00:01
In our question we are given a wire of circular cross -section that carries a current density which is not uniform but varies with distance from the center as given by the equation.
00:10
Now it varies for radius r in the range of 0 to r.
00:16
Here, a is a constant with its units in amper per square meter and the radius of the wire is of the magnitude 0 .42 millimeters.
00:27
We first need to determine that if the total current in the circle of the wire, is of the magnitude 6 .5 ampere what is the constant a in amper square meter? we have been given the equation for our current density.
00:45
Over here we have current equal to double integration of current density per unit area.
00:55
So, treating the upper and lower limit we have 0 to 2 5 0 to r and the equation.
01:13
Performing the integration and solving our above equation we get current i equal to 2 pi a r square by 2 minus r q 1 3 r from the limit 0 to r applying the upper and lower limit we have 2 pi a r square by 2 minus r square by for the solving we get current i equal to pi a r square by 3.
01:55
Rearranging the equation in order to get the constant a we have a equal to 3 i 1 by r square substituting the value of current given to us as constant a equal to 3 multiplied by current being 6 .5 aubier divided by 5 into 0 .42 into 10 to the power minus 3 meters the whole square solving which we get constant a equal to 3 .5 2 x0 to 10 to the power 7 ampere per square meter.
02:38
Now we also have to determine the expression for magnetic field outside of the wire which is in the region r being greater than radius.
02:52
Using ampier's law we have 2 pi r0 b equal to u .0 i.
03:05
Rearranging we have the magnetic field with respect to r being equal to mu not i on 2 pi r nod theta cap this is for the region r being greater than radius therefore the magnetic field outside the cylinder will be expressed in the form as determined by us now let us determine that for what what value of r in meters is the current that is enclosed maximum? and the current enclosed can be given as i i denoting current of r equal to 2 pi a r square by 2 minus r q upon 3r.
04:07
For current being maximum, d of i by d r should be 0...