Q. 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
A) 9 $cm$
B) 7 $cm$
C) 5 $cm$
D) 3 $cm$
Explanation
(a) For no force on wire $C$, force on wire $C$ due to wire $D =$ force on wire $C$ due to wire $B$
$\Rightarrow \frac{\mu_{0}}{4\pi} \times \frac{2 \times 15 \times 5}{x} \times l \times \frac{\mu_{0}}{4\pi} \times \frac{2 \times 5 \times 10}{(15 - x)} \times l \Rightarrow x = 9cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.