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(D) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_

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$\sqrt{2} \frac{\mu_0 I}{\pi a}$

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(D) The magnetic field $B$ due to a straight wire of length $L$ 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 loop of side $a$,the distance from the centre to each side is $r = a/2$.The angles subtended by the corners at the centre are $	heta_1 = 45^{\circ}$ and $	heta_2 = 45^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{2 \pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 I}{\sqrt{2} \pi a}$.Since there are $4$ identical sides,the total magnetic field is $B = 4 	imes B_1 = 4 	imes \frac{\mu_0 I}{\sqrt{2} \pi a} = 2\sqrt{2} \frac{\mu_0 I}{\pi a}$.Wait,re-evaluating: $B = 4 	imes \frac{\mu_0 I}{4 \pi (a/2)} (2 \sin 45^{\circ}) = \frac{\mu_0 I}{\pi a} 	imes 2 	imes \frac{1}{\sqrt{2}} = \sqrt{2} \frac{\mu_0 I}{\pi a}$.

The magnetic field due to a square loop of side $a$ carrying a current $I$ at its centre is:

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