00:01
All right, so in this problem, we have a length of wire of length d here that's carrying a current i.
00:07
And we want to find in part a what the magnetic field will be at point p, which is a distance r away from the middle of the wire here.
00:17
So we'll use bos of r for this.
00:20
So we set that up like b equals mu not i over 4 pi and the integral of dl.
00:30
And i'm going to write this as cross r vector and then over our vector cubed.
00:36
So sometimes you could write this just as a unit vector, but this will be a little easier just working with the full vector here.
00:42
So this is our equation.
00:43
So let's start filling in some of the variables and coordinates we're going to use here.
00:49
So i'll kind of mark in our arbitrary little spot dl here.
00:55
So let's say this is our little dl.
00:57
So that's dl there and that means that this vector here would be vector r so the distance from dl to point p and we'll set up our coordinates here so i'm going to call this middle parts we'll call that y equal zero and so that would put this the top part as negative d over 2 and the bottom as d over 2 i guess you could switch those if you wanted upwards to be positive here but shouldn't really make a difference at all.
01:33
Okay, so next then we can start filling in some of the steel cross -r here.
01:39
So we're going to have mu not i over 4 pi, and then our integral, well, our integral, we now know, should run the full length of this wire.
01:51
So that's going to be from d over 2 to negative d over 2.
01:58
And then i guess i've been calling this the y direction now.
02:03
So instead of dl, we'll say dy now.
02:06
So we have dy cross our r vector.
02:11
And to do this cross product, it'll be easier if we write the r vector in terms of its components, since that can make doing cross products easy since parallel ones will just be zero.
02:23
So it looks like then at this arbitrary point, our r vector, we could just write as whatever y positions it's at, so it has some vertical height.
02:37
So yj hat, and then plus r .i.
02:41
Hat.
02:42
So we're just saying it has this much vertical components and this much horizontal component.
02:51
And then we have this over r cubed.
02:55
So by pythagorean theorem here, r should just be the square root of big r.
03:02
Squared plus y squared.
03:06
So then we'd want this to the cubed.
03:10
So now when we look at this cross product, because we've written it as components, it'll be a little bit easier.
03:15
So d .y is in the j hat direction.
03:19
So then we can see pretty easily, d .y cross y, both them are in the j hat direction...