Two long straight wires $P$ and $Q$ carrying equal current $10\,A$ each were kept parallel to each other at $5\,cm$ distance. Magnitude of magnetic force experienced by $10\,cm$ length of wire $P$ is $F_1$. If distance between wires is halved and currents on them are doubled, force $F_2$ on $10\,cm$ length of wire $P$ will be :
JEE MAIN 2023, Medium
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Force per unit length between two parallel straight
$\text { wires }=\frac{\mu_0 i_1 i_2}{2 \pi d }$
$\frac{ F _1}{ F _2}=\frac{\frac{\mu_0(10)^2}{2 \pi(5\,cm )}}{\frac{\mu_0(20)^2}{2 \pi\left(\frac{5 cm }{2}\right)}}=\frac{1}{8}$
$\Rightarrow F_2=8 F_1$
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