MHT CET · Physics · Magnetic Effects of Current
Three long straight and parallel wires carrying currents are arranged as shown

The wire \(\mathrm{C}\) which carries a current of \(50 \mathrm{~A}\) is so placed that it experiences no force. The distance of wire \(\mathrm{C}\) from wire \(\mathrm{A}\) is
- A \(3 \mathrm{~cm}\)
- B \(5 \mathrm{~cm}\)
- C \(9 \mathrm{~cm}\)
- D \(7 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(9 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
The correct option is (C).
Concept: Using Ampear's circuital law the magnetic field due to a current carrying wire at a distance \(r\) is given by, \(B=\frac{\mu_0 I}{2 \pi r}\) where, \(\mathrm{I}\) is the current through the wire.
If the magnetic field due to the wire A cancels out the magnetic field due to the wire \(\mathrm{B}\) at the wire \(\mathrm{C}\), then
\(\frac{\mu_0 I_A}{2 \pi x}=\frac{\mu_0 I_B}{2 \pi(15-x)}\)
\(\frac{(15-x)}{x}=\frac{I_B}{I_A}\)
Therefore, for \(\mathrm{I}_{\mathrm{B}}=10 \mathrm{~A}\) and \(\mathrm{I}_{\mathrm{A}}=15 \mathrm{~A}\), we get \(\mathrm{x}=9 \mathrm{~cm}\).
Concept: Using Ampear's circuital law the magnetic field due to a current carrying wire at a distance \(r\) is given by, \(B=\frac{\mu_0 I}{2 \pi r}\) where, \(\mathrm{I}\) is the current through the wire.
If the magnetic field due to the wire A cancels out the magnetic field due to the wire \(\mathrm{B}\) at the wire \(\mathrm{C}\), then
\(\frac{\mu_0 I_A}{2 \pi x}=\frac{\mu_0 I_B}{2 \pi(15-x)}\)
\(\frac{(15-x)}{x}=\frac{I_B}{I_A}\)
Therefore, for \(\mathrm{I}_{\mathrm{B}}=10 \mathrm{~A}\) and \(\mathrm{I}_{\mathrm{A}}=15 \mathrm{~A}\), we get \(\mathrm{x}=9 \mathrm{~cm}\).
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