Two long, parallel wires carry currents i_1 and i_2 such that i_1 >...

Let the distance between the wires be $2d$. The magnetic field at the midway point due to a long straight wire carrying current $i$ is $B = \dfrac{\mu_0 i}{2\pi d}$.

When the currents are in the same direction, the magnetic fields at the midway point oppose each other. Since $i_1 > i_2$, the net magnetic field is:

$B_1 = \dfrac{\mu_0}{2\pi d}(i_1 - i_2) = 10 \mu\text{T}$

When the direction of $i_2$ is reversed, the currents are in opposite directions, and their magnetic fields at the midway point add up:

$B_2 = \dfrac{\mu_0}{2\pi d}(i_1 + i_2) = 30 \mu\text{T}$

Dividing the two equations, we get:

$\dfrac{i_1 + i_2}{i_1 - i_2} = \dfrac{30}{10} = 3$

$i_1 + i_2 = 3i_1 - 3i_2$

$2i_1 = 4i_2$

$\dfrac{i_1}{i_2} = 2$