The current of 5A flows in a square loop of sides 1m is placed in...

Magnetic field at the centre of the square loop is given by,

$B = 4\left[\dfrac{{\mu}_{0}I}{4\pid}\left(\sin 45 ° + \sin 45 °\right)\right]$, where $d$ is perpendicular distance of the centre to one of its sides.

$⇒ B = 4 \times \dfrac{{\mu}_{0} \times 5}{4\pi \times \dfrac{1}{2}}\left(\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right)$

$= 4 \times {10}^{- 7} \times 5 \times 2 \times \sqrt{2}$

$= 40\sqrt{2} \times {10}^{- 7}T$