Consider the arrangement given,



Force on a differentiable length element $\left(dl\right)$ of conductor $2$ due to field of conductor $1$ is

$dF=E_{1}dq$

where, $E_1=$ field of wire $1$ at location of $dl$

This force can be resolved into components $dF\cos \theta$ and $dF\sin \theta$

net force is sum of all $dF\cos \theta$ components, whereas $\mathrm{\Sigma }dF\sin \theta =0$.

So, net force on conductor $2$ is

$F=\int dF=\int E_{1}dq$

Here,

$E_1=\frac{2k{\lambda }_1}{R}=2k{\lambda }_1$

$dq={\lambda }_{2}dl$

$\frac{l}{r}=\tan \theta$ and $\frac{r}{R}=\cos \theta$

$\Rightarrow dl=r{\sec }^2\theta d\theta$

Also, $\theta$ varies from $\frac{-\pi }{2}$ to $\frac{\pi }{2}$ for conductor $2$.

So, $F={\int }_{-\pi /2}^{\pi /2}\frac{2k{\lambda }_1}{r\sec \theta }·{\lambda }_2·r{\sec }^2\theta d\theta$

$=2k{\lambda }_1{\lambda }_2{\int }_{-\pi /2}^{\pi /2}\sec \theta d\theta$

As, ${\int }_{-\pi /2}^{\pi /2}\sec \theta d\theta =\mathrm{a}$ constant

We can say that,

$F_{\mathrm{net}}=C·{\lambda }_1{\lambda }_2,$ where $C=$ some constant.

$\therefore F_{\mathrm{net}}\propto r^0$