A long thin copper wire has a mass of 100 grams / meter. If it is arranged near the surface of the earth pointing north in a 0.5 Tesla magnetic field pointing west, what current (in amps) is needed to levitate the rod (against gravity)?

Answer 1

Consider the gravitational and magnetic forces on a 1 meter section of wire. The magnetic force is I∗(1m)∗(0.5T) since the current and field are perpendicular, and the gravitational force is (0.1kg)∗(9.81m/s2). To levitate, these forces must be equal. Equate the expressions and solve for current.