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

Vedclass · Accepted Answer

$\frac{\sqrt{2} \mu_0 n i}{\pi l}$

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(A) The magnetic field due to a straight wire of length $a$ at a distance $r$ from its center is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For a square of side $a = 2l$,the distance from the center to the side is $r = l/2 = l$. The angles at the center are $	heta_1 = 	heta_2 = 45^\circ$.Thus,the field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi l} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{4 \pi l} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{4 \pi l} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 i}{2 \sqrt{2} \pi l}$.Since there are $4$ sides,the total field for one turn is $B = 4 	imes B_1 = 4 	imes \frac{\mu_0 i}{2 \sqrt{2} \pi l} = \frac{2 \sqrt{2} \mu_0 i}{\pi l}$.For $n$ turns,the total magnetic field is $B_{net} = n 	imes B = \frac{2 \sqrt{2} \mu_0 n i}{\pi l}$.Wait,re-evaluating the geometry: For a square of side $a=2l$,the distance from center to side is $d = l$. The formula $B = \frac{\mu_0 i}{4 \pi d} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 i}{4 \pi l} (\sqrt{2}) = \frac{\sqrt{2} \mu_0 i}{4 \pi l}$.Total field for $4$ sides is $4 	imes \frac{\sqrt{2} \mu_0 i}{4 \pi l} = \frac{\sqrt{2} \mu_0 i}{\pi l}$.For $n$ turns,$B_{net} = \frac{\sqrt{2} \mu_0 n i}{\pi l}$.

The magnetic field at the centre of a coil of $n$ turns,bent in the form of a square of side $2l$,carrying current $i$,is

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