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                                                                                                                                                                   

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 :

                 Force between BC and AB will be same in magnitude. \[{{F}_{BC}}={{F}_{BA}}=\frac{{{\mu }_{0}}{{l}^{2}}}{2\pi d}\] \[F=\sqrt{2}{{F}_{BC}}\] \[=\sqrt{2}\frac{{{\mu }_{0}}}{2\pi }\frac{{{l}^{2}}}{d}\] \[F=\frac{{{\mu }_{0}}{{l}^{2}}}{\sqrt{2}\pi d}\]