MHT CET · Physics · Magnetic Effects of Current
Three long, straight parallel wires carrying currents are arranged as shown. The wire C which carries a current of 5.0 A is so placed that it experiences no force. The distance of wire C from wire D is
- A 9 cm
- B 7 cm
- C 5 cm
- D 3 cm
Answer & Solution
Correct Answer
(A) 9 cm
Step-by-step Solution
Detailed explanation
Wire C experiences no force.
\(\begin{array}{ll}
\therefore & \mathrm{F}_{\mathrm{CD}}=\mathrm{F}_{\mathrm{CB}} \\
\therefore & \frac{\mu_0 \mathrm{i}_{\mathrm{C}} \mathrm{i}_{\mathrm{D}}}{2 \pi \mathrm{x}}=\frac{\mu_0 \mathrm{i}_{\mathrm{c}} \mathrm{i}_{\mathrm{B}}}{2 \pi(15-\mathrm{x})} \\
\therefore & \frac{5 \times 15}{\mathrm{x}}=\frac{5 \times 10}{15-\mathrm{x}} \\
\therefore & x=9 \mathrm{~cm}
\end{array}\)
\(\begin{array}{ll}
\therefore & \mathrm{F}_{\mathrm{CD}}=\mathrm{F}_{\mathrm{CB}} \\
\therefore & \frac{\mu_0 \mathrm{i}_{\mathrm{C}} \mathrm{i}_{\mathrm{D}}}{2 \pi \mathrm{x}}=\frac{\mu_0 \mathrm{i}_{\mathrm{c}} \mathrm{i}_{\mathrm{B}}}{2 \pi(15-\mathrm{x})} \\
\therefore & \frac{5 \times 15}{\mathrm{x}}=\frac{5 \times 10}{15-\mathrm{x}} \\
\therefore & x=9 \mathrm{~cm}
\end{array}\)
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