00:01
Okay, we're going to do a problem where we are given current coming in along the x -axis, taking a turn like this, along a hemisphere, a semicircle, and then coming back out this way.
00:30
And we've got to give the coordinates that we're given, origin right here at the center of that circle, x -axis coming out here and that's a 3d -ish sort of picture and we are given currents on this is going in these directions and we note that out here past the circle in this region where current goes both ways at the same place there's no contribution to a magnetic field out there so we we are trying to find the magnetic field due to segment one and segment two, this straight piece of wire, and this circular piece of wire.
01:35
All right, so the b field, what we're going to do at this point.
01:39
So we're given a point of symmetry, the x -axis coming out perpendicular to the y -z plane, forming a right -handed coordinate system, and we are looking for the components of the magnetic field at x00, and it may have three components in general, all at this point x.
02:23
Okay, so that's the positive x axis, but this is a point described by x.
02:32
All right now one of the things that we can note is this z component it's going to end up being zero.
02:42
We'll show that as part of the work that we do in the problem.
02:47
Now we're going to handle those segments, segment one and segment two separately.
02:53
So let's get a little sketch of that flat straight wire with current going along the z axis from minus a to a okay and positive v in that direction and we can look at a little piece of current distribution at some place called z and it has a length generically called d l that'll end up being d z in a minute and this is the x -axis, and we want to find what the magnetic field due to segment one, the flat straight piece of wire is at this point x.
03:49
Now, at this point, we have two different kinds of things that we can do.
03:53
One, we can note the generic, hey, here's how you find magnetic fields.
04:03
And you probably know this by now.
04:10
I .d .l.
04:14
Cross product r over r cubed.
04:23
And you integrate over all currents.
04:30
So that's the law of bos of r.
04:32
And in doing this integral, we have a lot of little bookkeeping to do.
04:38
We have to plug in all the different.
04:41
Pieces of this integral.
04:44
So before i try and write out the details, i like to make a list of all the pieces.
04:51
So we have a source point at, and that just means there's a little piece of current that is located there at z that i've pointed to, and it's in the positive z or k -hat direction.
05:19
And we have an observer vector that's at x i -hat and we write the vector out so the observer is at x i -hat okay and this vector r that points from the source to the observer is and you can see it's going to have a positive x component and a negative z component to it so that is always observer minus source point that's equal to x i hat minus z k hat okay so that's this vector r we also have d coming up in our integral and that's a vector and that's just a little vector that points in this piece over here that i've circled it's in the direction of the positive z axis and it has length dv and that's just something that points along our source distribution so when we do segment two the hem of the semi -circle we'll get a different result for that.
07:23
All right.
07:24
Now, we've got to continue with pieces of our integral.
07:29
So i've scrolled down to give you a little more room.
07:33
R is going to be the magnitude of this r vector, and that's simply taking the square root of the sum of the squares.
07:48
And i'll list it this way, because that'll fit in well with what we're doing in the rest of the problem.
07:59
Okay.
08:02
Now, there are some other pieces that go into the integral.
08:07
We have to put in a dl cross r.
08:17
We can see in our integral.
08:20
And you can check this by the right hand rule.
08:25
There's going to be a dz.
08:29
Okay, and let me get a pointer out for this.
08:32
It's going to be a dz.
08:32
It's going to be a dz and a k hat.
08:35
K hat cross i hat this term up here gives me a j hat and k hat cross k hat gives me zero so what we end up with here is actually pretty simple we end up with the x and the d z and j hat okay hat okay and that's going to be part of what goes in our integral.
09:09
We also have to note the limits on our integral.
09:18
The upper limit, of course, is a and lower limit is minus a, and we're integrating over z.
09:32
Okay, we have all the pieces at this point to go ahead and put our integral together.
09:40
Again, you may have to get some scraps paper out to do this if you want to be doing these steps on your own or you may want to go back up to this statement or at times you may want to go back to the full picture okay but we're going to move on i'm going to give you what you get when you plug in these pieces to the integral and remember we are doing what is the contribution to the b field due to only segment one of the wire.
10:19
So we're only doing part of the problem at this point.
10:22
We got u .0 .i over 4 pi.
10:30
The integral, remember, is going to be over the z direction, minus a to plus a.
10:37
Our cross product left these things behind.
10:45
And in the denominator, we had an r cubed.
10:49
Okay so we get and sometimes just for shorthand that thing in parentheses is always going to be the same i'll just leave it that way but i'll write it out full blown right now it's just x squared plus z squared the square root was to get the magnitude of that vector and the cube is because we had r cubed all right now i can see i've gotten integral to do and remember that our observation point x is constant with respect to z.
11:35
That just comes out of the integral.
11:37
It isn't even part of it.
11:40
So we can set this integral up separately and parse it out like this.
11:51
This integral for part one.
11:54
I'm going to leave my x with it that's out front.
11:58
It comes out of the integral.
12:00
Minus a plus a.
12:07
Dv and okay so let's talk about the integral for a moment first off i didn't leave out the j hat or the constants we're just going to take the integral this stuff right here and that's what i've done down here in this step when we put it back together i'll put everything with it no problem it's just keeping track of the cluttered this integral there are lots of different ways that you can do this integral by going back to your calculus books.
12:52
This is a physics course, not a calculus course.
12:56
So there's a trig substitution way to do this.
13:00
In fact, i like simply going to, if i need a refresher, a useful tool called wolfram alpha, or you may have mathematica or trig tables available.
13:12
But this is a really quick easy doable integral i'm going to keep the x with it okay we have x z over x squared and again this stuff to the one half power it's always x squared plus z squared in there and i have to evaluate at the limits okay but i treat integrals like this that are pretty straightforward to do as look up kind of things.
13:54
So the problem's long enough without having to reproduce your calculus course or something like that.
14:01
Okay, when i plug the limits in, we get, let's see, 2a, one of those x's cancels out, and x and x squared plus z squared.
14:20
To the one half and therefore the contribution to wire 1 u.
14:33
Not i over 4 pi and then a bunch of stuff that we can clean up later if we need to and the j half okay you've probably had some right -hand rule kind of stuff and if you go look at a straight wire along an axis and you point your thumb around that wire and wrap it around you find that when you're out in front popping out of the screen here of this wire if it were an infinite wire certainly the field would be pointing in the positive or up direction and that's what's indicated by the positive and j hat.
15:43
So i know i'm getting something right.
15:46
I like to check on my results here.
15:48
Magnetic field, the units are carried by things like mu not i, and there's always a whatever's left is a one over r in dimensions or a one over x and we see again that we have that.
16:05
So units, units cancel out and we're left.
16:09
We're left with a one over distance.
16:13
So i know that at least all the things that give me dimensions are working correctly at this point in the problem.
16:21
Now, what we have to do is go on to wire segment two.
16:27
Ok, so we have the b field, the full vector for wire segment one.
16:34
And now wire segment two is coming up.
16:42
And again, there's my.
16:44
Semicircle with current direction given and we have our positive z direction positive y and attempt to make the 3d picture positive x popping out of the screen and if we look at a piece of the current distribution here again a dl it's pointing off tangentially to this semicircle, and we're going to notice that the radius of the circle is a, and to get to a generic point on the x -axis, that's going to be our vector r.
17:43
Okay, and this point is going to be at theta.
17:49
I'm going to use some polar coordinates in a.
17:51
Sense to describe part of this problem.
17:56
So the angle stuff is going to get a little bit cluttery.
18:00
We have already written out the law of b .o...