A long, straight wire is fixed horizontally, carrying a current of...

Let $I_1$ be the current in the fixed wire and $I_2$ be the current in the second wire.

The magnetic force per unit length between two parallel current-carrying wires is given by:

$\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2 \pi d}$

For the second wire to remain suspended, the magnetic repulsion per unit length must balance its weight per unit length ($\lambda g$):

$\dfrac{\mu_0 I_1 I_2}{2 \pi d} = \lambda g$

Substituting the given values ($I_1 = 50.0 \text{ A}$, $d = 5.0 \times 10^{-3} \text{ m}$, $\lambda = 1.0 \times 10^{-4} \text{ kg m}^{-1}$, $g = 9.8 \text{ m s}^{-2}$):

$\dfrac{4 \pi \times 10^{-7} \times 50.0 \times I_2}{2 \pi \times 5.0 \times 10^{-3}} = 1.0 \times 10^{-4} \times 9.8$

$\dfrac{2 \times 10^{-7} \times 50.0 \times I_2}{5.0 \times 10^{-3}} = 9.8 \times 10^{-4}$

$2 \times 10^{-3} \times I_2 = 9.8 \times 10^{-4}$

$I_2 = \dfrac{9.8 \times 10^{-4}}{2 \times 10^{-3}} = 0.49 \text{ A}$

For the magnetic force to be repulsive and push the wire upwards, the currents in the two parallel wires must flow in opposite directions.