Two long wires PQR and MNP carry equal current I as shown such that QR and NP are parallel. Find the magnetic field at origin O.

Due to PQ and MN, magnetic field at O will be zero.

$\begin{array}{l}\mathrm{r}=\mathrm{a}\sin {60}^0 =\frac{\sqrt{3}}{2}\mathrm{a}\\ {\mathrm{B}}_{\mathrm{QR}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\mathrm{r}}\left[\sin {90}^0-\sin {30}^0\right]\otimes \\ =\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\left(\sqrt{3}\mathrm{a}/2\right)}\times \frac{1}{2}\otimes =\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }\sqrt{3}\mathrm{a}}\otimes \\ {\mathrm{B}}_{\mathrm{NP}}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{\mathrm{I}}{\mathrm{r}}\left[\sin {90}^0+\sin {30}^0\right]\odot =\frac{3{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }\sqrt{3}\mathrm{a}}\odot \end{array}$

Net field at O,

$\mathrm{B}={\mathrm{B}}_{\mathrm{NP}}-{\mathrm{B}}_{\mathrm{QR}}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\sqrt{3}\mathrm{\pi a}}\odot$