Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length $25\,cm$ is
JEE MAIN 2014, Diffcult
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Also given; length of wire $Q$
$=25\, \mathrm{cm}=0.25 \,\mathrm{m}$
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
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