\(=25\, \mathrm{cm}=0.25 \,\mathrm{m}\)
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
[$m _{p}=1.6 \times 10^{-27} kg , e =1.6 \times 10^{-19} C$ નો ઉપયોગ કરવો.]
${\mu _o}$$=4$$\pi $$ \times 10^{-7}$ $\frac{{Tm}}{A}$ લો. પૃથ્વીનું ચુંબકીયક્ષેત્ર અવગણો.