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Figure shows a square loop of the 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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\frac{2 μ_{0} i}{a} (\frac{1}{\sqrt{5}} + \frac{1}{3 \sqrt{13}}) Ä

\frac{μ_{0} i}{2a} (\frac{1}{\sqrt{5}} - \frac{1}{2 \sqrt{13}}) ⊙

\frac{2 μ_{0} i}{π a^{2}} (\frac{1}{\sqrt{3}} + \frac{1}{3 \sqrt{14}}) \otimes

\frac{2 μ_{0} i}{π a} (\frac{1}{\sqrt{5}} - \frac{1}{3 \sqrt{13}}) ⊙

answer is A.

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Detailed Solution

$r_{1} = \sqrt{\frac{a^{2}}{16} + \frac{a^{2}}{4}} = \frac{\sqrt{5} a}{4}$

$\sin θ_{1} = \sin θ_{2}$

$= \frac{a/2}{\frac{\sqrt{5} a}{4}} = \frac{2}{\sqrt{5}}$

$r_{2} = \sqrt{\frac{a^{2}}{4} + \frac{{9a}^{2}}{4}} = \frac{\sqrt{13} a}{4}$

$\sin θ_{3} = \sin θ_{4}$

$= \frac{a/2}{\frac{\sqrt{13} a}{4}} = \frac{2}{\sqrt{13}}$

$B = B_{1} - B_{2}$

$= \frac{μ_{0} i /2}{4 π r_{1}} (\sin θ_{1} + \sin θ_{2})$

$= \frac{μ_{0} i /2}{4 π r_{2}} (\sin θ_{3} + \sin θ_{4})$

$B = \frac{2 μ_{0} i}{π a} [\frac{1}{\sqrt{5}} - \frac{1}{3 \sqrt{13}}] ⊙$

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