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BlackZzzverrR [31]
3 years ago
11

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 clos

ely 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?
Physics
1 answer:
lukranit [14]3 years ago
4 0

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  

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