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(A) The square loop is divided into two parallel paths between points $A$ and $D$. Path $1$ consists of three sides $(AB, BC, CD)$ and path $2$ consis

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(A) The square loop is divided into two parallel paths between points $A$ and $D$. Path $1$ consists of three sides $(AB, BC, CD)$ and path $2$ consists of one side $(AD)$.Since the paths are in parallel,the potential difference across them is the same. The magnetic field at the center $O$ due to a straight wire segment of length $L$ carrying current $i$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 i}{4\pi d}(\sin 	heta_1 + \sin 	heta_2)$.For a square,the distance from the center to any side is $a/2$. Each side subtends an angle of $45^\circ$ at the center $(	heta_1 = 	heta_2 = 45^\circ)$.The magnetic field due to a single side of length $a$ carrying current $i$ is $B_s = \frac{\mu_0 i}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{\sqrt{2}\pi a}$.In the given circuit,the current $i$ splits into $i_1$ (through $ABC-D$) and $i_2$ (through $AD$). Since the resistance of path $ABC-D$ is $3R/4$ and path $AD$ is $R/4$,the currents are $i_1 = i/4$ and $i_2 = 3i/4$.The magnetic field due to the three sides $AB, BC, CD$ is $B_1 = 3 	imes \frac{\mu_0 i_1}{\sqrt{2}\pi a}$ (inward) and due to side $AD$ is $B_2 = \frac{\mu_0 i_2}{\sqrt{2}\pi a}$ (outward).$B_{	ext{net}} = B_1 - B_2 = \frac{\mu_0}{\sqrt{2}\pi a} (3i_1 - i_2) = \frac{\mu_0}{\sqrt{2}\pi a} (3(i/4) - 3i/4) = 0$.

$A$ wire of resistance $R$ is bent in the form of a square of side $a$ as shown in the figure. Find the magnetic induction at the center of the square $O$ due to the current flowing through it.

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