00:01
Let us begin with subpart gate.
00:04
Here, four long copper wires are used to form a square.
00:11
So these copper wires are parallel to each other.
00:15
Let us first draw the given square.
00:18
Here, the square is given in x y axis.
00:25
So the square has sides a equal size a.
00:30
A is given as 0 .20 meters and the cross sign represents into the page, where the magnetic field applied is into the page and the dot sign with a circle on it indicates that the magnetic field is out of the page.
00:51
So let us mark these sides as a, b, c and d and let the central point we denote as o.
01:02
Now, the current flowing to each wire, i is given as 1 .2 amputeous.
01:13
Now, from the figure we can write a -o, which is from a corner to the central point, a -o, which is equal to b -o equals c -o -e -o -equels b -o, equal to a -by -r -2, which can be determined using tritonometry.
01:31
So, a is given as 0 .20 meters divided by root 2, which is equal to 0 .1414 meters.
01:45
Now, the magnetic field at the center due to wire a, due to wire, a is given as b1, which is equal to mu -suro by 4 pi into 2i divided by a, o, the distance, a .o.
02:05
We have muo as 4 pi into 10 power minus 7 divided by 4 pi into 2 into 1 .2 ampiers divided by a .o is calculated as 0 .1414 meters.
02:22
So this will give a value of 1 .7 into 10 power minus 6 tesla x cap along o b.
02:34
So this is the magnetic field at the center due to wire a.
02:42
So the direction of this magnetic field is along od.
02:46
Now let us calculate the magnetic field at the center due to wire v.
02:53
Here we can write b2 equals mu zero by 4 pi into 2 i divided by bo.
03:03
Since a -o equals b -2 is also the same value which will be equal to 1 .7 into 10 power minus 6 tesla but which is along y -axis so y -cap along o -a.
03:23
So the direction of the magnetic field at the center due to wire b is along oa...