Two identical long parallel wires carry current '$I_1$' and '$I_2$' such that $I_1 > I_2$. When the currents are in the same direction, the magnetic field at a point midway between the wires is $6 \times 10^{-6}\ \text{T}$. If the direction of '$I_2$' is reversed, the field becomes $3 \times 10^{-5}\ \text{T}$. The ratio $\left(\dfrac{I_1}{I_2}\right)$ is

Let the distance between the two parallel wires be $2d$. The distance of the midway point from each wire is $d$.

The magnetic field due to a long straight wire carrying current $I$ at a distance $d$ is given by $B = \dfrac{\mu_0 I}{2 \pi d}$.

When the currents $I_1$ and $I_2$ are in the same direction, the magnetic fields at the midway point are in opposite directions. Since $I_1 > I_2$, the net magnetic field is:

$B_1 = \dfrac{\mu_0}{2 \pi d} (I_1 - I_2) = 6 \times 10^{-6}\ \text{T}$

When the direction of $I_2$ is reversed, the currents are in opposite directions, so their magnetic fields at the midway point add up. The net magnetic field becomes:

$B_2 = \dfrac{\mu_0}{2 \pi d} (I_1 + I_2) = 3 \times 10^{-5}\ \text{T} = 30 \times 10^{-6}\ \text{T}$

Taking the ratio of $B_2$ to $B_1$:

$\dfrac{I_1 + I_2}{I_1 - I_2} = \dfrac{30 \times 10^{-6}}{6 \times 10^{-6}}$

$\dfrac{I_1 + I_2}{I_1 - I_2} = 5$

$I_1 + I_2 = 5I_1 - 5I_2$

$4I_1 = 6I_2$

$\dfrac{I_1}{I_2} = \dfrac{6}{4} = \dfrac{3}{2}$

Answer: $\dfrac{3}{2}$