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Two very long, straight, parallel conductors A and $\mathrm{B}$ carry current of $5 \mathrm{~A}$ and $10 \mathrm{~A}$ respectively and are at a distance of $10 \mathrm{~cm}$ from each other. The direction of current in two conductors is same. The force acting per unit length between two conductors is $\left(\mu_0=4 \pi \times 10^{-7}\right.$ SI unit):
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Verified Answer
The correct answer is:
$1 \times 10^{-4} \mathrm{Nm}^{-1}$ and is attractive
As per question,
$\begin{aligned}
\frac{\mathrm{F}}{l} & =\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \mathrm{I}_1 \mathrm{I}_2}{r} \\
& =\frac{4 \pi \times 10^{-7} \times 2 \times 5 \times 10}{2 \pi \times 0.1} \\
& =10^{-4} \mathrm{Nm}^{-1}
\end{aligned}$
The nature of the force is attractive.
(where, the symbols have their usual meanings)
$\begin{aligned}
\frac{\mathrm{F}}{l} & =\frac{\mu_{\mathrm{o}}}{4 \pi} \times \frac{2 \mathrm{I}_1 \mathrm{I}_2}{r} \\
& =\frac{4 \pi \times 10^{-7} \times 2 \times 5 \times 10}{2 \pi \times 0.1} \\
& =10^{-4} \mathrm{Nm}^{-1}
\end{aligned}$
The nature of the force is attractive.
(where, the symbols have their usual meanings)
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