Two long wires carrying current ${{\rm{I}}_1}$ and ${{\rm{I}}_2}$ are arranged as shown in figure. The one carrying current ${{\rm{I}}_1}$ is along is the $\mathrm{y}$ - axis. The other carrying current ${{\rm{I}}_2}$ is along a line parallel to the yaxis given by ${\rm{x = 0}}$ and ${\rm{z = d}}$. Find the force exerted at ${{\rm{O}}_2}$ because of the wire along the ${\rm{x}}$ - axis.

We know that force on current $I $carrying conductor placed in magnetic field $B$ is,

$\mathrm{F}=\mathrm{B} \times \mathrm{I} \cdot d l$

$\mathrm{~F}=\mathrm{BId} l \sin \theta$

The direction of magnetic field at $\mathrm{O}_{2}$ due to the current $\mathrm{I}_{1}$ is parallel to $y$-axis and in $y$-direction.

As wire of current $\mathrm{I}_{2}$ is parallel to $y$-axis, current in $\mathrm{I}_{2}$ is also along $y$-axis. So, $\mathrm{I}_{2}$ and $\mathrm{B}_{1}$ (mag

netic field due to current $\mathrm{I}_{1}$ ) are also along $y$-axis. i.e. angle between $\mathrm{I}_{2}$ and $\mathrm{B}_{1}$ is zero.

So, magnetic force $\mathrm{F}_{2}$ on the wire of current $\mathrm{I}_{2}$ is,

$\mathrm{F}_{2}=\mathrm{B}_{1} \mathrm{I}_{2}$ dlsin0 $^{\circ}$

$\mathrm{F}_{2}=0$

Hence, force on $\mathrm{O}_{2}$ due to wire of current $\mathrm{I}_{1}$ is zero.