91Ó°ÊÓ

The expression of the resonant frequency is as follows:

$f=\frac{1}{2\mathrm{Ï€}\sqrt{LC}}$

Here, L and C are the inductance and capacitance of the circuit.

After rearranging and substituting values

$\begin{array}{l}L=\frac{1}{4{\mathrm{Ï€}}^2{f}^{2}C}\\ L=\frac{1}{4{\mathrm{Ï€}}^2{(1.0\text{Hz})}^2(1.0\text{F})}=0.0253\text{H}\end{array}$

Therefore, the inductance of the inductor is$0.0253\text{H}$ .

There are two layers of wires so that the number of turns gets doubled. Hence, the total number of turns in the inductor of two layers is as follows:

$N=\frac{2l}{d}$

Here, d is the diameter of the wire and N is the number of turns.

The area of the cross-section of the plastic cylinder is as follows:

$A=\mathrm{Ï€}{\left(\frac{d^{\text{'}}}{2}\right)}^2$

Here,$d^{\text{'}}$is the diameter of the plastic cylinder.

The expression of the inductance is as follows:

$L=\frac{{\mathrm{μ}}_{\text{o}}{N}^{2}A}{l}$

Here, is the number of turns, is the area of cross-section, is the length of the inductor, and is the permeability of free space.

$\begin{array}{l}L=\frac{{\mathrm{μ}}_{\text{o}}{N}^{2}A}{l}\\ L=\frac{{\mathrm{μ}}_{\text{o}}{\left(\frac{2l}{d}\right)}^2\left(\mathrm{π}{\left(\frac{d^{\text{'}}}{2}\right)}^2\right)}{l}\end{array}$

After rearranging the above equation, the length of the inductor as follows:

$\begin{array}{l}l=\frac{L}{{\mathrm{μ}}_{\text{o}}\mathrm{π}}\left(\frac{d^2}{d^{\text{'}2}}\right)\\ l=\frac{\left(0.0253\text{H}\right)}{\left(4\mathrm{π}×{10}^{−7}\text{H}/\text{m}\right)\mathrm{π}}{\left(\frac{\left(0.25\text{mm}\left(\frac{1\text{cm}}{10\text{mm}}\right)\right)}{(4.0\text{cm})}\right)}^2\\ =\frac{\left(0.0253\text{H}\right)}{\left(4\mathrm{π}×{10}^{−7}\text{H}/\text{m}\right)\mathrm{π}}{\left(\frac{(0.025\text{cm})}{(4.0\text{cm})}\right)}^2\\ =0.25\text{m}\end{array}$

Therefore, the length of the inductor is$0.25\text{m}$