Question

In recent years it has been possible to buy a 1.0 F capacitor. This is an enormously large amount of capacitance. Suppose you want to build a 1.1 Hz oscillator with a 1.0 F capacitor. You have a spool of 0.25-mm-diameter wire and a 4.0-cm-diameter plastic cylinder.

How long must your inductor be if you wrap it with 2 layers of closely spaced turns?

Answer 1

Answer:

0.8272 m

Explanation:

Given parameters are:

C = 1 F

f = 1.1 Hz

d = 0.25 mm = 0.00025 m

d' = 4 cm = 0.04 m

$f = \frac{1}{2\pi \sqrt {LC}}$ ⇒ $f^2 = \frac{1}{4\pi^2 LC}$

So, $L = \frac{1}{4\pi^2f^2C} = \frac{1}{4\pi^2*1.1^2*1} = 0.0209$ H

$L = \frac{\mu_0N^2A}{l}$

Here, $l = Nd$ ⇒ $N = l/d$

Then L will be simplified to $L = \frac{\mu_0lA}{d^2}$ ⇒ $l = \frac{Ld^2}{\mu_0A}$

So, $l = \frac{Ld^2}{\mu_0\pi r^2}$ where r = d/2 = 0.02 m

Thus,

$l = \frac{0.0209*0.00025^2}{4\pi*10^{-7}*\pi*0.02^2}=0.8272$ m

Answer 2

The length at which the inductor must be if you wrap it with 2 layers of closely spaced turns is;

l = 3.02 × 10^(-7) m

We are given;

Capacitance; C = 1 F

Frequency; f = 1.1 Hz

Diameter of wire; d = 0.25 mm = 0.00025 m

Diameter of plastic cylinder; d' = 4 cm = 0.04 m

Now, formula for frequency here is;

f = 1/(2π√LC)

Where;

L is inductance

C is Capacitance

Making L the subject gives;

L = 1/((2πf)²C)

L = 1/((2π × 1.1)² × 1)

L = 0.0209 H

Now, formula for Inductance is also;

L = (μ_o × N² × A)/l

Where;

μ_o = 4π × 10^(-7) H/m

N is number of ring turns = 2

A is area = π(d'²/4) = π × 0.04²/4 = 0.0004π

L is inductance = 0.0209 H

l is length of inductor

Making l the subject gives;

l = (μ_o × N² × A)/L

Plugging in the relevant values gives;

I = (4π × 10^(-7) × 2² × 0.0004π)/0.0209

l = 3.02 × 10^(-7) m

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