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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. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to ' $3 \mathrm{~d}$. The new value of the force between them is
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Verified Answer
The correct answer is:
$-\frac{2 \mathrm{~F}}{3}$
Force on each conductor is given by
$$
\mathrm{F}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~d}} \ell
$$
This force will be attractive.
If the direction of current is reversed in one conductor, the force will become repulsive.
$$
\therefore \mathrm{F}^{\prime}=-\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{3 \mathrm{~d}} \cdot \ell=-\frac{2}{3} \mathrm{~F}
$$
$$
\mathrm{F}=\frac{\mu_0}{4 \pi} \cdot \frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~d}} \ell
$$
This force will be attractive.
If the direction of current is reversed in one conductor, the force will become repulsive.
$$
\therefore \mathrm{F}^{\prime}=-\frac{\mu_0}{4 \pi} \cdot \frac{2 \mathrm{I}_1 \mathrm{I}_2}{3 \mathrm{~d}} \cdot \ell=-\frac{2}{3} \mathrm{~F}
$$
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