Figure shows a current carrying square loop ABCD of edge length is ' \(a\) ' lying in a plane. If the resistance of the \(A B C\) part is \(r\) and that of \(A D C\) part is 2 r , then the magnitude of the resultant magnetic field at centre of the square loop is

\(\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{BC}}+\overrightarrow{\mathrm{B}}_{\mathrm{CD}}+\overrightarrow{\mathrm{B}}_{\mathrm{DA}}\)
\(\begin{aligned} \overrightarrow{\mathrm{B}}= & {\left[\frac{-\mu_0(2 \mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}-\frac{\mu_0(2 \mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}\right.} \\ & \left.+\frac{\mu_0(\mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}+\frac{\mu_0(\mathrm{I} / 3)}{4 \pi(\mathrm{a} / 2)} \sqrt{2}\right] \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{B}}= & {\left[\frac{-2 \sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}+\frac{\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}}\right] \hat{\mathrm{k}} } \\ \overrightarrow{\mathrm{B}}= & \frac{-\sqrt{2} \mu_0 \mathrm{I}}{3 \pi \mathrm{a}} \hat{\mathrm{k}}\end{aligned}\)