00:01
B .dr is equal to mu .0 .i.
00:04
And we're asked to show, so wts means well, to show that b is equal to u .n .i divided by 2 .ir.
00:15
All right, and we're also given that c is x squared plus y squared is equal to r square.
00:22
So we have a circle.
00:24
And then what we do parameterize a circle, we just set x equal to r cosine, and y is equal to r sine t.
00:37
And then, so we want to change everything, so the integral along c of b .d .r, we want to write everything in terms of okay, so we take the derivative of r, which is r prime, and then we get the derivative of r cosine t is negative r s -t and the derivative of r -sign t is r -cosite.
01:03
All right, so now what we have to determine, is big b or the magnetic field? well, we have to think about this for a little bit.
01:11
So what we notice is that the magnetic field is going to be tangent to the curve c.
01:19
So basically what we have is a wire, then we have a plane, then we have circles around this plane.
01:27
And we know that the magnetic field b, just like r prime is going to be tangent to c.
01:34
So that means b is going to be a little.
01:35
The same form as r prime so our so our r prime was negative r sine t and our comma r cosy so since both r prime and b are going to be tangent to the curve c or the circle it's going to be of the same form so b is negative sign t and a comma b cosine right so now we plug everything in so we're going to plug in so now we're going to plug into integral over c of b .r sorry yeah see yeah b .dr and then we're going to rewrite it now in terms of t so it's going to be the integral from 0 to 2 pi of b .r prime all right so b is just negative b sine t comma b cosine and our r prime is going to be r -sign t, sorry, negative r -sign t, comma r -cosy.
02:40
And we take the dot product of that.
02:44
So what we get is br -sign -square -t plus br -cosin -square -t v -t...