Three infinite straight wires \(\mathrm{A}, \mathrm{B}\) and C carry currents as shown in figure. The resultant force on wire B is directed

\(\mathrm{F}=\mathrm{BIL}\)
Magnetic field due to \(A: \frac{\mu_0 I}{2 \pi d}=\frac{\mu_o}{2 \pi d}\)
Force on \(B\) due to \(A\),
\(\mathrm{F}_{\mathrm{AB}}=\frac{\mu_0}{2 \pi \mathrm{~d}} \times \mathrm{I} \times \mathrm{L}\)
Magnetic field due to \(\mathrm{C}: \frac{\mu_{\mathrm{o}} \mathrm{I}}{2 \pi \mathrm{~d}}=\frac{\mu_{\mathrm{o}} \times 3}{2 \pi \mathrm{~d}}\)
Force on B due to C,
\(\begin{aligned}
& \mathrm{F}_{\mathrm{CB}}=\frac{3 \mu_{\mathrm{o}}}{2 \pi \mathrm{~d}} \times \mathrm{I} \times \mathrm{L}\\
\therefore \quad & \mathrm{~F}_{\mathrm{AB}} \lt \mathrm{F}_{\mathrm{CB}}
\end{aligned}\)
\(\therefore \quad\) The resultant force will be towards A