A) perpendicular to the plane of paper and outward
B) perpendicular to the plane of paper and inward
C) towards A
D) towards C
Correct Answer: D
Solution :
Since, the currents in the three wires are flowing in same direction so, the wire B will experience a force of attraction due to both wires A and C, So, \[{{F}_{AB}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{A}}{{i}_{B}}}{d}\] \[=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\times 1\times 2}{d}\] \[=\frac{4{{\mu }_{0}}}{4\pi d}\] and \[{{F}_{CB}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{B}}{{i}_{C}}}{d}=\frac{{{\mu }_{0}}.}{4\pi }\times \frac{2\times 2\times 3}{d}\] \[=\frac{12{{\mu }_{0}}}{4\pi d}\] ?(2) As seen from eqs. (1) and (2) \[{{F}_{CB}}>{{F}_{AB}}\]hence, the net force of attraction will be directed towards wire C.You need to login to perform this action.
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