Two identical long parallel wires carry currents ' $\mathrm{I}_1$ ' and ' $\mathrm{I}_2$ ' such that $\mathrm{I}_1>\mathrm{I}_2$. When the currents are in the same direction, the magnetic field at a point midway between the wires is $8 \times 10^{-6} \mathrm{~T}$. If the direction of $\mathrm{I}_2$ is reversed, the field becomes $3.2 \times 10^{-5} \mathrm{~T}$. The ratio of $\mathrm{I}_2$ to $\mathrm{I}_1$ is

The distance from the midway point between two parallel wires to either wire is $r = \frac{d}{2}$. The magnetic field at this point due to each wire is $B = \frac{\mu_0 I}{2\pi r}$, giving $B_1 = \frac{\mu_0 I_1}{\pi d}$ and $B_2 = \frac{\mu_0 I_2}{\pi d}$.

When currents flow in the same direction, the magnetic fields oppose. With $I_1 > I_2$, the net field is $B_{\text{net1}} = \frac{\mu_0}{\pi d}(I_1 - I_2) = 8 \times 10^{-6}\, \text{T}$.

Reversing $I_2$ makes the fields additive: $B_{\text{net2}} = \frac{\mu_0}{\pi d}(I_1 + I_2) = 3.2 \times 10^{-5}\, \text{T}$.

Dividing these equations gives $\frac{I_1 + I_2}{I_1 - I_2} = 4$.

Solving yields $I_1 + I_2 = 4(I_1 - I_2)$, which simplifies to $5I_2 = 3I_1$.

The ratio is $\frac{I_2}{I_1} = \frac{3}{5}$.

Answer: C