00:01
High, in the given problem, radius of the cross section of the wire is given as, let it be capital r, and this is 3 .0 millimeter current density which is uniform throughout the cross section of the wire.
00:26
That is given as j is equal to 100 ampere per meter square.
00:32
Now in the first part of the problem we have to find magnetic field at a point which is at a distance of 2 .0 millimeter from the axis of the conductor.
00:45
So obviously this observation point is within the conductor.
00:49
So current passing through an amparian loop passing through this observation point.
01:10
If we make a cross section of this conductor which is having a radius of 3mm then the observation point is at a distance of 2mm so here this is the empirian loop this one this 2 millimeter this is 3 millimeter so the current passing through that current will be given by i -dash is equal to current density multiplied by the area of this ampereal loop means pi r square so applying ampires circuit a log to this amporeal loop we can say b dot d s where b is the magnetic field at this observation point b dot d s is equal to mu no times the current i -d -dash so for this b.
02:22
D s, it will be b into 2 pi r circumference of ampere in loop is equal to mu not times the current i dash which is j into pi r square.
02:36
So canceling this pi with the pi and 1 r will also be cancelled...