Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire $Q$ is ______ .
$(\mu_o = 4\pi\times10^{-7}\ \text{T.m/A})$
Answer: (B) $6\times10^{-6}$ N towards $R$
Force per unit length between wires: $\dfrac{F}{L} = \dfrac{\mu_0I_1I_2}{2\pi d}$. Parallel currents attract, antiparallel currents repel.
From $P$ (3 A up, Q is 1 A down → repulsion, pushes $Q$ towards $R$):
$$\frac{2\times10^{-7}\times3\times1}{0.03} = 2\times10^{-5}\ \text{N/m}$$
From $R$ (2 A down, same direction as $Q$ → attraction, pulls $Q$ towards $R$):
$$\frac{2\times10^{-7}\times2\times1}{0.02} = 2\times10^{-5}\ \text{N/m}$$
Both act towards $R$: total $4\times10^{-5}$ N/m. For $0.15$ m:
$$F = 4\times10^{-5}\times0.15 = 6\times10^{-6}\ \text{N towards } R$$
Solution by Sreeraj P, M.Sc Physics