Answer:
The inductor contains [tex]N = 523962.32[/tex] loops Â
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]
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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]
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 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 Â
