Question:
Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is:
Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is:
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Add the forces from the two outer wires using \(F=\frac{\mu_0 I_1 I_2\ell}{2\pi d}\), repulsion for opposite currents.
Updated On: Nov 11, 2026
- \(3\times10^{-4}\) N toward right
- \(6\times10^{-4}\) N toward right
- \(9\times10^{-4}\) N toward right
- Zero
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The Correct Option is A
Solution and Explanation
Figure: \(I_1=30\) A (up) at 3 cm to the left, \(I=10\) A (down) in the middle, \(I_2=20\) A (up) at 5 cm to the right. Parallel currents attract, antiparallel currents repel.
Force from \(I_1\): \(F_1=\frac{\mu_0}{2\pi}\frac{I_1I\ell}{d_1}=2\times10^{-7}\times\frac{30\times10\times0.25}{0.03}=5\times10^{-4}\) N, repulsive (antiparallel), so towards the right.
Force from \(I_2\): \(F_2=2\times10^{-7}\times\frac{20\times10\times0.25}{0.05}=2\times10^{-4}\) N, repulsive, so towards the left.
Net force \(=5\times10^{-4}-2\times10^{-4}=3\times10^{-4}\) N toward the right.
Answer: (1) \(3\times10^{-4}\) N toward right
Force from \(I_1\): \(F_1=\frac{\mu_0}{2\pi}\frac{I_1I\ell}{d_1}=2\times10^{-7}\times\frac{30\times10\times0.25}{0.03}=5\times10^{-4}\) N, repulsive (antiparallel), so towards the right.
Force from \(I_2\): \(F_2=2\times10^{-7}\times\frac{20\times10\times0.25}{0.05}=2\times10^{-4}\) N, repulsive, so towards the left.
Net force \(=5\times10^{-4}-2\times10^{-4}=3\times10^{-4}\) N toward the right.
Answer: (1) \(3\times10^{-4}\) N toward right
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