metal rod 0.150 m long and with a mass of 0.0500 kg is suspended from two thin, flexible wires,At right angles to the rod is a uniform magnetic field of 0.550 T pointing into the page

Question

find the direction and the magnitude of the electric current needed to levitate the rod

Answer 1

Say the rod runs along the x-axis from (0,0,0) to (L,0,0) and the magnetic field B points along the z-axis, but into the page B = (0,0,–B). The Lorentz force law is F = I(L,0,0)×B = (0,ILB,0) that must cancel the rod weight (0,–Mg,0). Therefore the current is I = Mg/LB.

Check the units: kg m s⁻²/m T = kg s⁻²/T = A.