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