Question

Suppose a straight 1.50 mm$\mathrm{m}\mathrm{m}$ -diameter copper wire could just "float" horizontally in air because of the force due to the Earth's magnetic field B⃗ $\overrightarrow{\mathbf{B}}$, which is horizontal, perpendicular to the wire, and of magnitude 4.1×10⁻⁵ T$\mathrm{T}$ . Find the current the wire would carry.

Answer 1

a) let the current needed is I

Now,density of copper, p = 8960 Kg/m^3

Now, for the wire to levitate

lamda * g = B * I

8960 * pi * (0.0015/2)^2 = 5.5 *10^-5 * I

solving for I

I = 287 A

the current needed is 287 A

b) No

c) as this current would causes too high heat losses