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Electromag. & Circuitcan't picture

  1. Dec 25, 2004 #1
    We just studied E.M. waves; May I assume that whenever we flip a switch, an E.M. wave is produced?

    -for example, flipping a light switch-

    If a wave is produced, how does it break off? We have a closed circuit. Does it break off & propogate near the Emf source?


    Ps. I am so so confused. I also wondered how the glow of a filament- which is electric energy transforming into thermal energy via collisions between moving electrons and the atoms of the wire- have anything to do with perpendic. electric and magnetic fields propogating through space.

    glow of filament = light = E.M. wave?

    Don't see how collisions of electrons & atoms translate into propogating perpendicular electric and magnetic fields.
     
  2. jcsd
  3. Dec 26, 2004 #2

    Integral

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    The main part of EM fields which you control to turn on and off the lights are contained within the wires so opening and closing the switch enables or disables current flow within the wires.

    As you state the current in the filament of a bulb causes atomic scale collisions, this results in the production of heat. The filament is so hot that it behaves like a blackbody radiator, thus the color produced is an indication of the temperature of the filament.

    At this point to the discussion would have to move to the photon description of light, I will not go there, other then to say a photon is composed of propagating perpendicular electric and magnetics. As soon as you have a changing electric field you have a perpendicular changing magnetic field. From the view point of relativity they are one and the same.
     
  4. Dec 27, 2004 #3
    Thanks for your reply. But I am still not seeing it. The fields propogate within the wire? Okay...yeah, we learned how magnetic fields are produced outside of a wire when current flows. (DC case). So I assume that in the case of AC, the magnetic field would also be changing because the current is changing. Or it would change the instant we flipped the switch. That takes care of the B-field part of the wave.

    Now the electric field is Force/Charge. Hmmm...I have now idea how I would represent the electric field in a circuit diagram. We never discussed how the electric field behaves in circuits. I guess it changes. Okay, so I assume it changes when we flip the switch; and it must be perpendicular to my B-field.

    Are the following assumptions correct?

    There are no E.M. waves breaking off then because we are not dealing with a capacitor? Right? Now is this E.M. wave in my wire related to the glowing filament's photons? Maybe I should research blackbody radiation. The only waves/things propogating through space, when we flip a switch, is those photons?
     
  5. Dec 27, 2004 #4

    Integral

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    The constantly changing current in a AC circuit means that both the magnetic and electric fields are changing. The main work done by a 60hz AC current in a wire is electron motion. Certainly there is a EM field around the wire. The common electrician AC amp meter uses this fact. It consists of a loose clamp around wire which measures the change EM field, this is the same as the current. The electron current creates the atomic scale vibrations in the filament of a light bulb which creates photons in a wide range of energies, mainly in the infrared range but also a significant amount in the visible range. Keep in mind that the electron is always at the center of an electric field, if the electron is vibrating so is the surrounding electric field, a changing electric field produces a changing magnetic field, thus a vibration electron must produce both electric and magnetic fields.
     
  6. Dec 27, 2004 #5

    pervect

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    You will generate a small amount of electromagnetic radiation when you turn on a light. Think of it as a long wire antenna, excited by a pulse. If you knew the rise-time of the pulse, you could do a fourier analysis of it to find the frequency components of it, and then knowing the antenna properties (ideally it would be determined by just the length of the wire if the wire were in , say, an open field somewhere) then you could figure out what sort of energy was radiated at what frequencies.

    If the wire is in a building, a full analysis becomes much more difficult, to the point of being impractical.

    If you consider a simpler case, for instant a coaxial cable which consists of an inner conductor, a dielectric, and an outer conducting layer, all in a nice standard circular geometry, you can see that the field witihin the wire is not the whole story. The speed of transmission of signals down a coaxial cable is constant, but slower than c, and is significantly impacted by the properties of the insulating dielectric. Modeling the cable as a lumped L-R-C circuit one can see why - a high dielectric constant increases the capacitance / unit length, while not affecting the inductance / unit length at all. This slows down the speed of signal propagation.

    Note that the surrounding outer conductor of coaxial cables also prevents them from radiating any electromagnetic energy to the environment - that's another reason they are simpler to analyze than a wire.
     
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