A thin copper rod 1m long has a mass of 50g. What should be the direction of magnetic field and the maximum current in the rod that would allow it to levitate above the ground in a magnetic field of 0.1T.

Answer 1

F = IBL
F = mg

mg = IBL

I = mg / BL

( (0.05 kg)(9.8 m/s^2) / (0.1 T)(1 m) )

= 4.9 A