Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length \(25\,cm\) is

\(=25\, \mathrm{cm}=0.25 \,\mathrm{m}\)

Force on wire \(Q\) due to wire \(R\)

\(F_{\mathrm{QR}}=10^{-7} \times \frac{2 \times 20 \times 10}{0.05} \times 0.25\)

\(=20 \times 10^{-5} \,\mathrm{N}\) (Towards left)

Force on wire \(Q\) due to wire \(P\)

\({F_{QP}} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{0.03}} \times 0.25\)

\(=50 \times 10^{-5} \,\mathrm{N}(\text { Towards right })\)

Hence, \(F_{\text {net }}=F_{Q P}-F_{Q R}\)

\(=50 \times 10^{-5}\, \mathrm{N}-20 \times 10^{-5}\, \mathrm{N}\)

\(=3 \times 10^{-4}\, \mathrm{N}\) towards right