00:01
Here we have a rotating cylinder with a total charge of q on its surface, spread out uniformly, and it's rotating about its long axis.
00:15
And we'd like to find the magnetic field right smack dab in the center of the cylinder, so the origin.
00:24
And the way we're going to do this is we're going to use a equation that was developed using the biote -savart law for the magnetic field a distance x along the axis.
00:40
So think of a single loop of wire that's carrying a current going around it, and we are finding the magnetic field up at observation point p.
00:58
So using the biote -savart law, the magnetic field at point.
01:06
Is given by mu -naut times i times the radius of the loop squared, divided by 2 times x squared plus r squared to the three -half's power.
01:34
Okay, and we can use this formula here because the o point, the origin point, is along the axis, of course.
01:43
And we can use x as our coordinate that measures the distance from the loops along the axis, but the loops themselves are going to be thin strips that i've shown one such thin strip with a thicknessy x that's going to shrink to infinitesimal when we do an integral.
02:07
But we can imagine that cylinder form broken into a series of loops, all of which are contributing to the magnetic field at point o being a different distance from that point, which is evaluated with coordinate x.
02:27
Okay, so what we want to do is imagine what a little differential magnetic field is due to a single loop.
02:37
So we're going to call that db, and it will be equal to, mu -0 over 2 r squared as constants out in front, and then d -i over the distance part.
02:57
And it's that distance x squared, x, i should say, in the function that we are going to have to be working with in our integral.
03:10
Let me see if i can get that x squared to look a little bit nicer.
03:14
Okay, so we'll have to come up with an expression for the differential current.
03:30
And what we mean by that is simply the current that is flowing in that loop due to the rotation.
03:40
It is current, but it's not the conventional idea of current, of powering that with a battery.
03:47
The current is coming from a little bit of differential charge.
03:54
And we'll get into that.
03:58
But d .i.
04:01
Is dq over the period of the circular motion.
04:08
So we can imagine that a small fraction of the entire surface charge is embedded on that little strip that's in blue.
04:19
And the period of the orbit, we can simply relate to the frequency, not the frequency, the angular velocity.
04:31
Through period is 2 pi over omega.
04:36
And it's a little awkward to write an omega versus a w.
04:42
So let me be clear about that.
04:46
And we can further write the dq as sigma times the differential area where sigma is the surface charge density charge per unit.
05:02
Surface area and that little strip has an area da that we can write down fairly quickly in a moment, but we want to put the period in now.
05:24
And i will go ahead and show off to the side what that surface charge density is.
05:30
It's the total charge over the surface area of a cylinder, which is q times the circumference of the enrapping circle times the full length of the cylinder, which is just fully w.
05:53
So that could be helpful at some point.
06:00
Okay, the d area is fairly easy to write down as well.
06:05
You can imagine taking that blue strip and laying it out and it will have a length of 2 pi r, and it will have a width of dx, and d area is length times width, so that's easy to write out as well.
06:30
And we can see that the thing we'll be integrating over is the x -coordinate.
06:38
So kind of putting all this together, yep, it's a lot of pieces to put all together, and it is easy to lose a piece, but let's go ahead what we have so far.
06:58
So db, yeah, we'll try not to lose the r squared.
07:07
And d .i is sigma constant.
07:14
2 pi over omega, another constant.
07:19
D area is 2 pi r.
07:29
And then we are left with the stuff in the integral.
07:40
So that is dx over x squared plus r squared...