Two identical circular wires of radius 20 cm and carrying current $\sqrt{2}A$ are placed in perpendicular planes as shown in figure.  The net magnetic field at the centre of the circular wires is ________ × 10^-8 T.

$\text{ (Take }\pi =3.14\text{ ) }$

Given radius of each circular wire is 20 cm current in the loops is $\sqrt{2}A$.

$\overrightarrow{B}\text{ due to }y\text{ at the centre }=\frac{{\mu }_{0}I}{2R}\widehat{i}$

$\begin{array}{l}=\frac{4\pi \times {10}^{-7}\times \sqrt{2}}{2\times 20\times {10}^{-2}}\\ =\frac{\pi }{10}\times {10}^{-5}\times \sqrt{2}\\ =\sqrt{2}\pi \times {10}^{-6}\widehat{i}\end{array}$

$\overrightarrow{B}\text{ due to }x\text{ at the centre }$

$\begin{array}{l}=\frac{{\mu }_{0}I}{2R}\widehat{j}\\ =\frac{4\pi \times {10}^{-7}\times \sqrt{2}}{2\times 20\times {10}^{-2}}\\ =\sqrt{2}\pi \times {10}^{-6}\widehat{j}\end{array}$

$\text{ Net }{\overrightarrow{B}}_x+{\overrightarrow{B}}_y$

$\begin{array}{l}=\sqrt{{\left(\sqrt{2}\pi \times {10}^{-6}\right)}^2+{\left(\sqrt{2}\pi \times {10}^{-6}\right)}^2}\\ =\sqrt{2}\times \sqrt{2}\pi \times {10}^{-6}\\ =2\pi \times {10}^{-6}\\ {\overrightarrow{B}}_{\text{net }}=628\times {10}^{-8}\mathrm{T}\end{array}$