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How do electrons start to travel though a wire?

  1. Jun 23, 2011 #1
    In trying to understand electricity I thought of the following: If I set up a wire one light year long, stretching out into space ½ a light year out and ½ a light year back, and connected a series of light bulbs all the way around the wire, and then connected a battery (large enough to handle the resistance obviously), how will the light bulbs light up? I thought of three different scenarios;

    a) All the bulbs light up instantaneously. (in which case I don’t understand anything)

    b) The bulbs on the negative side of the battery start lighting up one by one all the way around taking just over one year to light the last one next to the positive side. (I don’t like this idea either).

    c) Both bulbs on the positive side and on the negative side of the battery light up as electrons are being pushed from the negative and pulled by the positive, with the last bulb to light up being the furthest away from the battery at just over ½ year. (I’m going to go for this one)

    d) I just thought of a forth scenario that nothing happens for 1 light year and then all of them come on slowly, or something.

    Any insight would be helpful. I have a feeling I’m not thinking about this correctly to begin with. Thanks for any help in advance.
     
  2. jcsd
  3. Jun 23, 2011 #2

    Drakkith

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    Hrmm...I don't really know for sure. I thought about it for about 5-10 minutes and wasn't really able to come up with a good guess. I'm just gonna have to go with C.
     
  4. Jun 23, 2011 #3

    Vanadium 50

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    They would never light up.

    The resistance of 18 gauge wire is .0075 ohms/foot, so the resistance of the circuit is 2.3 x 1014 ohms. If I need 100 mV to light a bulb, my "battery" needs a voltage of 23 trillion volts. Of course that's impossible, but even if it were possible, the terminals would arc over.

    Basically, a light year is so long that you can always find a path of less resistance. So you won't be able to put any current in the circuit.
     
  5. Jun 23, 2011 #4
    If you used wires that could handle it the answer would be c) except that the wave might move slower than the speed of light. It depends on the arrangement of the wires.
     
  6. Jun 23, 2011 #5
    So basically what we are saying here is that the electrons closest to the battery are affected first?

    Forget the problem about the resistance; just think of it as an extremely short time period on a normal circuit.
     
  7. Jun 23, 2011 #6

    DaveC426913

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    I think the OP is simply interested in a circuit that is long enough to observe a sequence of events happening on a human scale.

    Say we shorten it dramatically and use high speed cameras to observe the events.

    The OP simply wants to understand if the lights all light up at once or in sequence.
     
  8. Jun 23, 2011 #7
    There is a voltage wave traveling c or slower from the battery/switch out to the end of the wire so sequential lightup is the answer.
     
  9. Jun 23, 2011 #8

    rcgldr

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    The potential between each battery terminal and wire is less than the potential between the terminals of the battery. Depending on the capacitance of the wire, it would seem that the lights would dimly light in sequence in parallel with distance from the battery, then full current would occur when the two potentials meet at the far end of the wire, and the lights would then go to full intensity in sequence in parallel starting at the far end of the wire back towards the battery.
     
  10. Jun 23, 2011 #9
    okay so this voltage wave will travel down the wire (from both +ve and -ve ends of the battery?) and if they meet (closed circuit) then electrons move all at once?

    In other words at the far side of the circuit put a switch. How long before we see a voltage difference and did any electrons next to the battery move even though the switch is open.

    Still other words if it is sequential how do the electrons next to the battery "know" that the circuit is closed and that they can start moving if the voltage hasn't completed the circuit yet?
     
  11. Jun 23, 2011 #10
    I was afraid this might be the case but I still have the problem of the lights lighting before the voltage has gone around the circuit. (and determined that it is closed)
     
  12. Jun 24, 2011 #11

    rcgldr

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    I forgot to mention inductance is also an issue, affecting the rate of potential propagation.

    The time it takes depends on how long it takes for the voltage to propagate thorugh the wire. Capacitance and inductance will be an issue here.

    They don't. The the far end of the wire could be open or closed and the initial reaction near the terminals would be the same.

    The initial voltage will be some fraction (probably 1/2) of the final voltage once the circuit fully completes. This may or may not be enough to cause the lights to get lit up.
     
  13. Jun 24, 2011 #12

    Drakkith

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    I THINK this is kind of how it works. Someone correct me if I'm wrong.

    The voltage source, in this case a battery, applies a force on the electrons, causing their random movements through the conductor to have a net movement away from the voltage source. Since the ones at the end of the wire cannot move further away the force equalizes and nothing happens. When the positive terminal is connected the electrons now have a destination to get to. The ones down towards the negative terminal don't immediately know that they now have a source, but over the course of a short amount of time the charge that had built up opposing their movement disappears and they again start their way towards the positive end. Note that a great many of these electrons aren't travelling away from the negative source. Most still take random paths through the conductor leading to no net direction overall, however as you increase the voltage more and more end up taking a net path towards the positive terminal. In the end you have a steady current through the whole conductor.
     
  14. Jun 24, 2011 #13

    davenn

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    yes exactly.....
    But the big Q Dave is.... do you have an answer ? :) was hoping to see your take on the problem

    I remember this same Q on a forum a very long time ago. Not sure if it ever did get answered satisfactorily.

    now the Voltage isnt travelling/flowing the path around the cct ? it doesnt in any normal cct. Current flows around the cct, voltage is just a potential difference between 2 points in the cct. its that potential difference that is "pushing" the electrons (current) out of the power source.
    pretty sure I have that right ;)

    I dont have an answer for sure, really looking forward to some insightful responses :)

    I suspect there's going to be a lag if the switch is closed immediately. We also have velocity factors of the wires to take into account. ie electron propagation in a wire or other conductor isnt at the speed of light.
    eg... 50 Ohm 1/4 inch coax 55% of c, hi quality 50 Ohm coax ~ 80 - 85% of c
    wire or aluminium dipole antenna ~ 95% of c.

    The other scenario I thought of is .....

    the power source is connected to the loop, the switch is still open, after any time lags taken into account, there is going to be + and - charges present on either side of the switch contacts. as per normal cct. will all lights then light up instantly when the switch is closed ?

    Dave
     
  15. Jun 24, 2011 #14

    rcgldr

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    You end up with the same propagation delays starting from the just closed switch.
     
  16. Jun 24, 2011 #15
    The change in electric field cannot travel faster than the speed of light, and the voltage on a charge depends on the distance the charge(electron) is in the field (integral of the field with respect to distance). When the switch is closed, this is a step change, or as close to as possible, which means that the spectrum of the EM wave generated should be made up of an infinite sum of sinusoids. This will look like a wave front of a new voltage level going through the conductor. If the voltage source is DC, then after the EM wave has looped, the magnetic component of the change in electric field goes away and only the magnetic field of the current is left, and a new static electric field is present.

    Also, think of high speed digital circuits, when a pulse is sent from one circuit to the next, there is a time delay before the information is transferred; there is not an instantaneous transfer of charge and data, and, circuits closer will see the signal before circuits further away which is a practical concern in high speed digital and RF circuit board layouts. If it were possible for all the lights to light up at the same time, high speed circuit designers would have already taken advantage of that effect to solve these issues where the data on the pins do not arrive at the same time due to conductor length differences.
     
    Last edited: Jun 24, 2011
  17. Jun 24, 2011 #16

    Vanadium 50

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    That's a different problem, and we need a different abstraction. In the original problem, you have a system that is big and slow. It doesn't work, for essentially the same reason you can't blow through a mile long straw. If you say, OK, let's scale it down so we have a system that is big and slow, now the abstraction that this is well modeled by DC theory breaks down. You can't think "wire" any more. You have to think "transmission line".
     
  18. Jun 24, 2011 #17
    Not necessarily a different problem, everything is ultimately a transmission line, even if it is in the big and slow scale, you still can apply the lumped circuit models, its just that the effects are so negligible that it looks the same as the "DC theory". The original question is actually asking about instantaneous changes which is very far away from a DC theory.
     
  19. Jun 24, 2011 #18
    when we connect a voltage source to the wire(assuming dc)..the electrons adjacent to the + polarity are attracted to the positive polarity and neutralizes the + charge..,so the atoms adjacent to this becomes positively charged and they attracts the electrons from their neighboring atoms and the process continues ...,similarly electrons of atoms which are adjacent to - polarity are repelled from the negative charge of the voltage source...and the process continues ...so i think lights adjacent to the battery terminals glow first...and in the conduction process electrons do not move with speed of light
     
  20. Jun 24, 2011 #19
    The electrons may not move at the speed of light, but the EM wave they generate when they move from the higher to lower potential will travel at or near the speed of light which is the field acting on the electrons further away from the source.
     
  21. Jun 24, 2011 #20
    It sounds like (in simpler language) we can think of the EM wave as a longitudinal pulse traveling down both sides of the wire (somewhere around c) away from the battery. This initially moves electrons but as the pulse reaches the far side if it can’t continue around the circuit the electrons do not move any further.

    It is almost as though we can treat the wire as a small capacitor that is filling one side while the EM wave is passing. This makes sense in that as the wave passes along the wire there will be a voltage difference in front of it and behind it. This means the electrons are moving without a complete circuit but only as much as the wire can “hold”. Whether this lights up the bulbs in the original thought depends on how long the wire is I suppose.

    Thank you everyone for your great responses so far.
     
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