00:01
So in this question, we've got a cylindrical conductor, and it's got a hollow centre, and walls of thickness b minus a.
00:13
So we've got this inner radius is a, and this outer radius is b, and it's a conductor, so it goes along like this.
00:25
It's carrying a current i, which is uniformly spread across the cross section.
00:31
So that means that we've got a uniform current density, and we can work out what that is.
00:37
So if i is the integral of j times 2 pi r, the r, the r from a to b, then what we're going to find, that 2 pi j times b squared minus a squared, is equal to i so that tells us that the current density is i over two pi one over b squared minus a squared and uh the current enclosed up to a radius r it's going to be the integral from a to r of i over two pi one over b squared minus a squared times two pi r d r and this is going to be i times r squared minus a squared over b squared minus a squared so this is the current enclosed by a a circle of radius r so now we can work out the magnitude of the magnetic field so we're told that i uh i is 38 milli amps uh mu nought is 4 pi times 10 to the minus 7 uh 10 meters per amp.
02:17
So the magnetic field, so we can use on pairs law, which tells us that 2 pi r times the magnetic field, so this is the magnetic field integrated around a loop in this direction, is going to be mu -naut times the current enclosed at that radius, which we've already said is mu -naut i r -square minus a squared on b -squared minus a squared.
02:44
Oh, a couple more parameters that we're being told.
02:50
So we're also told that a is 1 .5 centimeters, which is 1 .5 times 10 to the minus 2 meters.
03:01
And b is 4 .38 centimeters.
03:09
So b of r is going to be mu, mu, nought, a squared.
03:24
And this is for a less than r less than b.
03:32
For r greater than b, then the current enclosed is the whole current.
03:40
So this part goes to 1 and we get mu -nought i over 2 pi r.
03:46
And for r less than a, no current isn't closed, so the magnetic field is zero...