Suppose a straight 1.00-mm-diameter copper (density = 8.9 x103 kg/m3) wire could just "float" horizontally in air because of the force due to the earth's magnetic field, which is horizontal, perpendicular to the wire, and of magnitude 5.0 x 10-5 t. (a) what current would the wire carry?

We are given

diameter, d = 1.00 mm

density, ρ = 8.9 x10^3 kg/m3

magnetic field, B = 5.0 x 10^-5 T

We are asked to determine the current that would be carried by the wire

We use the formula

B = I R / A

where I is the current

R is the resistance and R = πρd / A

So,

B = I (πρd / A) / A

B = I πρd / A²

and

A is the area, A = πd²/4

So,

B = I πρd / (πd²/4) ²

B = I πρd / (π² d⁴ / 16)

B = 16 I ρ / πd²

Substituting the given and solving for I

5.0 x 10^-5 = 16 I (8.9 x10^3 kg/m3) / π (1 x 10^-3) ²

I = 1.1 x 10^-15 A