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Two long conductors separated by a distance 'd' carry currents $I_1$ and $I_2$ in the same direction. They exert a force ' $\mathrm{F}$ ' 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 also increased to $3 \mathrm{~d}$. The new value of force between them is
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Verified Answer
The correct answer is:
$\frac{-2 \mathrm{~F}}{3}$
The force per unit length of the conductors is given as:
$\mathrm{F}=\frac{\mu_4 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}$
When the value and direction of current in the first conductor and the distance between the conductors are changed,
$\begin{aligned}
\therefore \quad \mathrm{F}_2 & =\frac{-\mu_0 2 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times 3 \mathrm{~d}} \\
\therefore \quad & \frac{\mathrm{F}_2}{\mathrm{~F}}=\frac{-2}{3} \\
\mathrm{~F}_2 & =-\frac{2 \mathrm{~F}}{3}
\end{aligned}$
$\mathrm{F}=\frac{\mu_4 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \mathrm{d}}$
When the value and direction of current in the first conductor and the distance between the conductors are changed,
$\begin{aligned}
\therefore \quad \mathrm{F}_2 & =\frac{-\mu_0 2 \mathrm{I}_1 \mathrm{I}_2}{2 \pi \times 3 \mathrm{~d}} \\
\therefore \quad & \frac{\mathrm{F}_2}{\mathrm{~F}}=\frac{-2}{3} \\
\mathrm{~F}_2 & =-\frac{2 \mathrm{~F}}{3}
\end{aligned}$
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