A wire in the form of a square of side 'a' ca | vedclass

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(C) The magnetic field due to a straight wire of finite length at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4\pi r} (\sin 	heta_1

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

Vedclass · Answer

(C) The magnetic field due to a straight wire of finite length 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 any side is $r = a/2$.The angles subtended by the ends of each side at the centre are $	heta_1 = 45^\circ$ and $	heta_2 = 45^\circ$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{2\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2\pi a} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 i \sqrt{2}}{2\pi a} = \frac{\mu_0 i}{\sqrt{2} \pi a}$.Since there are $4$ identical sides,the total magnetic field at the centre is $B_{total} = 4 	imes B_1 = 4 	imes \frac{\mu_0 i}{\sqrt{2} \pi a} = \frac{2\sqrt{2} \mu_0 i}{\pi a}$.

$A$ wire in the form of a square of side '$a$' carries a current $i$. Then the magnetic induction at the centre of the square wire is (Magnetic permeability of free space = $\mu_0$)

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