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(B) Let the side of the equilateral triangle be $a$. The perimeter of the frame is $P = 3a$. Let $n$ be the number of turns per unit length. The total

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increases by a factor of $27$

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(B) Let the side of the equilateral triangle be $a$. The perimeter of the frame is $P = 3a$. Let $n$ be the number of turns per unit length. The total number of turns $N$ is given by $N = n 	imes P = n(3a)$.The area of the equilateral triangle is $A = \frac{\sqrt{3}}{4} a^2$.The self-inductance $L$ of a solenoid-like structure is given by $L = \mu_0 n^2 A l_{eff}$,where $l_{eff}$ is the effective length of the coil. Here,the total length of the wire is proportional to $N 	imes P$,but since $n$ (turns per unit length) is constant,the self-inductance of such a coil is proportional to the area $A$ and the number of turns $N$.Specifically,$L \propto N \cdot A$. Since $N = n(3a) \propto a$ and $A \propto a^2$,we have $L \propto a \cdot a^2 = a^3$.Wait,let's re-evaluate: The self-inductance $L$ of a coil is $L = \frac{\Phi}{I} = \frac{N B A}{I}$. For a coil with $n$ turns per unit length,$B = \mu_0 n I$. Thus $L = N (\mu_0 n I) A / I = \mu_0 n N A$.Substituting $N = n(3a)$ and $A = \frac{\sqrt{3}}{4} a^2$:$L = \mu_0 n (3na) (\frac{\sqrt{3}}{4} a^2) = \mu_0 n^2 \frac{3\sqrt{3}}{4} a^3$.If $a$ is increased by a factor of $3$ $(a' = 3a)$,then $L' \propto (a')^3 = (3a)^3 = 27a^3$.Therefore,the self-inductance increases by a factor of $27$.

$A$ copper wire is wound on a wooden frame,whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of $3$,keeping the number of turns of the coil per unit length of the frame the same,then the self-inductance of the coil:

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