Question

Two mutually perpendicular infinitely long straight conductors carry in uniformly distributed charges of linear densities ${\lambda }_1$ and ${\lambda }_2$ are positioned at a distance $r$ from each other. Force between the conductors depends on $r$ as

• A. $I/r$
• B. $I/r^2$
• C. $r$
• D. $r^0$

Answer 1

Answer: (B)

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 $\Sigma dF\sin \theta =0$ .
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 $\frac{-\pi }{2}to\pi \frac{\pi }{2}$ for conductor $2$.
so, $F=\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{2k{\lambda }_1}{r\sec \theta }\cdot {\lambda }_2\cdot r{\sec }^2\theta d\theta$
$=2k{\lambda }_1{\lambda }_2\underset{-\pi /2}{\overset{\pi /2}{\int }}\sec \theta d\theta$ $As\underset{-\pi /2}{\overset{\pi /2}{\int }}\sec \theta d\theta =$ a constant
we can say that
$F_{net}=C\cdot {\lambda }_1{\lambda }_2$ where $C=$ some constant
$\therefore F_{net\propto r^o}$