question_answer

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current \['I'\] along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire \['B'\] is given by [NEET-2017]

A)  \[\frac{{{\mu }_{0}}{{l}^{2}}}{\sqrt{2}\pi d}\]           

B)       \[\frac{{{\mu }_{0}}{{l}^{2}}}{2\pi d}\]

C)  \[\frac{2{{\mu }_{0}}{{l}^{2}}}{\pi d}\]        

D)       \[\frac{\sqrt{2}{{\mu }_{0}}{{l}^{2}}}{\pi d}\]

Correct Answer: A

Solution :

[a] Force between BC and AB will be same in magnitude.

\[{{F}_{BC}}={{F}_{BA}}=\frac{{{\mu }_{0}}{{l}^{2}}}{2\pi d}\]

\[F=\sqrt{2}{{F}_{BC}}\]

\[=\sqrt{2}\frac{{{\mu }_{0}}}{2\pi }\frac{{{l}^{2}}}{d}\]

\[F=\frac{{{\mu }_{0}}{{l}^{2}}}{\sqrt{2}\pi d}\]