Two parallel, long wires are kept $0.20\,m$ apart in vacuum, each carrying current of $x$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{-6}\,N$, then the value of $x$ is approximately
JEE MAIN 2022, Medium
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Force per unit length $=\frac{\mu_{0} i_{1} i_{2}}{2 \pi d}$
$=\frac{\mu_{0} \cdot x ^{2}}{2 \pi \times 0.2}$
$F =2 \times 10^{-6}=\frac{4 \pi \times 10^{-7} \times x ^{2}}{2 \pi \times 0.2}$
$\Rightarrow 10^{-6}=10^{-7} \frac{ x ^{2}}{0.2}$
$\Rightarrow x ^{2}=10 \times 0.2$
$=2$
$\Rightarrow x =\sqrt{2} \approx 1.4\,Amp$
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