- $B_2=\frac{\mu_0}{4\pi}\frac{2(3I)}{r}=\frac{\mu_0}{4\pi}\big(\frac{6I}{r}\big)$into the plane of the paper/$(\otimes)$.
$B_3=\frac{\mu_0}{4\pi}\frac{2(4I)}{3r}=\frac{\mu_0}{4\pi}\big(\frac{8I}{3r}\big)$out of the plane of the paper/$(\bigodot)$.
$B_A=B_2-B_3$ into the paper.
$=\frac{\mu_0}{4\pi}\Big(\frac{10I}{3r}\Big) into\text{ }the\text{ }paper.\text{} (\otimes)$
- $F_{21}=\frac{\mu_0}{4\pi}\frac{2I(3I)}{r}$ away from wire1 (/towards 3)
$F_{23}=\frac{\mu_0}{4\pi}\frac{2(3I)(4I)}{2r}$ away from wire 3 (towards 1)
$F_{\text{net}}=F_{23}-F_{21}$ towards wire1
$=\frac{\mu_0}{4\pi}\frac{6(I)^2}{r}$ towards wire1
