Question

Figure shows a square loop of edge a made of a uniform wire. A current $i$ enters the loop at the point $A$ and leaves it at the point $C$. Find the magnetic field at the point $P$ which is on the perpendicular bisector of $AB$ at a distance $a/4$ from it.

Answer 1

$A_0=\sqrt{\frac{a^2}{16}+\frac{a^2}{4}}=\sqrt{\frac{5a^2}{16}}=\frac{a\sqrt{5}}{4}$

$D_0=\sqrt{{\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{2}\right)}^2}=\sqrt{\frac{9a^2}{16}+\frac{a^2}{4}}=\sqrt{\frac{13a^2}{16}}=\frac{a\sqrt{13}}{4}$

Magnetic field due to AB

$B_{AB}=\frac{{\mu }_0}{4\pi }\times \frac{i}{2\left(a/4\right)}\left(\sin \right(90-i)+\sin (90-\alpha \left)\right)$

$=\frac{{\mu }_0\times 2i}{4\pi a}2\cos \alpha =\frac{{\mu }_0\times 2i}{4\pi a}\times 2\times \frac{\left(a/2\right)}{a\left(\sqrt{5}/4\right)}=\frac{2{\mu }_{0}i}{\pi \sqrt{5}}$

Magnetic field due to $DC$

$B_{DC}=\frac{{\mu }_0}{4\pi }\times \frac{i}{2\left(3a/4\right)}2\sin ({90}^{\circ }-B)$

$=\frac{{\mu }_{0}i\times 4\times 2}{4\pi \times 3a}\cos \beta =\frac{{\mu }_{0}i}{\pi \times 3a}\times \frac{\left(a/2\right)}{\left(\sqrt{13}/4\right)}=\frac{2{\mu }_{0}i}{\pi a3\sqrt{13}}$

The magnetic field due to $AD$ and $BC$ are equal and appropriate hence cancel each other.

Hence, net magnetic field is $\frac{2{\mu }_{0}i}{\pi \sqrt{5}}-\frac{2{\mu }_{0}i}{\pi a3\sqrt{13}}=\frac{2{\mu }_{0}i}{\pi a}[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}]$