The magnetic field at the centre O & of& the current-carrying...

Given, current through the square loop is $2A$.

$I = 2A$

The given situation is shown in the figure. Magnetic field due to wire $AB$ at the centre $O$ is

${B}_{AB} = \dfrac{\mu0}{4\pi} \cdot \dfrac{I}{r} \left[\sin 45 ° + \sin 45 °\right]$

$= {10}^{- 7} \times \dfrac{2}{2 \times {10}^{- 2}}\left[\dfrac{1}{2} + \dfrac{1}{2}\right]$

$= \sqrt{2} \times {10}^{- 5}T$ [Downward to the plane of square loop]

Since, magnetic field due to side $AB, BC, CD$ and $DA$ is same in magnitude and direction.

Hence, net magnetic field at $O$

$B = 4 \times {B}_{AB} = 4\sqrt{2} \times {10}^{- 5}T$