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The true speed of Electric and Magnetic force.

  1. Mar 20, 2009 #1
    To all the experts:

    Ok, in this example, I'm going to have an "ideal" circuit with no ohmic losses, etc. just for ease of calculation...

    The circuit is connected to a power source with it's own on/off switch.

    I turn on the power source and........

    1. How fast does the Electric Force transmit around the circuit?
    2. How fast does the Magnetic force generated by the moving current propogate away from the wire?

    I'm imagining that both are moving at either c, or close to that. In a perfect vaccuum, etc.

    These answers will no doubt lead to future questions.
  2. jcsd
  3. Mar 20, 2009 #2
    you need to introduce some realism to this ideal circuit.

    A circular loop where the switch and power supply are along the circle would help.
  4. Mar 20, 2009 #3

    ok, what if we observe a small section of straight wire within a circuit. No resistors, capacitors or anything like that. Just say point a to point b.

    Does that help?
  5. Mar 20, 2009 #4
    In my experience with signals in coax cables, "instantaneous" signals in RG-8 travel about 8 inches per nsec. If the cable had no dielectric, then the signals would travel about 12 inches per nsec (speed of light) . If you have no well defined layout for the wires, the impedance (meaning sqrt(L/C) where L and C are the inductance and capacitance per unit length) is not well defined. The signal will reflect back to the source, and because there is no termination or ohmic losses, you are going to have the signal rerflecting back and forth (oscillating) for ever and for ever, except that the impedance of free space is 377 ohms.....Is anyone out there listening?
  6. Mar 21, 2009 #5
    Most RG-XX coax cables have a velocity factor of 0.659, with the exception of RG-62, having a vf of 0.84.

    The magnetic flux travels at c where the current exists.

    If you have 1000 feet of RG-59, and apply a signal to both conductor and shield, at 3nS there will be a slight magnetic field around the first 1.5 feet, moving outwards at 12 in/nsec, and no magnetic field the rest of the length.

    If you've seen an image of a supersonic aircraft, looking at the compression of the air at the nose of the plane is "kinda" how the magnetic field precedes the electric current. The actual sizes/materials are wildly different though, as the airplane is the current moving at 0.6*c, and shock wave is the magnetic field moving at c.

    This isn't normally a concern when dealing with frequencies most DIY/hobby people can build circuits for, with exception of very high speed logic on solderless breadboards. It is a huge design issue with circuit layout and PC Boards running at speeds where the wavelength of the signal is approaching the length of a trace. Routing a PC Board is a high order art/science so that, for example, all 64 bits arrive at the same time over a distance of 3 inches.

    The actual velocity factor on fiberglass PC Boards is between 0.5 and 0.6.
  7. Mar 22, 2009 #6
    Thank you Bipolar. That's the kinda answer I was looking for. Do you know where I can find these velocity factors for other conductors??

    I believe Copper with no insulator is c right?

  8. Mar 22, 2009 #7


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    The speed of a light in a coaxial cable depends on the dielectric used; not the metal.
    But yes, if you were to use a cable with air as a dielectric (air is very similar to vacuum in this case) the speed of light will of course be c. The dielectric in most cables is made from some sort of plastic, e.g. PTFE.
  9. Mar 22, 2009 #8
    Remember, the math and reality of it NOT the same, just for visualization.

    The air compressed in front of the plane would be the magnetic field in front of the current, since the magnetic field is traveling at c, and the current is traveling at some fraction of c, usually between 0.5 and 0.8. Two conductor cables are simulated by capacitance per foot and inductance per foot to approximate reality.

    The actual equations for velocity factor are relatively straightforward, if you have the wavelength and measured parameters that aren't as easily obtained regarding the dielectric. This is further complicated by neighboring conductors, creating an electrical network at high frequencies.

    DC current when turning on is a "high frequency" in the form of the leading edge of a square wave.

    There are simplified "single wire" formulas on http://www.jenving.se/tables.htm#Formulas [Broken].

    Again, with most circuits, the vf doesn't change the rest of the theory.
    Last edited by a moderator: May 4, 2017
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