FABl=μ0iaib2πx towards C ….(1)
FCBl=μ0ibic2π(15−x) towards A ....(2)
Since, wire B experiences no force, FABl=FCBl
⇒μ0iaib2πx=μ0ibic2π(15−x)
Substituting the data given in the question we get,
⇒iax=ic(15−x)
15x=10(15−x)⇒45−3x=2x
⇒x=9 cm
Accepted answer : 9