Two identical long parallel wires carry currents $I_1$ and $I_2$ such that $I_1>I_2$. When the currents are…
Two identical long parallel wires carry currents $I_1$ 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 $6 \times 10^{-6} \mathrm{~T}$. If the direction of $I_2$ is reversed, the field becomes $3 \times 10^{-5} \mathrm{~T}$. The ratio $\left(\frac{I_1}{I_2}\right)$ is-
$\frac{3}{4}$
$\frac{1}{2}$
$\frac{2}{3}$
$\frac{3}{2}$
Solution
Using ampere's law: $B \propto I$ and $B \propto 1 / d$.
For two wires separated by a distance $d$, if $B_1$ is the field when the currents are in the same direction and $B_2$ is the field when currents are in opposite directions, then
$\frac{B_1}{B_2}=\frac{\frac{\mu_0}{2 \pi}\left(\frac{i_1}{d}-\frac{i_2}{d}\right)}{\frac{\mu_0}{2 \pi}\left(\frac{i_1}{d}+\frac{i_2}{d}\right)}=\frac{1}{5}$
On solving:
$\begin{aligned} & \frac{\left(\frac{i_1}{i_2}-1\right)}{\left(\frac{i_1}{i_2}+1\right)}=\frac{1}{5} \\ & \therefore \frac{i_1}{i_2}=\frac{3}{2}\end{aligned}$