A) rotate about an axis parallel to the wire
B) move away from the wire
C) move towards the wire
D) remains stationary
Correct Answer: C
Solution :
Two current carrying conductors attract each other when current is flowing in the same direction. The force of attraction is given by \[\frac{F}{l}=\frac{{{\mu }_{0}}}{2\pi }\frac{{{I}^{2}}}{a}\]where a is distance between the wires. In the given case, force of attraction between wires PQ and AD is \[{{F}_{A}}=\frac{{{\mu }_{0}}}{2\pi }\frac{{{I}^{2}}}{a}\] Force of repulsion (currents in opposite directions) between PQ and BC is \[{{F}_{R}}=\frac{{{\mu }_{0}}}{2\pi }\frac{{{I}^{2}}}{a}\] Since, \[b>a\] \[\therefore \] \[{{F}_{A}}>{{F}_{R}}\] Hence, loop moves towards the wire.You need to login to perform this action.
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