Two parallel long wires carry currents, $i_1$ and $i_2$ with $i_1>i_2$. When the currents are in same direction, the magnetic field at a point midway between them is 10 μT. If the direction of $i_2$ is reversed, the field becomes 30 μT. The ratio of $i_1/i_2$ is:

2 : 1

The correct answer is Option (3) → 2 : 1

The magnetic field $(\vec B)$ at a point at distance (r) from a long straight wire -

$B=\frac{μ_0i}{2πr}$

$B_{same}=B_1+B_2=\frac{μ_0i_1}{2πr}+\frac{μ_0i_2}{2πr}$

$⇒10×10^{-6}=\frac{μ_0}{2πr}(i_1+i_2)$

$B_{opposite}=B_1-B_2=\frac{μ_0}{2πr}(i_1-i_2)$

$⇒30×10^{-6}=\frac{μ_0}{2πr}(i_1-i_2)$

$∴\frac{30×10^{-6}}{10×10^{-6}}=\frac{K(i_1-i_2)}{K(i_1+i_2)}$

$⇒3(i_1+i_2)=i_1-i_2$

$⇒\frac{i_1}{i_2}=2$