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

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

$\frac{2 μ_{0} i}{a} (\frac{1}{\sqrt{5}} + \frac{1}{3 \sqrt{13}}) ↑ ↑$

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

answer is A.

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Detailed Solution

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

$r_{2} = \sqrt{\frac{a^{2}}{4} + \frac{9 a^{2}}{16}} = \frac{\sqrt{13} a}{4}$

$\sin θ_{3} = \sin θ_{4} = \frac{a / 2}{\sqrt{13} a / 4} = \frac{2}{\sqrt{13}}; B = B_{1} - B_{2}$

$= \frac{μ_{0}}{4 π} \frac{i / 2}{r_{1}} (\sin θ_{1} + \sin θ_{2}) - \frac{μ_{0}}{4 π} \frac{i / 2}{r_{2}} (\sin θ_{3} + \sin θ_{4})$

$\Rightarrow B = \frac{2 μ_{0} i}{πa} [\frac{1}{\sqrt{5}} - \frac{1}{3 \sqrt{13}}]$

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