00:01
Hi, here in this given problem first of all this is a straight current carrying conductor carrying a current i in downward direction.
00:11
Close to it there is a rectangular current carrying loop like this.
00:20
It is carrying a current i 1 in counter clockwise direction like this, like this.
00:31
Length of this rectangular loop this is given as l, its width that is b and distance of the left side of the loop from the conductor that is given as a.
00:49
The magnitudes of these current i is equal to 2 ampere, i 1 is equal to 4 ampere, length of the rectangular loop 0 .9 meter, width of the coil b is equal to 0 .15 meter and its distance from the straight conductor a is equal to 0 .1 meter.
01:16
In the first part of the problem we have to find magnetic field, magnitude along with the direction at the location of the left side of the rectangular loop.
01:28
So, magnetic field at the location of left side arm of the loop that should be outward, out of the plane of paper and that is using right hand thumb rule and this magnetic field will be given by the expression using biot -einstein law mu naught upon 4 pi into 2 i by the distance a, plugging in the known values 10 raised to power minus 7 for mu naught upon 4 pi 2 times of current i which is 2 ampere divided by a which is 0 .1.
02:44
So, this magnetic field comes out to be equal to 2 i by distance a 4 into 10 to the power minus 6 tesla directed outward, out of the plane of paper perpendicularly and this is the answer for the first part of this given problem here.
03:06
Then in the second part of the problem we have to find net force experienced by this rectangular loop.
03:12
So, as the current i and i 1 are in the same direction for the left side arm of the coil...