Question

Calculate the magnetic field at the centre of a coil of a form of square of side 2a carrying current I.

Answer 1

A coil in the form of a square of side $2a$ carries a curent $I$.

Consider wire $AB$.

distance from $C=a$ [Ref. image 1]

Now, magnetic field due to a finite wire carrying current $I$ at a point is given by (at a distance $r$)

$B=\frac{{\mu }_{0}I}{2\pi r}(\sin \propto +\sin B)$ [Ref. image 2]
So magnetic field at B due to AB:-

$B_1=\frac{{\mu }_{0}I}{4\pi a}(\sin {45}^o+\sin {45}^o)$

$B_1=\frac{{\mu }_{0}I}{4\pi a}\left(\sqrt{2}\right)$ [From geometry, $\propto =\beta ={45}^o$]

Similarly, other wires contribute to same field in same direction given as:-

$B=4B_1=4\times \frac{{\mu }_{o}I}{4\pi a}\left(\sqrt{2}\right)$

$\Rightarrow B=\sqrt{2}\frac{{\mu }_{0}I}{\pi a}$