+3 Votes

Two thin long parallei 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

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

in Physics by | 47 Views

1 Answer

+2 Votes
Best Answer

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


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


Related Questions

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]