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(A) The magnetic field at the centre of a square loop of side length $b$ carrying current $I$ is given by $B = 4 	imes \frac{\mu_0 I}{4\pi (b/2)} (\s

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$\frac{2{\mu _0}I}{\pi a}$

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(A) The magnetic field at the centre of a square loop of side length $b$ carrying current $I$ is given by $B = 4 	imes \frac{\mu_0 I}{4\pi (b/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{\pi (b/2)} 	imes 2 	imes \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}\mu_0 I}{\pi b}$.From the geometry,the radius of the circumscribed circle is $a$. The distance from the centre to a corner is $a$,and the distance from the centre to the midpoint of a side is $b/2$. In the right-angled triangle formed by the centre,the midpoint of a side,and a corner,we have $\sin 45^{\circ} = \frac{b/2}{a}$.Thus,$b/2 = a \sin 45^{\circ} = a / \sqrt{2}$,which implies $b = \sqrt{2} a$.Substituting $b = \sqrt{2} a$ into the magnetic field formula:$B = \frac{2\sqrt{2}\mu_0 I}{\pi (\sqrt{2} a)} = \frac{2\mu_0 I}{\pi a}$.

$A$ thin circular frame of radius $a$ is made of insulating material. $A$ square loop is constructed within it. If the loop carries current $I$,then the magnetic induction at the geometrical centre $O$ will be:

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