KCET · Physics · Magnetic Effects of Current
\(A\) and \(B\) are two infinitely long straight parallel conductors. \(C\) is another straight conductor of length \(1 \mathrm{~m}\) kept parallel to \(A\) and \(B\) as shown in the figure. Then the force experienced by \(C\) is

- A towards \(A\) equal to \(06 . \times 10^{-5} \mathrm{~N}\)
- B towards \(B\) equal to \(54 . \times 10^{-5} \mathrm{~N}\)
- C towards \(A\) equal to \(54 . \times 10^{-5} \mathrm{~N}\)
- D towards \(B\) equal to \(06 . \times 10^{-5} \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) towards \(B\) equal to \(06 . \times 10^{-5} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
The force between the conductors is attractive if the currents in them are in the same direction and repulsive if the currents are in opposite direction Mutual force between conductors \(A\) and \(C\)
\(\begin{aligned}F_{1} &=\frac{\mu_{0}}{2 \pi} \frac{I_{1} I_{2} l}{r} \\&=\frac{\mu_{0}}{2 \pi} \frac{2 \times 3 \times 1}{0.05} \\&=2.40 \times 10^{-5} \mathrm{~N} \text { (towards } A \text { ) }\end{aligned}\)
Mutual force between conductors \(B\) and \(C\)
\(=\frac{\mu_{0}}{2 \pi} \times \frac{4 \times 3 \times 1}{0.08} \)
\(=3 \times 10^{-5} \mathrm{~N} \text { (towards } B \text { ) }\)
Hence, the resultant force experienced by \(C\)
\(=(3-2.4) \times 10^{-5} \mathrm{~N} \)
\(=0.6 \times 10^{-5} \mathrm{~N}(\text { towards } B)\)
\(\begin{aligned}F_{1} &=\frac{\mu_{0}}{2 \pi} \frac{I_{1} I_{2} l}{r} \\&=\frac{\mu_{0}}{2 \pi} \frac{2 \times 3 \times 1}{0.05} \\&=2.40 \times 10^{-5} \mathrm{~N} \text { (towards } A \text { ) }\end{aligned}\)
Mutual force between conductors \(B\) and \(C\)
\(=\frac{\mu_{0}}{2 \pi} \times \frac{4 \times 3 \times 1}{0.08} \)
\(=3 \times 10^{-5} \mathrm{~N} \text { (towards } B \text { ) }\)
Hence, the resultant force experienced by \(C\)
\(=(3-2.4) \times 10^{-5} \mathrm{~N} \)
\(=0.6 \times 10^{-5} \mathrm{~N}(\text { towards } B)\)
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