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(D) The magnetic field due to a straight wire of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin 	heta_1 + \s

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

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(D) The magnetic field due to a straight wire of length $L$ 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 coil of side $a$,the distance from the centre to any side is $d = a/2$.The angles subtended by the corners at the centre are $	heta_1 = 	heta_2 = 45^\circ$ (or $\pi/4$ radians).The magnetic field due to one side is $B_{	ext{side}} = \frac{\mu_0 I}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi a}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2\pi a}(\frac{2}{\sqrt{2}}) = \frac{\sqrt{2}\mu_0 I}{2\pi a} = \frac{\mu_0 I}{\sqrt{2}\pi a}$.Since there are $4$ identical sides,the total magnetic field at the centre is $B_{	ext{total}} = 4 	imes B_{	ext{side}} = 4 	imes \frac{\mu_0 I}{\sqrt{2}\pi a} = \frac{4}{\sqrt{2}} \frac{\mu_0 I}{\pi a} = \frac{2\sqrt{2}\mu_0 I}{\pi a}$.

$A$ square coil of side $a$ carries a current $I$. The magnetic field at the centre of the coil is

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