Three long, straight and parallel wires carrying currents are arranged as shown in the figure. The wire $C$ which carries a current of $5.0\, amp$ is so placed that it experiences no force. The distance of wire $C$ from wire $D$ is then
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(a) For no force on wire $C$, force on wire $C$ due to wire $D$ = force on wire $C$ due to wire $B$
$ \Rightarrow \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2 \times 15 \times 5}}{x} \times l = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2 \times 5 \times 10}}{{\left( {15 - x} \right)}} \times l$
$ \Rightarrow \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2 \times 15 \times 5}}{x} \times l = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2 \times 5 \times 10}}{{\left( {15 - x} \right)}} \times l$
$ \Rightarrow x = 9\,cm.$
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