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

B at P due to AD = $\frac{{\mathrm{\mu }}_0}{4\pi }.\frac{i}{2}.\frac{4}{d^2}.a\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{a}{4}}\right)}^2}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{3a}{4}}\right)}^2}}\right]$along•
$\frac{{\mathrm{\mu }}_{0}i}{4\pi a}\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{a}{4}}\right)}^2}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{3a}{4}}\right)}^2}}\right]$along•

B at P due to AC=$\frac{{\mathrm{\mu }}_0}{4\pi }.\frac{i}{2}.\frac{16}{9a^2}.a.2\left[\frac{\left({\displaystyle \frac{3a}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{3a}{4}}\right)}^2+{\left({\displaystyle \frac{a}{2}}\right)}^2}}\right]$
$=\frac{4{\mathrm{\mu }}_{0}i}{9\pi a}\left[\frac{\left({\displaystyle \frac{3a}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{4}}\right)}^2+{\left({\displaystyle \frac{3a}{2}}\right)}^2}}\right]$along•
B at P due to AB = $\frac{{\mathrm{\mu }}_0}{4\pi }.\frac{i}{2}.\frac{16}{9a^2}.a.2\left[\frac{\left({\displaystyle \frac{3a}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{3a}{4}}\right)}^2+{\left({\displaystyle \frac{a}{2}}\right)}^2}}\right]$along•
B at P due to BC = $\frac{{\mathrm{\mu }}_0}{4\pi }.\frac{i}{2}.\frac{4}{a^2}.a.\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{a}{4}}\right)}^2}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{3a}{4}}\right)}^2}}\right]$ along•
$=\frac{{\mathrm{\mu }}_{0}i}{2\pi a}\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{a}{4}}\right)}^2}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^2+{\left({\displaystyle \frac{3a}{4}}\right)}^2}}\right]\mathrm{along}\otimes$
So, net magnetic field at point P.
$$\begin{gathered} \mathrm{B}=\frac{4{\mathrm{\mu }}_{0}i}{\mathrm{\pi }a}\left[\frac{\left({\displaystyle \frac{\mathrm{a}}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^2+{\left({\displaystyle \frac{\mathrm{a}}{4}}\right)}^2}}\right]-\frac{4{\mathrm{\mu }}_{0}i}{9\mathrm{\pi }a}\left[\frac{\left({\displaystyle \frac{3\mathrm{a}}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^2+{\left({\displaystyle \frac{3\mathrm{a}}{4}}\right)}^2}}\right] \\ =\frac{4{\mathrm{\mu }}_{0}i}{\mathrm{\pi }a}\frac{1}{4}\left[\frac{1}{\sqrt{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{16}}}}\right]-\frac{4{\mathrm{\mu }}_{0}i}{9\mathrm{\pi }a}.\frac{3}{4}\left[\frac{1}{\sqrt{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{9}{16}}}}\right] \end{gathered}$$
$=\frac{4{\mathrm{\mu }}_{0}i}{4\pi a}\left[\frac{4}{\sqrt{5}}\right]-\frac{{\mathrm{\mu }}_{0}i}{3\pi a}\left[\frac{4}{\sqrt{13}}\right]\left[\frac{4}{\sqrt{13}}\right]$ along•
$=\frac{4{\mathrm{\mu }}_{0}i}{2\pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}\right]$ along•
$=\frac{2{\mathrm{\mu }}_{0}i}{\pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}\right]$ along•