Question

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with closely packed coils made from 12.0 mm of insulated wire 1.1 mmmm in diameter. The permeability of free space is 4π×10−7T⋅m/A4π×10−7T⋅m/A.

How many loops will the inductor contain?

Answer 1

The inductor contains
`N = 523962.32` loops

Step-by-step explanation:

From the question we are told that

The capacitance of the capacitor is
`C = 286nF = 286 * 10^(-9) \ F`

The resonance frequency is
`f = 18.0 kHz = 18*10^(3) Hz`

The diameter is
`d = 1.1 mm = (1.1 )/(1000) = 0.00011 \ m`

The of the air-core inductor is
`l = 12 \ m`

The permeability of free space is
`\mu_o = 4 \pi *10^(-7) \ T \cdot m/A`

Generally the inductance of this air-core inductor is mathematically represented as

`L = (\mu_o * N^2 \pi d^2)/(4 l)`

This inductance can also be mathematically represented as

`L = (1)/(w^2)`

Where
`w` is the angular speed mathematically given as

`w = 2 \pi f`

So

`L = (1)/(4 \pi ^2 f^2)`

Now equating the both formulas for inductance

`(\mu_o * N^2 \pi d^2)/(4 l) = (1)/(4 \pi ^2 f^2)`

making N the subject of the formula

`N = \sqrt{(1)/((2 \pi f)^2) * (4 * l )/(\mu_o * \pi d^2 C) }`

`N = (1)/(2 \pi f) * (2)/(d) * \sqrt{(l)/(\pi * \mu_o * C) }`

Substituting value

`N = (1)/( 3.142 * 18*10^(3) * 0.00011 ) \sqrt{(12)/( 3.142 * 4 \pi *10^(-7)* 286 *10^(-9)) }`

`N = 523962.32` loops