question_answer

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

A)  \[\frac{{{\mu }_{0}}{{i}^{2}}}{{{b}^{2}}}\]

B)  \[\frac{{{\mu }_{0}}i}{2\pi {{b}^{2}}}\]

C)  \[\frac{{{\mu }_{0}}i}{2\pi b}\]

D)  \[\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi b}\]

Correct Answer: D

Solution :

 Let two long parallel thin wires X and Y carry current i and separated by a distance b apart. The magnitude of magnetic field\[\overrightarrow{B}\]it at any point on\[Y\]due to current\[{{l}_{1}}\]in X is given by \[B=\frac{{{\mu }_{0}}}{2\pi }\frac{{{i}_{1}}}{b}\] The magnitude of force acting on length\[l\]of Y is  \[F={{i}_{2}}Bl={{i}_{2}}\left( \frac{{{\mu }_{0}}}{2\pi }\frac{{{i}_{1}}}{b} \right)l\] 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, \[\frac{F}{l}=\frac{{{\mu }_{0}}{{i}^{2}}}{b}\]