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

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$- \frac{F}{3}$

$\frac{2 F}{3}$

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

$\frac{F}{3}$

answer is D.

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Detailed Solution

Force between two conductors, $F \propto \frac{I_{1} I_{2}}{d}$

When $I_{1}$ is changed to $2 I_{1}$ and $d$ is changed to $3 d$ .

$∴ F' \propto \frac{(2 I) (I_{2})}{3 d} \propto \frac{2}{3} F$

As direction of current is reversed so,

$F' = \frac{- 2 F}{3}$

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