00:01
Hi, in the given problem, first of all, there is a rectangular current carrying coil.
00:10
The directions are given as upward in the plane of paper.
00:15
This is y, x, and towards right, this is the x -axis.
00:24
Magnetic field passing through this coil is perpendicular to the plane of paper and coming out of it means simply we can say magnetic field is along positive xxas in the first part of the problem the values given are for magnetic field this is 2 .5 tesla and sorry in this problem the magnetic field is not upward out of the plane of paper actually this magnetic field is two words left means along negative x x x.
01:08
So the magnetic field is having the same magnitude as 2 .5 tesla, but this is to the left.
01:22
Means if we draw the magnetic field lines here this is the magnetic field which is directed along negative xxas and having the magnitude 2 .5 tesla.
01:42
Now as and one more thing the length of this loop is given as 30 centimeter and its width is given as 20 centimeter.
02:00
Now using the expression for magnetic force experienced by a straight current carrying conductor kept in a magnetic field that is given as f is equal to i l cross b cross product of current element with the magnetic field or it will be given by in magnitude it is i lb sine theta so for the upper arm as the direction of current is clockwise this is the direction of current and the magnitude of current and the magnitude of current current that is also given to us and this is just 1 .0 ampere.
03:00
So for the upper arm, the angle between current element which is towards right means along positive x -axis and the magnetic field which is towards left means along negative x -axis.
03:14
So for the upper arm angle will be 180 degree.
03:19
So the force experienced by the upper arm will be i.
03:26
L .b...