Answer
The angle between the net force and the vertical line is less than $45^{\circ}$
Work Step by Step
We can write an expression for the force on a current-carrying wire due to another current-carrying wire:
$F = \frac{\mu_0~L~i_1~i_2}{2\pi~d}$
According to the text on page 854: "Parallel wires carrying currents in the same direction attract each other, whereas parallel wires carrying currents in opposite directions repel each other."
The force due to the closer wire is directed upward while the force due to the farther wire is directed to the left.
$F_y = \frac{\mu_0~L~i^2}{2\pi~d}$
$F_x = -\frac{\mu_0~L~i^2}{2\pi~D}$
Since $d \lt D$, the magnitude of $F_y$ is greater than the magnitude of $F_x$
Therefore, the angle between the net force and the vertical line is less than $45^{\circ}$
