A wire in the form of a square of side a carr | vedclass

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(C) The magnetic field produced by one side of the square (e.g.,side $AB$) at the centre is given by the formula for a finite wire:$B_1 = \frac{\mu_0

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The magnetic field produced by one side of the square (e.g.,side $AB$) at the centre is given by the formula for a finite wire:$B_1 = \frac{\mu_0 i}{4 \pi r} (\sin \phi_1 + \sin \phi_2)$Here,$r = \frac{a}{2}$ and $\phi_1 = \phi_2 = 45^{\circ}$.Substituting these values:$B_1 = \frac{\mu_0 i}{4 \pi (a/2)} (\sin 45^{\circ} + \sin 45^{\circ})$$B_1 = \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:$B = 4 	imes B_1 = 4 	imes \frac{\mu_0 i}{\sqrt{2} \pi a} = \frac{2 \sqrt{2} \mu_0 i}{\pi a}$

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

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