00:01
So in this problem, we have two pieces of copper wire.
00:07
So the first piece is 1 .2 meters long, and the other piece is 1 .8 meters long.
00:15
And we also have different diameter on the two pieces.
00:21
So the first piece has diameter, d equal, let me see, 1 .6.
00:31
Yeah.
00:33
So let's see this is d1.
00:34
So it's 1 .6 millimeters.
00:37
And for the second piece, it's 0 .8 millimeters.
00:44
And so in part a, and we also know that there is a current in this wire.
00:53
So the current is 2 .5 mdemean.
00:59
So part a is we want to know the current, we want to know the current inside the 1 .8 piece of the wire.
01:10
So as you can see that these two wires, they're in serious connection, right? so in that sense, the current are the same on its wires.
01:18
So the current is simply 2 .5 mnipers.
01:24
And part b, i want to find out the magnitude of electrofield in the 1 .6 millimeter diameter section.
01:32
So to do that, we need to firstly determine the resistance of this part.
01:37
So the resistance has the expression r1.
01:45
We have expression r1 over a, right? so row is the resistivity of the copper, and the row one is, and l1 is the length of the section, and a is the cross section.
01:58
And the resistivity of copper is, you can find the value in the table, in a textbook, it's 1 .72.
02:09
Times 10 to the negative 8th omometer.
02:15
So we also have the lens and we can find out the cross section...