0:00
Hello.
00:01
So to solve the drift velocity, we have the formula v -sub -d is equal to 1 over n -a -e, where a, excuse me, where i is equal 10 amps that's given to us, n is the number density of conduction electrons for aluminum, and that is approximately 3 .5 times 10 to the power of 28 electrons per meters cubed.
00:25
A is the cross -sectional area of the wire segment, and e is just your standard 1 .6 times 10 to the negative 19c.
00:34
This is the elementary charge.
00:37
So step one is to calculate the cross -section.
00:39
We know that the cross -section is our a, and that is equal to pi r squared, in this case.
00:45
Segment 1, the top, we'll just write each segment as one instead of...
00:58
We have three segments.
00:59
Segment one, two, and three, the top, middle, and bottom.
01:02
So we'll just label it as one, two, and three.
01:04
Our r is equal to two millimeters, which is equal to two times ten to negative third meters.
01:12
And so our a, we'll call this a1, is equal to pi times 2 .0 times 10 to negative third squared, is equal to 1 .256 times 10 times 10 to the negative fifth meters squared.
01:29
Our second radius, is equal to 1 millimeter, and from now on i'm just going to write them in exponential form, or sorry, scientific notation, and we have a2 is equal to pi times 1 times 10 to negative 3rd, which will give us 3 .142 times 10 to negative 6 meters squared, and then for segment 3, we have a radius of 2 again.
02:00
So we already found the answer to that.
02:02
We have 1 .2 .2.
02:04
256 times 10 to negative 5th meters squared because they share the same radius.
02:09
So now we have to compute v of d for each segment...