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Two long conductors, separated by a distance $\mathrm{d}$ carry current $\mathrm{I}_1$ and $\mathrm{I}_2$ in the same direction. They exert a force $F$ on each other. Now the current in one of them increased to two times and its direction 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:
$-2 F / 3$
$-2 F / 3$
Force between two long conductor carrying current
$$
F=\frac{\mu_0}{2 \pi} \frac{l_1 l_2}{d} \ell
$$
According to question
$$
F^{\prime}=\frac{\mu_0}{2 \pi} \frac{\left(-2 I_1\right)\left(I_2\right)}{d} \ell
$$
From equation (i) and (ii), $\mathrm{F}^{\prime}=-\frac{3}{2} \mathrm{~F}$.
$$
F=\frac{\mu_0}{2 \pi} \frac{l_1 l_2}{d} \ell
$$
According to question
$$
F^{\prime}=\frac{\mu_0}{2 \pi} \frac{\left(-2 I_1\right)\left(I_2\right)}{d} \ell
$$
From equation (i) and (ii), $\mathrm{F}^{\prime}=-\frac{3}{2} \mathrm{~F}$.
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