00:01
Hello and welcome to this video solution of numerate.
00:04
So here it's given that a filamentary conductor is formed into an equilateral triangle.
00:10
So let me just draw this triangle kind of like this right with sides of length l carrying current ie right.
00:26
Now you have to find the magnetic field intensity at the center of the triangle right.
00:31
So what i do is i draw this centroids okay passing through the point o let's say right.
00:46
Now here if the current is somewhat moving like this is a current i.
00:55
Let's say i name this as abc right.
00:59
Now what we have here is the magnetic field produced at o due to any one segment let's say bc right.
01:12
So for that what you need to have is you need to have this distance from o to the point here let's say i have got m here this is small d right and also with that you need to have two angles that is which is subtending it in the other point right.
01:36
So these are the two angles right.
01:37
Now since this is a equilateral triangle these angles are 60 degrees right.
01:43
This is 60 degrees.
01:45
So what we have is we can write the magnetic field b due to bc at the center will be equal to it's it's a mu naught i by 4 pi by 4 pi.
02:12
Here you will be having sine theta 1 plus sine theta 2 right...