question_answer

Two thin long parallel wires separated by a distance \[r\] carry equal currents\[i\]. The magnitude of the force per unit length exerted by one wire .on the other wire will be

A) \[{{\mu }_{0}}i/2\pi r\]                 

B) \[{{\mu }_{0}}{{i}^{2}}/{{r}^{2}}\]

C) \[{{\mu }_{0}}{{i}^{2}}/2\pi r\]                  

D) \[{{\mu }_{0}}i/{{r}^{2}}\]

Correct Answer: C

Solution :

When a wire is placed in a magnetic field\[B\], carrying current\[i\], separated by a distance\[r\]from another wire carrying the same current, then magnitude of force\[(F)\]is                 \[F=iBl=i\left( \frac{{{\mu }_{0}}}{2\pi }\frac{i}{r} \right)l\] Force per unit length\[\frac{F}{l}=\frac{{{\mu }_{0}}}{2\pi }\frac{{{i}^{2}}}{r}\] When currents are in same direction, then from Flemings left hand rule, force of attraction acts between them.