Questions
A current carry square loop ABCD is placed at a distance a from an infinitely long current carrying long straight conductor PQ as shown in the figure. The conductor PQ lies in the plane of the loop. Then
detailed solution
Correct option is C
F→DA=F→BCF→AB=i AB→×B→1FAB=i×a×μoI2πaF→CD=i CD→×B→2=i×a×μoI2π2a∴FAB>FCDTalk to our academic expert!
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