Question

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?

Answer 1

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