A straight segment $OC$$\left(of length L\right)$of a circuit carrying a current $I$ is placed along the $x-axis$. Two infinitely long straight wires $A$ and $B$, each extending from $z=-\infty$ to $+\infty$, are fixed at $y=-a$ and $y=+a$  respectively, as shown in the figure. If the wires $A$ and $B$ each carry a current $I$ into the plane  of the paper, obtain the expression for the force acting on the segment $OC$. What will be the force on $OC$ if the current in the wire$B$ is reversed?

Let us assume a segment of wire $OC$ at a point $P$, a distance $x$ from the centre of length $dx$ as shown in the figure.

Magnetic field at $P$ due to current in wires $A$ and $B$ will be in the directions perpendicular to $AP$ and $BP$ respectively as shown.

$\left|B\right|=\frac{{\mu }_0}{2\mathrm{\pi }}\frac{I}{AP}$

Therefore, net magnetic force at $P$ will be along negative $y-$axis as shown below

$$\begin{gathered} B_{net}=2\left|B\right|\cos \theta \\ =2\left(\frac{{\mu }_0}{2\mathrm{\pi }}\right)\frac{I}{AP}\left(\frac{x}{AP}\right) \\ {B}_{net}=\left(\frac{{\mu }_0}{\mathrm{\pi }}\right)\frac{Ix}{{\left(AP\right)}^2} \\ {B}_{net}=\frac{{\mu }_0}{\mathrm{\pi }}.\frac{Ix}{\left(a^2+x^2\right)} \end{gathered}$$

Therefore, force on this element will be 

$dF=I\left\{\frac{{\mu }_0}{\mathrm{\pi }}\frac{Ix}{a^2+x^2}\right\}dx$  [in negative $z$-direction]

$\therefore$ Total force on the wire will be $F={\int }_{x=0}^{x=L}dF=\frac{{\mu }_0{I}^2}{\mathrm{\pi }}{\int }_0^L\frac{xdx}{x^2+a^2}$

                                                        $=\frac{{\mu }_0{I}^2}{2\mathrm{\pi }}\ln \left(\frac{L^2+a^2}{a^2}\right)$   [in negative $z$-axis]

Hence, $F=\frac{{\mu }_0{I}^2}{2\mathrm{\pi }}\ln \left(\frac{L^2+a^2}{a^2}\right)\hat{k}$

(b) When direction of current in $B$ is reversed net magnetic field is along the current. Hence,force is zero.