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Two thin long parallel conductors separated by a distance ‘d’ carry currents $(I_{1})$ and $(I_{2})$ in same direction. They exert a force $\vec{F}$ on each other. Now, the current in one of them is increased to two times and its direction is reversed. The distance between them is increased to 3d. The new value of force between them is

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$- \frac{\vec{F}}{3}$

$\frac{\vec{F}}{3}$

$2 \vec{F}$

$- \frac{2 \vec{F}}{3}$

answer is D.

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Detailed Solution

$F = \frac{μ_{0} i_{1} i_{2} L}{2 π d}$

$F^{1} = \frac{μ_{0} i_{1} 2 i_{2} L}{2 π (3 d)} (direction reversed)$

$\frac{F}{- F^{1}} = \frac{3}{2} \Rightarrow F^{1} = \frac{- 2 F}{3}$

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