Force on a differentiable length element $(d l)$ of conductor 2 due to field of conductor 1 is

$d F=E_{1} d q$

where, $E_{1}=$ ficld of wire 1 at location of $d l$.

'This force can be resolved into components $d F \cos \theta$ and $d F \sin \theta$ net force is sum of all $d F \cos \theta$ components, whereas $\Sigma d F \sin \theta=0$ So, net force on conductor $2$ is

$F=\int d F=\int E_{1} d q$

Here,

$E_{1}=\frac{2 k \lambda_{1}}{R}=2 k \lambda_{1}$

$d q=\lambda_{2} d l$

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

$d l =r \sec ^{2} \theta d \theta$

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

So, $F=\int \limits_{-\pi / 2}^{\pi / 2} \frac{2 k \lambda_{1}}{r \sec \theta} \cdot \lambda_{2} \cdot r \sec ^{2} \theta d \theta$

$=2 k \lambda_{1} \lambda_{2} \int \limits_{-\pi / 2}^{\pi / 2} \sec \theta d \theta$

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

We can say that,

$F_{\text {net }}=C \cdot \lambda_{1} \lambda_{2}$, where $C=$ some constant.

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