00:01
In this problem on the topic off sources of magnetic fields, we want to find the magnetic field at the center off a semi circle with a given radius.
00:11
Now we're told that the straight wires extremely great distance outward to the left and also carry carry a given current.
00:19
We'll take the original coordinates to be at the center off the same circle.
00:23
The magnetic field at the center of the semi circle is then the vector sum off the magnetic fields from each of the two long wires.
00:30
We'll use the right hand girl and show that these fields point into the page.
00:34
So hence weaken some.
00:35
The magnitude of these fields and the magnetic field for each of the long segments is obtained as follows.
00:44
So we'll call this magnetic field the straight for these straight portions of wire.
00:51
And this is equal to the integral off new, not high over for pie d l cross okay, are over our square.
01:15
So the idea across our had traveled and this weaken right as the integral oh, new not i about for pie multiplied by d l cross victor are over are cute where we replace our hat with bette r r we are resisting magnitude so we can bring out the constants, which is you, not times i or more for pie.
02:03
And we integrate from minus infinity to the center of the semi circle which is at zero.
02:11
And this is d x.
02:16
An infinite has more distance along the x axis times in vector i cross with our which is minus x in the eye direction minus are devious the same misako along the j direction.
02:36
This is all divided by the radius off the semi circle which is x squared plus r squared to the power three over to so this simplifies to minus you not i times are over for pie multiplied by the integral minus infinity to zero off the x over x squared plus r squared to the poetry over to.
03:25
So if we solve this integral, we get this to b minus.
03:29
You're not i times are over for pie along in vector k times x over r squared into x squared plus r squared to the power half and we evaluate this from minus infinity 20 and if we substitute these values in, we get this to simplify to minus knew not i over four pi r that's alone ineffective k...