Inductance

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Question

A coil of Cu wire (radius r, self-inductance L) is bent in two concentric circular turns each having radius $\frac{r}{2}$ . The self-inductance now

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Solution

The self-inductance of a coil is
$\begin{array}{l} L = \frac{μ_{0} π N^{2} r}{2} \\ \frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} \frac{r_{1}}{r_{2}} \\ \frac{L}{L_{2}} = {(\frac{1}{2})}^{2} \frac{r}{r / 2} = \frac{1}{2} \\ L_{2} = 2 L \end{array}$

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