Two infinitely long parallel wires having linear charge densities ${\lambda }_1$ and ${\lambda }_2$ respectively are placed at a distance of R metres. The force per unit length on either wire will be $(K=\frac{1}{4\pi {ϵ}_0})$

Step 1: Given that:

Two infinitely long straight wires.

The linear charge density of the first wire = ${\lambda }_1$

The linear charge density of the second wire = ${\lambda }_2$

Distance between the wires = R

$K=\frac{1}{4\pi {\epsilon }_0}$

Step 2: Formula used:

The electric field due to a current carrying straight wire is given as;

$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$

Where $\lambda$ is the linear charge density of the wire and r is the distance of the point at which the electric field is found.

The force on a charge q in an electric field = qE

Step 3: Calculation of the force per unit length of either wire:

The force on the second wire if we consider that the wire is placed in the elctric field of the second wire,

$F_2=Q{E}_1$

$F_2=Q\frac{{\lambda }_1}{2\pi {\epsilon }_{0}R}$.................(1)

Now,

We know that,

Linear charge density = $\frac{Totalcharge}{lengthofthewire}$

For wire 2,

${\lambda }_2=\frac{Q}{l_2}$

$Q={\lambda }_2{l}_2$

Putting the value of Q in equation (1) we get;

$F_2={\lambda }_2{l}_2\frac{{\lambda }_1}{2\pi {\epsilon }_{0}R}$

$F_2=\frac{{\lambda }_1{\lambda }_2{l}_2}{2\pi {\epsilon }_{0}R}$

If the length of both the wire are taken as

$l$ that is $l_1=l_2=l$ then,

Force per unit length of either wire will be, $\frac{F}{l}$ .

Now,

$\frac{F_2}{l}=\frac{{\lambda }_1{\lambda }_2}{2\pi {\epsilon }_{0}R}\times \frac{2}{2}$

$\frac{F_2}{l}=\frac{2{\lambda }_1{\lambda }_2}{4\pi {\epsilon }_{0}R}$

$\frac{F_2}{l}=\frac{2{\lambda }_1{\lambda }_{2}K}{R}$

Taking,

$F_2=F$ Then,

$\frac{F}{l}=\frac{2{\lambda }_1{\lambda }_{2}K}{R}$

Thus,

The force per unit length of either wire is $\frac{2{\lambda }_1{\lambda }_{2}K}{R}$ .

Hence,

Option A) $\frac{2{\lambda }_1{\lambda }_{2}K}{R}$ is the correct option.