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 = \[\frac{\mu_0}{4\pi} . \frac{i}{2} . \frac{4}{d^2} . a\left[ \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{4} \right)^2}} + \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{3a}{4} \right)^2}} \right]\] along•

\[\frac{\mu_0 i}{4\pi a}\left[ \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{4} \right)^2}} + \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{3a}{4} \right)^2}} \right]\] along•

B at P due to AC= \[\frac{\mu_0}{4\pi} . \frac{i}{2} . \frac{16}{9 a^2} . a . 2\left[ \frac{\left( \frac{3a}{4} \right)}{\sqrt{\left( \frac{3a}{4} \right)^2 + \left( \frac{a}{2} \right)^2}} \right]\]

\[= \frac{4 \mu_0 i}{9\pi a}\left[ \frac{\left( \frac{3a}{4} \right)}{\sqrt{\left( \frac{a}{4} \right)^2 + \left( \frac{3a}{2} \right)^2}} \right]\] along•

B at P due to AB = \[\frac{\mu_0}{4\pi} . \frac{i}{2} . \frac{16}{9 a^2} . a . 2\left[ \frac{\left( \frac{3a}{4} \right)}{\sqrt{\left( \frac{3a}{4} \right)^2 + \left( \frac{a}{2} \right)^2}} \right]\] along• 

B at P due to BC = \[\frac{\mu_0}{4\pi} . \frac{i}{2} . \frac{4}{a^2} . a . \left[ \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{4} \right)^2}} + \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{3a}{4} \right)^2}} \right]\]  along•

\[= \frac{\mu_0 i}{2\pi a}\left[ \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{4} \right)^2}} + \frac{\left( \frac{a}{2} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{3a}{4} \right)^2}} \right]\text{ along } \otimes\]

So, net magnetic field at point P.

 

\[B = \frac{4 \mu_0 i}{\pi a}\left[ \frac{\left( \frac{a}{4} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{a}{4} \right)^2}} \right] - \frac{4 \mu_0 i}{9\pi a}\left[ \frac{\left( \frac{3a}{4} \right)}{\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{3a}{4} \right)^2}} \right]\]
\[ = \frac{4 \mu_0 i}{\pi a}\frac{1}{4}\left[ \frac{1}{\sqrt{\frac{1}{4} + \frac{1}{16}}} \right] - \frac{4 \mu_0 i}{9\pi a} . \frac{3}{4}\left[ \frac{1}{\sqrt{\frac{1}{4} + \frac{9}{16}}} \right]\]

\[= \frac{4 \mu_0 i}{4\pi a}\left[ \frac{4}{\sqrt{5}} \right] - \frac{\mu_0 i}{3\pi a}\left[ \frac{4}{\sqrt{13}} \right]\left[ \frac{4}{\sqrt{13}} \right]\]along•

 

\[= \frac{4 \mu_0 i}{2\pi a}\left[ \frac{1}{\sqrt{5}} - \frac{1}{3\sqrt{13}} \right]\]  along•

 

\[= \frac{2 \mu_0 i}{\pi a}\left[ \frac{1}{\sqrt{5}} - \frac{1}{3\sqrt{13}} \right]\] along•