Electrons Moving in Conductors
Most recent answer: 10/22/2007
Q:
We know that electrons are free to move about in a conductor ..they have a drift velocity of 1cm/s , yet when we see any conductor (antenna for eg.), the electron actually does not move along the whole length of the conductor,it just vibrates about its mean position and its enery moves ahead. How is it that the energy the electron produces moves ahead of itself ?
Secondly , in an AC circuit, the voltage changes its polarity. Then how does current flow in its circuit ? If I assume that the electron moves , then the moment the current reverses its polarity , the electron should move in the reverse direction, ie a step forward and a step backward. How does it move at all ? If I assume that the electron does not move and the energy it produces moves , then also, how does it move ? How do we actually get any current ?
- Ria (age 21)
Kolkata,India
- Ria (age 21)
Kolkata,India
A:
Hi Ria,
Fun questions!
Conductors have many, many mobile electrons. For many metals, it is about one electron per atom which is free to move. These electrons have rather large velocities, due to the fact that they fill up energy levels just as the electrons in atoms fill up available energy levels. When an electric field is applied, they are mobile and quickly move (average net motion, on top of their random motion) to cancel out the field, and so a good low-frequency approximation is that electric fields are always zero inside of conductors.
As you say, when a conductor carries a current, the electrons have a net drift velocity which is often quite small. The actual drift velocity depends a lot on the geometry of the conductor, the amount of current flowing, and the density of mobile charge carriers (the drift velocity is proportional to the current, and inversely proportional to the cross-sectional area and the density of mobile charge carriers). As you correctly say, in an AC circuit (with frequencies of 60 Hz here or 50 Hz where you are, or at radio frequencies), the electrons don’t drift very far. But what happens is that a large number of electrons all collectively shift their positions together. When you apply an electric field to a conductor, each electron only has to move a little bit, but all of them move together, and so the net current can be quite high.
Signals propagate along wires at very high speeds. If a wire is perfectly conducting, then the speed of a signal propagating along depends on the insulating material around the wire. This is because the energy transfer is actually in the electric and magnetic fields. Poynting’s vector is E cross B, and is proportional to the energy flow per unit area per unit time, and since the electric field vanishes inside a conductor, no energy flows inside a conductor (!), just outside, in the immediate vicinity. The speed of propagation of signals then depends on the dielectric properties of the insulation, and typically is about 70% of the speed of light. The electrons just flow in the conductor to satisfy the conditions of the electric and magnetic fields obeying Maxwell’s equations on the surface of the conductor.
About AC current -- the time average of the current in an AC circuit should be zero (unless there is some DC offset) -- current flows back and forth constantly, in response to alternating voltages. Many devices are perfectly happy to consume energy provided by AC circuits -- like light bulbs and toasters. Energy is deposited in the resistive elements of these items, independent of which way the current is flowing. The instantaneous power dissipated is V*I where V is the voltage between the light bulb's contacts and I is the instantaneous current. Since V=IR where R is the resistance of the light bulb, the power then is V2/R. To get the average power (wattage) for the light bulb, we just average V2 over time. Even if V is constantly switching sign, the power dissipated remains positive.
This is all a very classical answer to your question. In reality, all electrons are indistinguishable. It is not a fair question to ask about where a single electron goes as it does not retain its identity separate from the other electrons. But the classical picture is sufficient to explain the answers to these questions. One question it does not explain is why the electrons in metals are so mobile, and why they don’t bump into the atoms more, and quantum mechanics is needed for that one.
Tom
Fun questions!
Conductors have many, many mobile electrons. For many metals, it is about one electron per atom which is free to move. These electrons have rather large velocities, due to the fact that they fill up energy levels just as the electrons in atoms fill up available energy levels. When an electric field is applied, they are mobile and quickly move (average net motion, on top of their random motion) to cancel out the field, and so a good low-frequency approximation is that electric fields are always zero inside of conductors.
As you say, when a conductor carries a current, the electrons have a net drift velocity which is often quite small. The actual drift velocity depends a lot on the geometry of the conductor, the amount of current flowing, and the density of mobile charge carriers (the drift velocity is proportional to the current, and inversely proportional to the cross-sectional area and the density of mobile charge carriers). As you correctly say, in an AC circuit (with frequencies of 60 Hz here or 50 Hz where you are, or at radio frequencies), the electrons don’t drift very far. But what happens is that a large number of electrons all collectively shift their positions together. When you apply an electric field to a conductor, each electron only has to move a little bit, but all of them move together, and so the net current can be quite high.
Signals propagate along wires at very high speeds. If a wire is perfectly conducting, then the speed of a signal propagating along depends on the insulating material around the wire. This is because the energy transfer is actually in the electric and magnetic fields. Poynting’s vector is E cross B, and is proportional to the energy flow per unit area per unit time, and since the electric field vanishes inside a conductor, no energy flows inside a conductor (!), just outside, in the immediate vicinity. The speed of propagation of signals then depends on the dielectric properties of the insulation, and typically is about 70% of the speed of light. The electrons just flow in the conductor to satisfy the conditions of the electric and magnetic fields obeying Maxwell’s equations on the surface of the conductor.
About AC current -- the time average of the current in an AC circuit should be zero (unless there is some DC offset) -- current flows back and forth constantly, in response to alternating voltages. Many devices are perfectly happy to consume energy provided by AC circuits -- like light bulbs and toasters. Energy is deposited in the resistive elements of these items, independent of which way the current is flowing. The instantaneous power dissipated is V*I where V is the voltage between the light bulb's contacts and I is the instantaneous current. Since V=IR where R is the resistance of the light bulb, the power then is V2/R. To get the average power (wattage) for the light bulb, we just average V2 over time. Even if V is constantly switching sign, the power dissipated remains positive.
This is all a very classical answer to your question. In reality, all electrons are indistinguishable. It is not a fair question to ask about where a single electron goes as it does not retain its identity separate from the other electrons. But the classical picture is sufficient to explain the answers to these questions. One question it does not explain is why the electrons in metals are so mobile, and why they don’t bump into the atoms more, and quantum mechanics is needed for that one.
Tom
(published on 10/22/2007)