Question

Two thin long parallel wires separated by a distance \(b\) are carrying a current \(I\) ampere each. The magnitude of the force per unit length exerted by one wire on the other, is

\(\begin{array}{llll}\text { (a) } \frac{\mu_{0} I^{2}}{b^{2}} & \text { (b) } \frac{\mu_{0} I}{2 \pi b^{2}} & \text { (c) } \frac{\mu_{0} I}{2 \pi b} & \text { (d) } \frac{\mu_{0} I^{2}}{2 \pi b}\end{array}\)

Answer 1

The correct option of this question will be (d).

Solution —

Let two long parallel thin wires $X$ and $Y$ carry current $I$ and separated by a distance $b$ apart.

The magnitude of magnelic field $B$ at any point on $Y$ due to current $I_{1}$ in $X$ is given by

$B=\frac{\mu_{0}}{2 \pi} \frac{I_{1}}{b}$

The magnitude of force acting on length $I$ of $Y$ is

$F=I_{2} B l=I_{2}\left(\frac{\mu_{0}}{2 \pi} \frac{I_{1}}{b}\right) l$

Force per unit length is

$\frac{F}{l}=\frac{\mu_{0}}{2 \pi} \frac{I_{1} I_{2}}{b}$

Given $I_{1}=I_{2}=I$,

therefore

$\frac{F}{l}=\frac{\mu_{0}}{2 \pi} \frac{I^{2}}{b}$