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Q.
Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force experienced by $10\, cm$ length of wire $Q$ is
Solution:
Force on wire $Q$ due to wire $P$ is
$F_{P}=10^{-7} \times \frac{2 \times 30 \times 10}{0.1} \times 0.1$
$=6 \times 10^{-5}\, N$
(Towards left)
Force on wire $Q$ due to wire $R$ is
$F_{R}=10^{-7} \times \frac{2 \times 20 \times 10}{0.02} \times 0.1$
$=20 \times 10^{-5} N$
(Towards right)
Hence
$F _{ net }= F _{ R }- F _{ P } $
$=14 \times 10^{-5} \,N$
$=1.4 \times 10^{-4} \,N$
(Towards right)
