00:01
Hi, here in this given problem you have asked for three quite lengthy different problems.
00:07
So out of them, i'm solving the first one only, the 10th, number 10, in which we have been given two circular loops, current carrying loops, the first one, the inner one, and then outer one.
00:24
The outer one is having a radius r1 carrying a current in clockwise direction i1, the inner one, the smaller loop that is having a radius r2 and carrying a current in counterclockwise direction i2.
00:45
And these values are i1 is equal to 5 .30 ampere, i2, that is equal to 2 .80 a.
00:57
Then radii r1 is equal to that is 12 .0 centimeter and r2 this is 8 .50 centimeter.
01:15
Now in the first part of the problem we have to find net magnetic field at the center.
01:23
So here, magnetic field at the center that will be into the plane of paper, into the page, due to the outer loop.
02:00
As it is carrying current in clockwise direction, so it will become south pole and we know magnetic field enters into the south pole.
02:09
And it will be out of the page due to inner loop as current there is counterclockwise so it will behave like north pole and magnetic field comes out of the north pole so net magnetic field here will be that will be given by b1 minus b2 as two of them are in opposite direction so that will be b1 minus b2 means for b1 this is mu not i 1 by 2 r1 minus mu not i 2 by 2 r2 so taking this mu not by 2 as a common out here it is i 1 minus i 2 by r 1 now plugging in all the known values mu not that is 4 pi into 10 xx per minus 7 tesla meter per amper divided by 2 and here i1 that is 5 .30 ampere divided by r1 which is 12 centimeter or 12 into 10 to minus 2 meter then minus 5 this is 2 .80 and the radius is 8 .50 centimeter or into 10 dash per minus 2 meter.
03:46
Then here we can take this 10 -1 minus 2 also as a common out.
03:52
So out of the bracket it will become 2 pi into 10 -drich to the power minus 5.
04:02
And the first term that will be calculated to be equal to 0 .442 and then it is minus 0 .329...