00:01
So in this problem, so in part a, we know the current i is 5 ampires.
00:12
And in one second, the total charge flow into the lighting bulb will be i times t, which equal 5 columns, right? and part b, so i want to find out the charge density.
00:29
Yeah, hold on, yeah.
00:30
We still need to find out the number of electrons.
00:32
So n is simple equal to q divided by e, where it is the charge of the single electron.
00:38
And this gives you 3 .1 to 5 times 10 to the 19 electrons, right? and part b, i want to find out the charge density, the current density.
00:54
So it's simply j -equal i divided by the cross -section area.
00:59
So because we already have the diameter, right? so we can find out the area.
01:04
So this gives you 1 .51 times 10 to the 6 ampers meter squared.
01:14
And in part c, so basically we want to find out the velocity of the electron inside this wire.
01:24
So that is the drift velocity...