How do electrons start to travel though a wire?

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SUMMARY

This discussion centers on the behavior of electrons in a long wire circuit when connected to a battery. The consensus is that when a circuit is closed, a voltage wave travels through the wire, causing electrons to move sequentially rather than instantaneously. The resistance of the wire, specifically 18 gauge wire with a resistance of 0.0075 ohms/foot, results in an impractically high voltage requirement to light bulbs at the far end of a one-light-year circuit. The final conclusion is that while the lights may not light up simultaneously, they will illuminate in sequence as the voltage wave propagates through the circuit.

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  • Concept of voltage wave propagation and its effects in long circuits.
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  • #31
hanii said:
hey guys i got a doubt...! do electrons emit or produce EM wave while they are in conduction?? i thought that only magnetic field is associated with them...! (according to amperes law i think)

When a charge accelerates, it produces an EM wave. This is how they create waves with antennas.
 
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  • #33
but in a conductor ...electrons travel with a constant speed called drift speed...they don't accelerate...this is due to resistivity of the material used as conductor
 
  • #34
hanii said:
but in a conductor ...electrons travel with a constant speed called drift speed...they don't accelerate...this is due to resistivity of the material used as conductor

The drift speed depends upon the current and that is not the same all along the wire at the same time..

If you apply a step function to the end of the wire then the nearby electrons will start to move, producing a charge imbalance which will be producing an electric field which will start to move the next electrons in line. All this takes time because 1. The electrons take time to get moving and 2. Time is needed for the electric field change to propagate across each step along the wire (speed of light). The process gives a moving wavefront along the wire. The distant parts of the wire are undisturbed until this wavefront reaches them.

As the wire is long, the electrons will move 'easily' because the potential on their far side starts off at zero. The pulse will travel along the wire until it reaches a discontinuity (say an unconnected end). At that point, the potential at the end will reach that of the pulse and electrons will stop flowing - things will have 'backed up' (in mechanical terms) and no more current can flow from the power supply. This is very much like the effect of charging a capacitor - except that the capacitance is distributed along the length of line and there is also an inductance involved, which has current flowing through it. At the open end this current will stop and that change in current in the inductance will generate a Voltage which will then propagate back to the source (an echo).
All this is a rather naff way of describing some of what goes on in a transmission line. The theory may be too mathematical for some tastes but it predicts accurately what happens in real life transmission lines. They can make open circuits look like short circuits and vice versa.
 
  • #35
There is another point. If the wire is in a loop, the same thing is happening at the negative terminal of the battery - a negative step function sets of from the negative terminal at just under c and at the same time as the positive step starts off from the + terminal. The two waves will meet half way round.
 
  • #36
i want to ask one more doubt...,please spare me...you said that there are two effects traveling opposite to one another...nearly at c...,then what would be their relative speed with respect one another??
 
  • #37
Zero, because they are traveling side by side until they meet at the other end At the same time.
Now ask a difficult one. ;-)
 
  • #38
hanii said:
but in a conductor ...electrons travel with a constant speed called drift speed...they don't accelerate...this is due to resistivity of the material used as conductor

I re-read this and there is one big flaw here. Electrons in a metal are constantly moving in random directions at massive speeds due to the temperature. Their average speed (the drift speed) is only a few mm per second, which is a tiny fraction of their RMS velocity.
 
  • #39
sophiecentaur said:
There is another point. If the wire is in a loop, the same thing is happening at the negative terminal of the battery - a negative step function sets of from the negative terminal at just under c and at the same time as the positive step starts off from the + terminal. The two waves will meet half way round.

Are you sure about this? I don't think that's true. Current/voltage does not travel in both directions at the same time, only to cancel in the middle.

I think your point is mixing together sign convention and voltage level, where you're using both sign conventions at the same time to get two voltages traveling in opposite directions and meeting in the center. You should only use one sign convention at a time, and then you only have one voltage step, instead of 2 from opposite directions.
 
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  • #40
Where did my reply go? Maybe it will show up, but I'll say it again.
I scanned the posts and here is as far as I have gotten with this interesting question.
You all have made me think (correct me if I am wrong) that at first the bulbs close to the source would brighten and then dim as the charge carriers quickly spread out as water from a broken dam. Then as the charge builds the lights would slowly brighten.
 
  • #41
DragonPetter said:
Are you sure about this? I don't think that's true. Current/voltage does not travel in both directions at the same time, only to cancel in the middle.

I think your point is mixing together sign convention and voltage level, where you're using both sign conventions at the same time to get two voltages traveling in opposite directions and meeting in the center. You should only use one sign convention at a time, and then you only have one voltage step, instead of 2 from opposite directions.

I'm not sure what "cancel(ing) in the middle" means.
Current flowing out of one of the connections to the cable will be matched by current flowing into the other. There will be a PD across / between the two wires. If they started off at Zero potential wrt 'Earth' then there WILL be a positive pulse traveling along one and a negative pulse traveling along the other (a balanced signal). If there is an appropriate load at the other end, current will flow through the load and dissipate. If the load is of the wrong impedance, there will be waves reflected back at the end (a short circuit in the case of the earlier models).
 
  • #42
Think of your wire as a tube completely filled with marbles. Think of your battery as a marble pump inserting marbles into the "negative" end of the tube as it retrieves marbles from the "positive" end of the tube. Then you can envision how all of your lights would illuminate simultaneously.
 
  • #43
NightSwimmer said:
Think of your wire as a tube completely filled with marbles. Think of your battery as a marble pump inserting marbles into the "negative" end of the tube as it retrieves marbles from the "positive" end of the tube. Then you can envision how all of your lights would illuminate simultaneously.

Nope. For all the lights to light up at the same time, the marbles would have to move in unison, with zero time for propagation of the wave. That implies an instantaneous signal i.e. faster than light.
 
  • #44
NightSwimmer said:
Think of your wire as a tube completely filled with marbles. Think of your battery as a marble pump inserting marbles into the "negative" end of the tube as it retrieves marbles from the "positive" end of the tube. Then you can envision how all of your lights would illuminate simultaneously.

Even in this analogy, you'd still expect the modulus and density of these marbles to impose a speed at which the impulse would propagate along the tube. Nothing would be instantaneous.
 
  • #45
True enough. "Simultaneously" is an exaggeration.
 

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