Question

Equal currents are flowing in four infintely long wires.Distance between two wires is same and directions of currents are shown in figure.Match the following two columns.

Answer 1

Current in same direction means attraction and current in opposite direction means repulsion.
$\overrightarrow{F_{ij}}$ is force on wire i due to wire j.

$\overrightarrow{F_{12}}=\frac{{\mu }_0{i}^2}{2\pi \times d}(-\hat{i})$

$\overrightarrow{F_{13}}=\frac{{\mu }_0{i}^2}{2\pi \times 2d}\left(\hat{i}\right)$

$\overrightarrow{F_{14}}=\frac{{\mu }_0{i}^2}{2\pi \times 3d}(-\hat{i})$

Force on wire 1 due to wires 2,3 and 4:

$F_1=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}=\frac{{\mu }_0{i}^2}{2\pi \times d}(-\hat{i})+\frac{{\mu }_0{i}^2}{2\pi \times 2d}\left(\hat{i}\right)+\frac{{\mu }_0{i}^2}{2\pi \times 3d}(-\hat{i})=\frac{{\mu }_0{i}^2}{2\pi \times d}(-\frac{5}{6})\hat{i}$

$\therefore \overrightarrow{F_1}$ is leftward.
By symmetry, force on wire 4 is $\overrightarrow{F_4}$ is rightward.
$\overrightarrow{F_{21}}$ and $\overrightarrow{F_{23}}$ are equal and opposite.

$\overrightarrow{F_{24}}=\frac{{\mu }_0{i}^2}{2\pi \times 2d}\hat{i}$

$\therefore \overrightarrow{F_2}=\frac{{\mu }_0{i}^2}{2\pi \times 2d}\hat{i}$
Again by symmetry, $\overrightarrow{F_{31}}=\frac{{\mu }_0{i}^2}{2\pi \times 2d}(-\hat{i})$

$\therefore \overrightarrow{F_3}=\frac{{\mu }_0{i}^2}{2\pi \times 2d}(-\hat{i})$