Answer:
The inductor contains [tex]N = 523962.32[/tex] loops
Explanation:
From the question we are told that
The capacitance of the capacitor is [tex]C = 286nF = 286 * 10^{-9} \ F[/tex]
The resonance frequency is [tex]f = 18.0 kHz = 18*10^{3} Hz[/tex]
The diameter is [tex]d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m[/tex]
The of the air-core inductor is [tex]l = 12 \ m[/tex]
The permeability of free space is [tex]\mu_o = 4 \pi *10^{-7} \ T \cdot m/A[/tex]
Generally the inductance of this air-core inductor is mathematically represented as
[tex]L = \frac{\mu_o * N^2 \pi d^2}{4 l}[/tex]
This inductance can also be mathematically represented as
[tex]L = \frac{1}{w^2}[/tex]
Where [tex]w[/tex] is the angular speed mathematically given as
[tex]w = 2 \pi f[/tex]
So
[tex]L = \frac{1}{4 \pi ^2 f^2}[/tex]
Now equating the both formulas for inductance
[tex]\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}[/tex]
making N the subject of the formula
[tex]N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }[/tex]
[tex]N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }[/tex]
Substituting value
[tex]N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }[/tex]
[tex]N = 523962.32[/tex] loops
