I had a question about the veritasium video, at some point they say that the electrons barely move, that confused me since that doesn't match my understanding of amperage, flow of electrons being for example 6.24x10^18 per second.
Anyone care to help me out? Maybe that just isn't much?
Calculate the volume of wire that would contain 6e18 free electrons, assuming 1 electron per atom.
You can intuit it by knowing Avogadro's constant is 6e23.
So 1e-5 mol, 1 mol of (atomic) Copper is 63g, so that's 6e-4g of copper. Density of copper is 9g/cm3 so let's say 1e-4 cm3 = 1e-1 mm3 = 0.1 cubic mm of copper.
So one amp in a copper wire is equivalent to 0.1 cubic mm of that wire's electrons flowing out each second. I'd say that's "barely moving".
Just to validate then, these barely moving electrons will create a field that induces a current on the electrons on the other side of the wire that is 1m away, which would turn on the bulb, that if the lightbulb was an ideal current detector that ignores all other sources of electric fields?
Jabbles|4 years ago
You can intuit it by knowing Avogadro's constant is 6e23.
So 1e-5 mol, 1 mol of (atomic) Copper is 63g, so that's 6e-4g of copper. Density of copper is 9g/cm3 so let's say 1e-4 cm3 = 1e-1 mm3 = 0.1 cubic mm of copper.
So one amp in a copper wire is equivalent to 0.1 cubic mm of that wire's electrons flowing out each second. I'd say that's "barely moving".
(very roughly)
griffinheart|4 years ago
Just to validate then, these barely moving electrons will create a field that induces a current on the electrons on the other side of the wire that is 1m away, which would turn on the bulb, that if the lightbulb was an ideal current detector that ignores all other sources of electric fields?