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
Diffcult
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(a) 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$$ = 6 \times {10^{ - 5}}\,N$ (Towards left)
Force on wire $Q$ due to wire $R$ is
${F_R} = {10^{ - 7}} \times \frac{{2 \times 20 \times 10}}{{0.02}} \times 0.1 = 20 \times {10^{ - 5}}\,N$ (Towards right)
Hence ${F_{net}} = {F_R} - {F_P} = 14 \times {10^{ - 5}}\,N = 1.4 \times {10^{ - 4}}\,N$
(Towards right)
${F_P} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{0.1}} \times 0.1$$ = 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)
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