A uniform conducting wire of length 24a and r | vedclass

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Question

(A) The total length of the wire is $L = 24a$. For the equilateral triangle, the number of turns $N_{t} = \frac{L}{3a} = \frac{24a}{3a} = 8$. The area

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The total length of the wire is $L = 24a$. For the equilateral triangle, the number of turns $N_{t} = \frac{L}{3a} = \frac{24a}{3a} = 8$. The area of the equilateral triangle is $A_{t} = \frac{\sqrt{3}}{4}a^{2}$. The magnetic moment is $M_{t} = N_{t} I A_{t} = 8 	imes I 	imes \frac{\sqrt{3}}{4}a^{2} = 2\sqrt{3} I a^{2}$. For the square, the number of turns $N_{s} = \frac{L}{4a} = \frac{24a}{4a} = 6$. The area of the square is $A_{s} = a^{2}$. The magnetic moment is $M_{s} = N_{s} I A_{s} = 6 	imes I 	imes a^{2} = 6 I a^{2}$. The ratio is $\frac{M_{t}}{M_{s}} = \frac{2\sqrt{3} I a^{2}}{6 I a^{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$. Comparing this with $1 : \sqrt{y}$, we get $y = 3$.

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