STD 12 - 4. Moving charges and magnetism — NEET STD 11 Science — Question

$=25\, \mathrm{cm}=0.25 \,\mathrm{m}$

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