00:01
Our question says to use figure 49 as reference, and it wants us to show that for points along the positive x -axis, the axis of the wire, the magnetic field b is zero.
00:11
Okay, so we can go ahead and do that, and we'll indicate this as part a.
00:19
So consider a small element on the rod.
00:24
Small element of change, we'll call this element dl, at a distance r from the point q, as shown in the figure.
00:31
So from the biaz -s -sav -art law, the magnetic field at point q is going to be equal to, well, of course it's b, the magnetic field, is going to be equal to, mu -not over 4 pi, times the integral of the current i times dl, l is that infinitesimal length change that we were talking about, cross vector r divided by r cubed.
01:07
Cubed, okay? and again, r is the distance that we're considering from the point q on the figure.
01:21
And so since the point q is along the length of the wire, then dl and r are parallel.
01:27
So we can tell that dl is, we can say parallel.
01:32
This is the symbol for parallel with r.
01:35
Since that's true, dl cross r is zero, because the cross product of two things that are parallel, is zero.
01:49
Well, you plug zero into that integral, and anything times zero is zero, so therefore the magnetic field b is zero.
01:57
And that is our proof.
01:58
We can box that in as our solution for a.
02:02
Part b says to determine a formula for the field at points along the y -axis, such as point p.
02:11
Okay, so consider the small element d, dx on the rod.
02:18
So we're going to let the distance between the element dl and the point p, b little r.
02:24
So from the right -hand rule, the magnetic field at point p due to the current in the wire is out of the page as shown.
02:29
So we're going to call out of the page the positive k -hat direction.
02:33
So what do we have? we have b.
02:37
We'll indicate this as part b.
02:39
We have the magnetic field b.
02:44
Sorry, the cursor got stuck there.
02:45
Let's rewrite that.
02:47
There we go.
02:49
Is equal to, we'll pull everything out of the integral that's a constant here.
02:52
So we have mu not times i divided by 4 pi.
03:03
And then the integral here is dx.
03:07
That's what we're integrating with respect to times sine theta over little r squared.
03:19
And then this is all going in the out of the page direction or the k had direction.
03:24
So we're going to use some trigonometric identities to simplify this a little bit.
03:27
So from looking at the graph, we have r is equal to y over sine theta.
03:32
Because sine theta is equal to y over r or opposite over hypotenuse...