Q 12-04-113JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
Two long straight wires $P$ and $Q$ carrying equal current $10\ \text{A}$ each were kept parallel to each other at $5\ \text{cm}$ distance. Magnitude of magnetic force experienced by $10\ \text{cm}$ length of wire $P$ is $F_1$. If distance between wires is halved and currents on them are doubled, force $F_2$ on $10\ \text{cm}$ length of wire $P$ will be
Answer: (A) $8F_1$
$F=\dfrac{\mu_0I_1I_2l}{2\pi d}\propto\dfrac{I^2}{d}$. Doubling both currents gives $\times4$, halving $d$ gives $\times2$: $F_2=8F_1$.
Solution by Sreeraj P, M.Sc Physics