Two identical circular wires of radius 20cm and carrying current 2...

The formula to calculate the magnetic field $\left(B\right)$ at the centre of a current carrying circular loop is given by

$B = \dfrac{{\mu}_{0}i}{2R} . . . ( 1 )$

where, ${\mu}_{0}$ is the permeability of free space, $i$ is the current and $R$ is the radius of the loop.

${B}_{V}, {B}_{H}$ are the magnetic field due to the vertical and the horizontal loop, which by magnitude are equal.

The resultant magnetic field $\left({B}_{net}\right)$ at the centre is, then, given by

${B}_{\text{net}} = \sqrt{{B}^{2} + {B}^{2}} = \sqrt{2}B = \dfrac{\sqrt{2} \times 4\pi \times {10}^{- 7} \times \sqrt{2}}{2 \times 0 . 2}$

$= 628 \times {10}^{- 8}T$