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प्रश्न
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.
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उत्तर
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•
\[ = \frac{4 \mu_0 i}{\pi a}\frac{1}{4}\left[ \frac{1}{\sqrt{\frac{1}{4} + \frac{1}{16}}} \right] - \frac{4 \mu_0 i}{9\pi a} . \frac{3}{4}\left[ \frac{1}{\sqrt{\frac{1}{4} + \frac{9}{16}}} \right]\]
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