Physics - Magnetism Question with Solution | TestHub

Given, current through the square loop is $2 \mathrm{A}$.

$I=2 A$

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

$B_{AB}=\frac{{\mu }_0}{4\pi }·\frac{I}{r}\left[\sin 45°+\sin 45°\right]$

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

$=\sqrt{2}\times {10}^{-5} \mathrm{T}$ [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 $O$

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