Two identical wires A and B,each of length l, | vedclass

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(B) For wire $A$ (circle): The circumference is $l = 2\pi R$,so $R = \frac{l}{2\pi}$. The magnetic field at the center is $B_A = \frac{\mu_0 I}{2R} =

Vedclass · Accepted Answer

$\frac{\pi^2}{8\sqrt{2}}$

Vedclass · Answer

(B) For wire $A$ (circle): The circumference is $l = 2\pi R$,so $R = \frac{l}{2\pi}$. The magnetic field at the center is $B_A = \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2(l/2\pi)} = \frac{\mu_0 I \pi}{l}$.For wire $B$ (square): The perimeter is $l = 4a$,so $a = \frac{l}{4}$. The distance from the center to a side is $d = \frac{a}{2} = \frac{l}{8}$. The magnetic field at the center due to one side is $B_1 = \frac{\mu_0 I}{4\pi d} (\sin 45^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{4\pi (l/8)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{2\mu_0 I}{\pi l} \sqrt{2}$.Since there are $4$ sides,the total field is $B_B = 4 	imes B_1 = \frac{8\sqrt{2}\mu_0 I}{\pi l}$.Calculating the ratio: $\frac{B_A}{B_B} = \frac{\mu_0 I \pi / l}{8\sqrt{2}\mu_0 I / \pi l} = \frac{\pi^2}{8\sqrt{2}}$.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{2r} \odot$	$(A). \frac{\mu_0}{2\pi} \frac{i}{r}(\pi - 2)$	$(A). \frac{\mu_0}{2r} \frac{2i}{r}(\pi + 1) \otimes$
$(B). \frac{\mu_0 i}{2r} \otimes$	$(B). \frac{\mu_0 i}{4\pi} \frac{i}{r}(\pi + 2) \otimes$	$(B). \frac{\mu_0 i}{4r} \frac{2i}{r}(\pi - 1) \otimes$
$(C). \frac{3\mu_0 i}{8r} \otimes$	$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \text{Zero}$
$(D). \frac{3\mu_0 i}{8r} \odot$	$(D). \frac{\mu_0 i}{4r} \odot$	$(D). \text{Infinite}$

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