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

Answer:

The inductor contains $N = 523962.32$ loops  

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  $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

       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

              $L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

               $L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

             $w = 2 \pi f$

So

            $L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

         $\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of  the formula

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

              $N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

             

 Substituting value

            $N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

              $N = 523962.32$ loops