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, \[{{F}_{AB}}=\frac{{{\mu }_{0}}}{4\,\pi }\,.\,\frac{2{{i}_{A}}\,{{i}_{B}}}{d}\] \[=\frac{{{\mu }_{0}}}{4\,\pi }.\frac{2\times 1\times 2}{d}\] \[=\frac{4{{\mu }_{0}}}{4\,\pi d}\] ?. (i) and \[{{F}_{CB}}=\frac{{{\mu }_{0}}}{4\,\pi }.\frac{2{{i}_{B}}\,{{i}_{C}}}{d}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\times 2\times 3}{d}\] \[=\frac{12{{\mu }_{0}}}{4\,\pi \,d}\] ?. (ii) As seen from eqs. (1) and (2) \[{{F}_{CB}}>{{F}_{AB}}\] hence, the net force of attraction will be directed towards wire C.
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