00:01
Okay, we're looking at question 28 .68, where we have this weirdly bent wire that's essentially, it's a third of a semicircle that's bent around, and we have two different.
00:15
There's an inner radius here, which i've labeled r1, that's 20 centimeters, which i've converted to meters, 0 .2 meters.
00:21
And then the outer radius, r2, is 30 centimeters.
00:25
I've converted that to 0 .3 meters.
00:28
And this angle here, is 120 degrees.
00:32
So again that's since one whole circle is 360 degrees, this is one third of a circle.
00:40
And the current here is going along this direction, you can follow it, and the magnitude is 12 amps.
00:46
And we want to figure out what the magnitude and direction of the magnetic field is due to this current loop at point p right here.
00:54
So to do that, we need to use the bos of our law, which i've written up right here.
01:02
And one key thing to notice first is that we have this cross product, the dl, which is in the direction of the current, cross r, which is the vector that points from the point to the section of wire that we're considering.
01:15
So if we look at these two little, this vertical point and then this little end cap over here, when we think about the direction of the current that's going to be pointing straight down, and then r is also going to be pointing straight down.
01:29
So these two vectors are parallel, which means there will be no contribution from the magnetic field from these endpoints, because the same thing is going to apply for this little point over here.
01:40
So the contribution from these two little cap points is going to be zero.
01:44
So we just need to think about the inner part and the outer part.
01:48
So let's just split those up into two parts, and we can just add them together.
01:54
So since there's going to be a factor of mu knot over four pi in both of the, those i'm going to pull that out and just leave it out here for now.
02:03
So we have this vector, me not over 4 pi.
02:06
Now let's think about this inner circle with this radius.
02:10
We have, we're going to use r1 as our radius here.
02:14
We have dl cross r.
02:16
Now this is the dl at every point is going to be tangent to this inner semicircle and r is going to be the radius that's going to be pointing straight to the point of wire here.
02:30
So those two vectors are going to be perpendicular.
02:33
So dl cross r is really just going to be, actually this i should notice that this is over here these are actually just unit vectors.
02:43
So this is indicating just the direction of r.
02:47
This isn't really indicating any sort of magnitude.
02:49
This is just going to tell us the direction of r here.
02:53
And since the direction of r is perpendicular to dl, this dl cross r is just going to turn into regular dl because we dl cross r hat will just be dl times r hat times sine of theta.
03:10
The magnitude of r hat is just one and sine of theta when both vectors are perpendicular will be equal to one because if theta is 90, sign of theta is one.
03:22
So we're going to have this contribution from the the first part, this inner and then we're going to have the same thing from the second wire except now this i the current is now going in the other direction so i'm going to put turn this to a minus sign because the current is now going in a different direction so now at this point we're going to need to integrate this equation so let's think about what here is a constant and what is what needs to be integrated over so mu not and four pi these are constant i is just the magnitude of the current, so that's also a constant.
04:04
R1 and r2 are also just constants.
04:07
We know those.
04:08
Those aren't changing.
04:09
So the only thing that we have to integrate over here in each of these cases is this dl.
04:15
So i'm going to just pull that out.
04:19
So it's clear that that's what we're integrating over.
04:22
I'll write this in one second.
04:26
So we want to be integrating over this.
04:32
Dl because everything else is a constant.
04:36
So let's think about when we integrate over this dl, this is the integral here is really just going to be the length of each piece of wire.
04:45
So the length of this inner piece of wire here, like this inner piece of wire here, like i said, this is just one third of this entire circle...