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(D) The square loop is divided into four segments $AB, BC, CD,$ and $DA$. According to the Biot-Savart law,the magnetic field produced by a finite wir

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Zero

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(D) The square loop is divided into four segments $AB, BC, CD,$ and $DA$. According to the Biot-Savart law,the magnetic field produced by a finite wire segment at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4\pi r} (\sin 	heta_1 + \sin 	heta_2)$. For a square of side $a$,the distance from the center to each side is $r = a/2$. The angles subtended at the center by the ends of each side are $	heta_1 = 	heta_2 = 45^\circ$. Thus,the field due to one side is $B_1 = \frac{\mu_0 I}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{\sqrt{2}\pi a}$. In the given circuit,the current $I$ enters at $A$ and leaves at $C$. The current splits into two parallel paths $ABC$ and $ADC$. By symmetry,the magnetic fields produced by segments $AB$ and $AD$ at the center $O$ are equal in magnitude but opposite in direction,thus cancelling each other. Similarly,the fields from $BC$ and $DC$ cancel each other. Therefore,the net magnetic field at the centre $O$ is zero.

The magnetic field at the centre $O$ of a square loop of side $a$ carrying current $I$ as shown in the figure is:

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