Question

Two long conductors separated by a distance ' d ' carry currents ' $\text{I}_1$ ' and ' $\text{I}_2$ ' in the same directions. They exert a force ' $F$ ' on each other. The distance between them is increased to ' $3 \text{ d}$ '. If new repulsive force of magnitude ' $\frac{2}{3} \text{ F}$ ' is found between these conductors, the required change in the magnitude and direction of one of the currents in the conductor is respectively}

• A. same, reversed.
• B. twice, reversed.
• C. thrice, same.
• D. twice, same.

Hint:

Same direction current = Attraction; Opposite direction current = Repulsion.

Answer 1

The Correct Option is B

Step 1: Concept
Force between parallel wires is $F = \frac{\mu_0 I_1 I_2 L}{2\pi d}$.

Step 2: Meaning
Initially, $F \propto \frac{I_1 I_2}{d}$ (attractive). New force $F' = \frac{2}{3} F$ (repulsive).

Step 3: Analysis
For the force to be repulsive, the direction of one current must be reversed.
$F' = \frac{k I_1 I_2'}{3d} = \frac{2}{3} \left(\frac{k I_1 I_2}{d}\right) \implies \frac{I_2'}{3} = \frac{2 I_2}{3} \implies I_2' = 2 I_2$.

Step 4: Conclusion
The current magnitude is doubled and the direction is reversed. Final Answer: (B)