A coil of Cu wire (radius r,self-inductance L | vedclass

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(A) The self-inductance $L$ of a circular coil is given by the formula $L \propto N^2 r$,where $N$ is the number of turns and $r$ is the radius of the

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$2L$

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(A) The self-inductance $L$ of a circular coil is given by the formula $L \propto N^2 r$,where $N$ is the number of turns and $r$ is the radius of the coil.Initially,$N_1 = 1$ and $r_1 = r$,so $L_1 = L$.Finally,the wire is bent into $N_2 = 2$ turns,each with radius $r_2 = r/2$.Using the ratio formula: $\frac{L_1}{L_2} = \left( \frac{N_1}{N_2} ight)^2 	imes \left( \frac{r_1}{r_2} ight)$.Substituting the values: $\frac{L}{L_2} = \left( \frac{1}{2} ight)^2 	imes \left( \frac{r}{r/2} ight) = \frac{1}{4} 	imes 2 = \frac{1}{2}$.Therefore,$L_2 = 2L$.

$A$ coil of $Cu$ wire (radius $r$,self-inductance $L$) is bent into two concentric turns,each having a radius of $r/2$. What is the new self-inductance?

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