Question

81. III In recent years it has been possible to buy a $1.0\text{F}$capacitor. This is an enormously large amount of capacitance. Suppose you want to build a $1.0\text{Hz}$oscillator with a $1.0\text{F}$capacitor. You have a spool of 0.25-mm-diameter wire and a $4.0-\text{cm}$-diameter plastic cylinder. How long must your inductor be if you wrap it with 2 layers of closely spaced turns?

Short answer

The length of the inductor is

Step-by-step solution

Step 1. Introducttion

The expression of the resonant frequency is as follows:

$f=\frac{1}{2\pi \sqrt{LC}}$

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

After rearranging and substituting values

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

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

Step 2. Explanation

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:

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

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

The expression of the inductance is as follows:

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

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.

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

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

$\begin{array}{l}l=\frac{L}{{\mu }_{\text{o}}\pi }\left(\frac{d^2}{d^{\text{'}2}}\right)\\ l=\frac{\left(0.0253\text{H}\right)}{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\pi }{\left(\frac{\left(0.25\text{mm}\left(\frac{1\text{cm}}{10\text{mm}}\right)\right)}{(4.0\text{cm})}\right)}^2\\ =\frac{\left(0.0253\text{H}\right)}{\left(4\pi \times {10}^{-7}\text{H}/\text{m}\right)\pi }{\left(\frac{(0.025\text{cm})}{(4.0\text{cm})}\right)}^2\\ =0.25\text{m}\end{array}$

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