magnetic field of the Earth is 5.0 x 10^-5T. What is the magnitude of the current in the wire that keeps it levitated just above the ground? Answer in units of A, GOOD LUCK!!!

Answer 1

The net force on a 1 meter segment of the wire would be the upward force due to the current, countered by the downward force due to gravity. The upward force on a 1 meter length of the wire will be BIL = (5.0*10^-5)(A)(1.0)=(5.0*10^-5)A. The downward force due to gravity will be mg = (0.003)(9.8) = 0.0294 N. These forces will be equal when the wire "floats" so:

A = 0.0295/0.00005 = 590 Amps (more or less)

Answer 2

I figured this out once, just for fun - a lot of years ago - I don't remember the answer, however what I did come up with is that since the Earth's magnetic field is so weak, the amount of current needed would melt any wire that it would lift. This was a function of resistance and density of copper.

Answer 3

poop A