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A typical value for the electron drift velocity in a copper wire is $10^3\ \mathrm{m\ s^{-1}}$. In the circuit below, the length of the copper wire joining the negative terminal of the batter to the lamp is $0.50\ \mathrm{m}$.

circuit diagram

(i) The switch S is closed. Calculate the time it would take for an electron to move from the negative terminal of the battery to the lamp.

(ii) The lamp lights in a time much less than that calculated in (e)(i). Explain this observation.

In the second part, I can't imagine the situation. I reckon that not all electrons travel with a drift velocity, so the faster ones reach the bulb and make it glow.

But how exactly does this lighting happen? The electron comes in contact with the circuitry and lights up the bulb, or is it because of the electric field?

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A single electron takes some time to move from the battery to the bulb but the lamp lights up faster than that. The reason is due to the fact that it is not the electrons travelling from the battery that light up the lamp when it is first lit, rather due to nearby electrons. The electric field is set up almost instantaneously in the circuit due to movement of electrons from their initial position all over the wire,(at the speed of nearly c, depending on the medium) and the electrons nearer to the lamp pass through the circuit lighting it. So even if the drift velocity of the electrons is so small, the lamp gets lit up.

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  • $\begingroup$ "The electric field is set up almost instantaneously in the circuit due to movement of electrons from their initial position all over the wire,(at the speed of nearly c, depending on the medium) and the electrons nearer to the lamp pass through the circuit lighting it." So does this imply that the nearby electrons pass the electric field to the bulb and thus, light it? $\endgroup$ – model_checker Jan 7 '16 at 13:05
  • $\begingroup$ Not exactly, the electric field created by the battery is set up all over the wire due to the movement of electrons from their initial positions. As a result, the electrons near the bulb are given a push to move through the bulb which lights the bulb. It is not the electrons coming all the way from the battery that light up the bulb rather the electrons near the bulb which light it up. $\endgroup$ – Bruce Lee Jan 7 '16 at 13:10
  • $\begingroup$ Pardon me if I restate your explanation. I'm just trying to fully understand the concept. The nearby electrons are pushed by the electric field created by the battery, which along with their drift velocity allows them reach the bulb and hence light it up. Is that correct? $\endgroup$ – model_checker Jan 7 '16 at 13:16
  • $\begingroup$ The drift velocity of the nearby electrons is created due to the electric field of battery itself. That is how the bulb is lit. $\endgroup$ – Bruce Lee Jan 7 '16 at 13:18
  • $\begingroup$ So then who lights the bulb, the drift velocity or the electric field? $\endgroup$ – model_checker Jan 7 '16 at 13:22

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