An infinite horizontal square grid of side ‘a’ made of uniform material has current i fed into it at node A and this current leaves at B. The semi-infinite wires used for the entry and exit of current into this grid are shown in the figure -2 which is an enlarged version (not drawn to scale) of the grid in the locality of AB when observed along its plane. The feeding wires are parallel to AB and just above the plane of the grid by a small distance $δ (δ \to 0)$ . These feeding wires are thus not visible in figure-1. Consider an Amperian circular loop with AB as axis, midpoint of AB as its centre and R as its radius as shown in figure 2. C is the midpoint of square formed by AB as its side shown in figure -1. Take $R = \frac{a}{2}$

Magnitude of circulation $\int \vec{B} . d \vec{l}$ around the shown Amperian loop if $\vec{B}$ is taken due to grid and feeding wires is x.
Magnitude of circulation $\int \vec{B} . d \vec{l}$ around the shown Amperian loop if $\vec{B}$ is taken due to entire grid excluding the feeding wires is y.
Magnitude of $\vec{B}$ at point C is z. Then

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$x = \frac{μ_{0} i}{2}$

$y = μ_{0} i (\frac{3}{2} - \sqrt{2})$

$y = μ_{0} i (\frac{1}{\sqrt{2}} - \frac{1}{2})$

$z = \frac{x}{π a}$

answer is A, C.

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Detailed Solution

Using super position principle $i_{A B} = \frac{i}{2}$
$x = μ_{0} \frac{i}{2} .....$ (i)
Let’s find circulation due to only feeding wires.
………. __________A B ________ ………
If is equal to circulation due to infinite long wire minus the circulation due to the missing segment.
$μ_{0} i {\frac{μ_{0} i}{4 π \frac{a}{2}} (\sin 45^{0} + \sin 45^{0})} 2 π \frac{a}{2} = μ_{0} i (1 - \frac{1}{\sqrt{2}}) .... (i i) y = | Q u a n t i t y (i) - Q u a n t i t y (i i) | = μ_{0} i (\frac{1}{\sqrt{2}} - \frac{1}{2})$
Magnetic field on the entire Amperean loop is not the same (Symmetry is 2D but not 3D)
So, $z \neq \frac{x}{2 π \frac{a}{2}}$