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Frequency in a wire

  1. Apr 6, 2009 #1
    I am wondering what exactly is happening when you transmit a frequency through a wire at an electron level. How do the electrons make a sine wave for example. I understand you can make an oscillator then what seems to be magic at this point a sine wave is transmitted down the wire. What are the electrons doing to make the frequency. I don't know if the wire is the best place to work for this but that's where I thought of it. Would I need to go back to oscillators to try and figure this out or did I miss something easy or am I over complicating it?
     
  2. jcsd
  3. Apr 6, 2009 #2

    Born2bwire

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    Basically what you are doing is sending an electromagnetic wave down the wire. You excite the wave at the terminals and the wave travels along with electric and magnetic fields that oscillate with the frequency of the signal. The electric field exerts a force on the electrons in the wire, as does the magnetic field, due to the Lorentz force. These forces cause the electrons on the surface of the wire (with a good conductor the fields do not penetrate very deep into the wire, on a perfect conductor they do not penetrate at all) to oscillate locally, creating currents. These currents in turn create sinusoidal electric and magnetic fields. The secondary fields will in turn support the propagation and guiding of the original wave. Since the electric and magnetic fields simply oscillate sinusoidally about a fixed point, the electrons will oscillate as well. So for an AC signal, you are mainly propagating the electromagnetic wave, not the electrons down the wire.
     
    Last edited: Apr 6, 2009
  4. Apr 6, 2009 #3
    Signals are actually sent on two wires; one for the signal and the other for the return current. If you don't have a physical return wire, then the current will find a way back through the ground.

    If you are sending (conducting) a signal through the wire:
    There is a voltage gradient along the wire and a current in the wire. Per convention, electrons move in the opposite direction to the current. The current is dQ/dt, which means that charges are moving across a point in the wire, but not how many and fast individual electrons are moving..The instantaneous current is proportional to the voltage gradient along the wire. If this is in air, the signal travels at the speed of light (1/sqrt(permittivity times permeability of free space)). The ratio of the voltage and current is the impedance, which is probably somewhere between 10 ohms and 377 ohms (impedance of free space). If the signal is terminated in a matching resistive load at the far end, the voltage waveform and the current waveform are in phase. The Poynting vector integral will give both the power flow and direction of the signal. The power, the E field and the H field, are both BETWEEN the wires and not IN the wires.

    Transmitting signal into space:
    The wire is terminated so that there reflections at the ends and standing waves (both current and voltages) on the wire. If the wire circuit is a resonant antenna, the accelerated electrons radiate energy into space..
     
    Last edited: Apr 6, 2009
  5. Apr 7, 2009 #4
    Ok, that makes sense. At least I have a better idea to follow now. Before I didnt really know how to ask the question but I got the answer that I wanted. Thanks.
     
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