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Electron speed in wire in circuit

  1. Feb 25, 2009 #1
    I do not agree with the "answer" below. What do you think about it?
    Question:
    When you turn the key in an automobile, you complete a circuit from the negative terminal through the electric starter and back to the posstive battery terminal. This is a DC circuit and electrons migrate through the circuit in a direction from the negative battery terminal to the posstive terminal. About how long must the key be in the ON position for the electrons starting from the negative terminal to reach the positive terminal?
    a) Atime shorter than that of the human reflex turning a switch on or off.
    b)1/4 seconds
    c/4 seconds
    d)4 minutes
    e)4 hours.
    The answer: is e
    Although the electrical signal travels through the closed circuit at about the speed of light, the actual speed of electron (drift velocity) is much less. Although electrons in the open circuit (key in OFF position) at normal temperatures have an average velocity of some millions of kilometers per hours., they produce no current because they are moving in all possible directions. There is no net flow in any preferred direction. But when the key is turned to the ON position, the circuit is completed and the electric field between that battery terminals is directed through the connecting circuit. It is this electric field that is established in the circuit at about the speed of light.
    The electrons all along the circuit continue their random motion, but are accelerated by the impressed electric field in a direction toward the end of the circuit connected to the positive terminal. The accelerated electrons cannot gain appreciable speeds because of collisions witht eh anchored atoms in their paths. These collions continually interrupt the motion of the electrons so that their net average speed is extremely slow--less than a tiny fraction of a centimeter per second. So some hours are require for the electrons to migrate from one battery terminal through the circuit to the other terminal.
     
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  3. Feb 25, 2009 #2

    mgb_phys

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    Why don't you agree?
     
  4. Feb 25, 2009 #3
    I do not agree because the electron would have so great speed from the electric force. I would move through the wire at any length in very little time. Furthermore, the electron would not have to collide with the atoms along the wire.
     
  5. Feb 25, 2009 #4

    mgb_phys

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    In a 12V circuit an electron has an energy of 12eV, thats 2x10^-18 Joules not a great deal of force.
     
  6. Feb 26, 2009 #5
    Think of it this way.

    If you have a pipe full of water and put a drop of water in the top, a drop instantly comes out of the bottom. A circuit behaves much the same way. The electron in the circuit doesn't have to flow all the way from the battery to the load to turn it on. The copper wire is already full of electrons and all they have to do is propogate through. This is the drift velocity and it is usually pretty slow.
     
  7. Feb 26, 2009 #6
    The electrons DO bump into stuff, like atoms. That's why wires get warm from current.
     
  8. Feb 26, 2009 #7

    alxm

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    Yes. But with a pipe, you can put some dye in it and see how long it take to come out.

    Electrons in a fermi sea are quantum-mechanically indistinguishable. If you can't know the electron coming out is the one you put in, how could you ever really know the velocity? :)
     
  9. Feb 27, 2009 #8
    Can't dye electrons green? By "know", it really depends on what level of inference you are willing to accept.
     
  10. Feb 27, 2009 #9
    Maybe an ammeter? And, of course, a set of micrometers and a meter stick.
     
    Last edited: Feb 27, 2009
  11. Feb 27, 2009 #10
    But according to kinetic energy this small force would provide an enormous velocity (E = 1/2mv^2) to the electron that has mass m= 9. X 10^-31kg.
    Nonetheless, I would interest in in drift current. This may answer my question. So please give some more explanation if you can.
    Other argument, though. the electrons in the conducting wire only on the outer of the wire, not in the inner side of the wire. It would move very freely at this layer, wouldn't it?
    I acknowledge that electron do collide with atoms in the wire, so that the wire get hot so as speak. But some electron would move with out colliding. (A little probability needed).
     
  12. Feb 27, 2009 #11
    Thank you you all for your contribution to my question.
     
  13. Feb 27, 2009 #12
    Wikipedia,
    http://en.wikipedia.org/wiki/Electric_current#The_drift_speed_of_electric_charges

    (so in four hours the drift would be about 4 x 60 x 60 seconds or 14,400 mm, 14 meters)...about right!!!!
     
  14. Feb 27, 2009 #13

    f95toli

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    No, that is not how it works. The problem here is that in order to give a full answer to this question one needs to use some rather messy quantum mechanical calculations (which you can find in more advanced books on solid state physics dealing with many-particle physics), and those calculations will still only be valid for an idealized material (no defects etc).

    The "simple answer" (which still involves some QM) is that in a good conductor the resistivity at room temperature will mostly be due to electron-phonon scattering (a phonon is a quantized lattice vibration) although in a real material there is also contribution from defects (grain boundaries, vacancies, impurities etc). Hence, the fact that there are so many phonons around (i.e. the ion lattice is "vibrating") means that electrons are constantly "colliding" with something.

    It important to realize that you can't think of electrons as small balls that are traveling through the lattice and sometimes just happens to "hit" an ion, it is quite easy to show that a perfectly ordered conductor (no defects) held at zero temperature(no phonons) would have zero resistivity, the reason being that electrons are NOT scattered by a perfect periodic potential.



    Btw- before someone asks- yes I do mean phonons with an "n"; not photons.
    It is a bit unfortunate that the name chosen for quantized lattice vibrations is so similar to name for the quanta of the electromagnetic field.
     
  15. Feb 28, 2009 #14
    Thank you very much for your reply, I am sorry I have apply my introductory physics into something that much more complex. Obviously, I have never had quantum machanic so pardon me. However, it is quite interesting that I now could see how Newtonian physics fail at the quantum level. Thanks all!
     
  16. Mar 20, 2009 #15
    Ok,

    This is kinda related but: in AC, is there any time lag between one end of the circuit changing direction as opposed to the other? (on the order of the speed of light )
     
  17. Mar 21, 2009 #16
    First, review what happens to signals in coaxial cables, such as RG-58 or RG-59. The signals travel at about 66% or 80% the speed of light. Why? Because the signal velocity is given by

    v = sqrt[ 1/(epsilon mu)] where epsilon is permittivity of the dielectric in the cable, and mu is the permeability.

    If there were only air (or vacuum), it would be epsilon zero (8.85 x 10^-12 Farads per meter, and mu zero (4 pi x 10^-7 henrys per meter). Calculate the velocity and you get 2.99 x 10^8 meters per sec. Even if you do not have coaxial transmission lines, the properties of free space limit the propagation of signals, because for example, there are always magnetic fields around a wire carrying current..

    Also calculate SQRT(mu zero/epilon zero). You get 377 ohms, which is the impedance of free space. So why does free space have an impedance?
     
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