00:01
All right, so we have a wire with a current density that looks like some, it's called some constant j.
00:07
Knot.
00:07
This is not included in the problem, but you have to include it for the units to work out.
00:11
So some constant j .0 times r, and this is all in the z direction, right? because, you know, the current density has to have units of current per square meter.
00:23
So anyway, we want to know what's the magnetic field of this, you know, wire with this current density both inside and outside the wire.
00:32
So let's use amper as law, which says that the closed loop integral of b .d .l around some ampyrian path is equal to mu not times the current enclosed.
00:45
So inside the wire, the magnetic field is going to be constant in all directions along this empirian path, which we can just take to be a circle of radius like little r inside the wire.
00:57
And the current enclosed there is going to be the.
01:01
The integral from 0 to r of this current density we have.
01:04
So j -not, let's call us r -prime squared, d -r -r -prime, because we're integrating over an area, and also we'll get a 2 -pi here because we're integrating over an angle theta.
01:17
So this is going to give us b, let's just do the integral on the right -hand side.
01:24
So we're going to get 2 -pi -mu -not times j -not times 1 -3rd, r cubed.
01:33
And so if we calculate b by dividing out 2 pi r from both sides of the equation, we're going to get mu not, j, 0 over 3 times r squared.
01:46
That's our magnetic field.
01:48
And then for the case of outside the wire, we'll just have b times 2 pi r is equal to mu not times the total current.
01:58
And the total current is just going to be the integral.
02:01
You know, 2 pi, that the integral over theta becomes 2 pi, from 0 to, let's say, the radius of the wire, let's call it r.
02:11
So, j, not, r prime, squared, dr prime.
02:16
And so this will be 2 pi j0 over 3r cubed.
02:20
And so that means our magnetic field.
02:20
In terms of this current density is going to be mu not, j.
02:20
J .0 over 3r cubed.
02:23
And so that means our magnetic field.
02:25
In terms of this current density is going to be mu not, j0 over 3r times r so that's outside the wire so for r greater than r and then this is for r less than r and you can check because we just substitute this expression back in here you'll get the familiar mu not i over 2 pi r so those are both equivalent next up we're given several examples of a magnetic field we want to know which one of those are actually realistic so one of the conditions we require for magnetic field is that the divergence of the magnetic field is zero and so any any of these choices that does not satisfy this cannot be a real magnetic field and if we look at the very first choice that we're given where this putative magnetic field is like some constant times r in the r hat direction this does not have a vanishing divergence the divergence of this is going to be you know something like 1 over r squared times the partial with respect to r of a .r.
03:40
Cube.
03:41
If we're looking at what the divergence looks like in those different coordinates.
03:50
And so obviously that's not going to be zero.
03:53
And so, well, that's in spherical coordinates.
03:57
If we're doing this in cylindrical coordinates, i'm kind of assuming this is in cylindrical coordinates because we're talking about wires and things like that.
04:04
This divergence would just be a times r hat, which is still not zero, so either way it doesn't work...
