00:01
So in this question, you are giving a non -cylinder, you know, cable, right, that consists of a conduct cylindrical share, which is basically this is a share, right? this is a share.
00:13
It has inner radius a and out radius b.
00:16
And this current come out of this paper page, right? so the current way it go, come out of the page.
00:23
So it will be, let me just try to indicate it.
00:26
I used to dot to indicate, right? there's a currents everywhere, you know.
00:30
And this current has a density that that is actually proportioned to the radius within this share right now, but zero elsewhere.
00:42
So you ask actually find the magnetic flux density b, right? i call it b in these three regions in this region and this region and also outside this region.
00:54
So let's begin with the inner region.
00:58
Let's look at the inner region.
01:00
First, we should notice that because of the symmetry of this syninder, you know that the magnetic field lines, the magnetic field lines where, you know, everywhere, i mean, you just circle, right? it's just concentric circle.
01:15
So it's going to like this.
01:15
Let me try to indicate these lines.
01:20
So it would be something like, roughly would be something like this, i mean, sure, with a different color.
01:26
It's yellow.
01:28
It would be something like this, right? it would be something like this.
01:32
Circle, right? concentry circles.
01:38
And inside, i also should be like this.
01:40
Oh, inside, let me use a different color.
01:43
And let's use this color.
01:47
So inside it would be something like this, if there's any one.
01:50
And then also outside.
01:53
Let me try to do with this here.
01:56
So the magnet field lines must go round, must circulate in circles, right? i run the same thing.
02:04
Because of the symmetry.
02:06
Now you can use the amphosal, actually define the magnet flux.
02:11
That's the, that is the b, right? you know that according to ample's law, according to ample's law, you know that if you do a circle, for example, if you look at a circle like this, on this line, and if you integrate b along that a circle, along that circle.
02:40
And this must be equal to the current enclosed in this circle...