A current i flows through a square loop of si | vedclass

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(C) The magnetic field due to a finite straight wire of length $a$ at a perpendicular distance $r = a/2$ from its center is given by:$B_1 = \frac{\mu_

Vedclass · Accepted Answer

$\frac{{2\sqrt 2 {\mu _0}i}}{{\pi a}}$

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(C) The magnetic field due to a finite straight wire of length $a$ at a perpendicular distance $r = a/2$ from its center is given by:$B_1 = \frac{\mu_0 i}{4\pi r} (\sin 	heta_1 + \sin 	heta_2)$Here,$	heta_1 = 	heta_2 = 45^\circ$ and $r = a/2$.$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{\mu_0 i \sqrt{2}}{2\pi a} = \frac{\mu_0 i}{\sqrt{2} \pi a}$Since there are $4$ such sides,the total magnetic field at the center is:$B_{net} = 4 	imes B_1 = 4 	imes \frac{\mu_0 i}{\sqrt{2} \pi a} = \frac{2\sqrt{2} \mu_0 i}{\pi a}$

$A$ current $i$ flows through a square loop of side $a$. What is the magnetic field at its center?

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