Question

The magnetic field due to a square loop of side L and current I at a point at a distance L from each vertex is $n\frac{{\mu }_0}{4\pi }\frac{I}{L}$. Here n is

Answer 1

ab = bc = cd = da = ap = bp = cp = dp = L
Field at P due to side cb is
$B=\frac{{\mu }_0}{4\pi }\frac{I}{\left(eP\right)}(sin\theta +sin\theta )$
$B=\frac{{\mu }_0}{4\pi }\frac{I}{\frac{\sqrt{3}L}{2}}(\frac{1}{2}+\frac{1}{2})(\theta ={30}^{\circ })$
$B_{net}=Bsin\alpha \times 4$
$Sin\alpha =\frac{oe}{pe}=\frac{L}{2\times \frac{\sqrt{3}}{2}L}=\frac{1}{\sqrt{3}}$
$B_{net}=\frac{{\mu }_0}{4\pi }\frac{2I}{\sqrt{3}L}\times \frac{1}{\sqrt{3}}\times 4=\frac{8}{3}\frac{{\mu }_0}{4\pi }\frac{I}{L}$