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  4. Two long straight wires P and Q carrying equal current 10 A ench were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1 If distance between wires is halved and currents on them are doubled, force F2 on 10 cm length of wire P will be:

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Q. Two long straight wires $P$ and $Q$ carrying equal current $10\, A$ ench 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:

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Solution:

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$