Question

Figure (35-E6) 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

B at P due to AD=$\frac{{\mu }_0}{4\pi }.\frac{i}{2}.\frac{4}{a^2}.4.a[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{4}\right)}^2}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{3a}{4}\right)}^2}}]$ along $⨂$

$=\frac{{\mu }_0}{2\pi a}[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{4}\right)}^2}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{3a}{4}\right)}^2}}]$ along $⨂$

B at P due to AC=$\frac{{\mu }_0}{4\pi }.\frac{i}{2}.\frac{16}{9a^2}.a.2\left[\frac{\left(\frac{3a}{4}\right)}{\sqrt{{\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{2}\right)}^2}}\right]$

=$\frac{4{\mu }_{0}i}{9\pi a}\left[\frac{\left(\frac{3a}{4}\right)}{\sqrt{{\left(\frac{3a}{4}\right)}^2+{\left(\frac{a}{2}\right)}^2}}\right]$ along $⨂$

B at P due to AB $=\frac{{\mu }_{0}i}{4\pi }.\frac{i}{2}.\frac{16}{9a^2}.a.2\left[\frac{\left(\frac{a}{4}\right)}{\sqrt{{\left(\frac{a}{4}\right)}^2+{\left(\frac{a}{2}\right)}^2}}\right]$ along $⨂$

B at P due to BC=$\frac{{\mu }_0}{4\pi }.\frac{i}{2}.\frac{4}{9a^2}.a.[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{4}\right)}^2}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{3a}{4}\right)}^2}}]$ along $⨂$

$=\frac{{\mu }_{0}i}{2\pi a}[\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{4}\right)}^2}}+\frac{\left(\frac{a}{2}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{3a}{4}\right)}^2}}]$ along $⨂$

So, net B=$\frac{4{\mu }_{0}i}{\pi a}\left[\frac{\left(\frac{a}{4}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{4}\right)}^2}}\right]-\frac{4{\mu }_{0}i}{\pi a}\left[\frac{\left(\frac{3a}{4}\right)}{\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{3a}{4}\right)}^2}}\right]$

=$\frac{4{\mu }_{0}i}{\pi a}\frac{1}{4}\left[\frac{1}{\sqrt{\frac{1}{4}+\frac{1}{16}}}\right]-\frac{4{\mu }_{0}i}{9\pi a}.\frac{3}{4}\left[\frac{1}{\sqrt{\frac{1}{4}+\frac{1}{16}}}\right]$

$=\frac{4{\mu }_{0}i}{4\pi a}\left[\frac{4}{\sqrt{5}}\right]-\frac{{\mu }_{0}i}{3\pi a}\left[\frac{4}{\sqrt{13}}\right]\left[\frac{4}{\sqrt{13}}\right]$ along $⨂$

=$\frac{4{\mu }_{0}i}{2\pi a}[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{3}}]$ along $⨂$=$\frac{2{\mu }_{0}i}{\pi a}[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{3}}]$ along $⨂$