Two long conductors separated by a distance ' d ' carry currents ' I₁ ' and ' I₂ ' in the same directions. They exert a force ' F ' on each other. The distance between them is increased to ' 3 d '. If new repulsive force of magnitude ' 2/3F ' is found between these conductors, the required change in the magnitude and direction of one of the currents in the conductor is respectively [length of the conductors is constant] (a) same, reversed. (b) twice, reversed. (c) thrice, same. (d) twice, same.

Answer: (b). F ∝ (I₁ I₂)/(d). Let I₂ → kI₂ and d → 3d: F' = k/3F |F'| = 2/3F ⇒ k = 2 Force is now repulsive, so the current direction must be reversed. So the answer is (b) twice, reversed.

Two long conductors separated by a distance ' d ' carry currents ' I₁… | MHT CET 2025 (19 Apr, Shift 2)

Q13MHT CET 2025 · 19 APR · SHIFT 2Medium

Force on Current-Carrying Conductors

Two long conductors separated by a distance ' d ' carry currents ' $I_1$ ' and ' $I_2$ ' in the same directions. They exert a force ' $F$ ' on each other. The distance between them is increased to ' 3 d '. If new repulsive force of magnitude ' $\frac{2}{3}F$ ' is found between these conductors, the required change in the magnitude and direction of one of the currents in the conductor is respectively [length of the conductors is constant]

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Answer: (b)

$F \propto \frac{I_1 I_2}{d}$ . Let $I_2 \to kI_2$ and $d \to 3d$ : $F' = \frac{k}{3}F$
$|F'| = \frac{2}{3}F \Rightarrow k = 2$
Force is now repulsive, so the current direction must be reversed.
So the answer is (b) twice, reversed.

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Force on Current-Carrying Conductors question from MHT CET 2025 (19 Apr, Shift 2)

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