Question

Suppose a straight 1.60mm -diameter copper wire could just "float" horizontally in air because of the force due to the Earth's magnetic field B⃗ , which is horizontal, perpendicular to the wire, and of magnitude 4.2×10−5T .

What current would the wire carry?

Answer 1

Fb = Fg

ILB = mg

IkB = 2pi (d/2)^2 g (density copper)

I = pi (0.0016/2)^2(9.8)(8.9x10^3)/(4.2x10-5) = 4175.38 A

Hope this helps