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Invisible fence for dogs. In this type of system, a wire is buried under the surface and a current at a given frequency passes through the wire. The dog wears a small unit made of a pickup coil and electronics that delivers a short high-voltage pulse to the dog through a couple of electrodes pressed against its skin. The pulse is not harmful, but it does provide a correction that encourages the dog to keep away. In an invisible fence, the wire carries a 10kHz sinusoidal current with an amplitude of 0.5 Amps. The dog carries a sensoron its collar made as a coil with 150 turns and 30 mm in diameter.If the detection level at the coil inside the collar is set at 200 μV RMS (i.e., the level at which the dog will receive a correction pulse), what is the furthest distance from the wire the dog will "feel" the presence of the fence?

Respuesta :

okpalawalter8

Answer:

The distance is  [tex] s =  2.3 \  m [/tex]

Explanation:

From the question we are told that

   The  frequency of the current is  [tex]f =  10\ kHz  = 10 *10^{3} \  Hz[/tex]

   The magnitude of the  current  is  [tex]I_o  =  0.5\ Amps[/tex]

     The number of turns is  [tex]N =  150\  turns[/tex]

     The  diameter is  [tex]d =  30 mm  = 0.03 \ m[/tex]

      The detection level is  [tex]V_{rms}  =  200 \mu V  =  200 * 10^{-6} \  V[/tex]

Generally the radius is mathematically represented as

      [tex]r =  \frac{d}{2}[/tex]

=>    [tex]r =  \frac{0.03}{2}[/tex]

=>      [tex]r 0.015 \ m [/tex]

Generally the magnetic field  generated by the fence is mathematically represented as

      [tex]B  =  \frac{\mu_o *  N * I}{2 *  \pi * s}[/tex]

Here s is the point where the do will feel the magnetic field

      [tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o =   4\pi * 10^{-7} N/A^2 [/tex]    

   So

         [tex]B  =  \frac{ 4\pi * 10^{-7}  *  150 *  I}{2 * 3.142  * s}[/tex]

Generally the magnetic flux is mathematically represented as

     [tex]\Phi =  B *  \pi * r^2[/tex]

=>   [tex]\Phi =   \frac{ 4\pi * 10^{-7}  *  150 * I}{2 * 3.142  * s} *  \pi * r^2[/tex]

=>   [tex]\Phi  =  \frac{2.121 *10^{-8} *  I }{s}[/tex]

Generally the induced emf is mathematically represented as

      [tex]\epsilon  =- \frac{d \Phi}{dt}[/tex]

=>    [tex]\epsilon  =- \frac{ 2.121 *10^{-8} }{ s } *  \frac{d I}{dt}[/tex]

Generally the angular frequency is mathematically represented as

   [tex]w =  2 \pi f[/tex]

=>        [tex]w =  2 *  10*10^{3} \pi [/tex]

=>       [tex]w =  20000 \pi [/tex]

So the current is mathematically represented as

        [tex]I = I_o sin (wt)[/tex]

=>     [tex]I = I_o sin ( 20000 \pi * t)[/tex]

So

  [tex]\epsilon  =- \frac{ 2.121 *10^{-8} }{ s } *  \frac{d [ I_o sin ( 20000 \pi * t)]}{dt}[/tex]

  [tex]\epsilon  =- \frac{ 2.121 *10^{-8} }{ s } *  I_o * 20000 \pi   [  cos ( 20000 \pi * t) [/tex]  

   [tex]\epsilon  =- \frac{ 2.121 *10^{-8} }{ s } * 0.5 * 20000 \pi   [  cos ( 20000 \pi * t) [/tex]    

    [tex]\epsilon  =- \frac{ 6.6*10^{-4} }{ s } [  cos ( 20000 \pi * t) [/tex]  

Here the  

    [tex]\epsilon_{rms} = \frac{ \frac{ 6.6*10^{-4} }{ s }}{\sqrt{2} }[/tex]

=>     [tex]\epsilon_{rms} = \frac{ 6.6*10^{-4} }{ s\sqrt{2}  }[/tex]

But from the question we are told that  [tex]V_{rms}  =  200 \mu V  =  200 * 10^{-6} \  V[/tex]

So

    [tex]\frac{ 6.6*10^{-4} }{ s\sqrt{2}  } = 200 * 10^{-6}[/tex]

=>  [tex]6.6*10^{-4}= 2.8284 *10^{-4} * s[/tex]

=>  [tex] s =  2.3 \  m [/tex]

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