Example- Current Density
All right, let’s do an example related to the current density. Let’s say the current density across a cylindrical conductor, the current density across a cylindrical conductor of radius big R, varies in magnitude according to J is equal to J0 times 1 minus little r, over big R. Where, little r is the distance from the central axis of the wire. Okay, so, according to this then, at little r is equal to 0, on the axis, in other words, the current density is J0, since this total is going to be 0. And, when little r is equal to big R, we will have 1 over here, 1 minus 1 will make 0. And, therefore, at the surface of this cylindrical conductor, current density is 0. We’d like to calculate the current in terms of, let’s say, the conductor’s cross sectional area.
Okay, now, we have a cylindrical conductor, something like this, and the current is flowing through this conductor, and the current density is changing, the radius of this conductor is big R, and the current density is changing according to this function. Therefore, if we look at the axis of this conductor, along the axis, current density is J0, and as we go away from the axis and approach the surface of the conductor, when little r becomes big R, then the current density is 0. And we’d like to determine the current flowing through this conductor as a function of the cross-sectional area of this conductor. That’s the question.
Now, let’s look at this from the top view. And, I will draw the cross-section in a larger diagram, like this. So, here is the top view, the radius is R and little r represents the distance from this center, from the axis. Now, since current density is changing, and as you recall, current density was defined as current per unit area, and if the current density were constant, therefore, we could have easily calculated the current by taking the dot product of these two vectors. But since J is changing in radial direction, then we will apply the same procedure that we had applied earlier for the variable charge density cases.
Here, since the change is in radial direction, we’re going to choose an incremental ring. In other words, a ring concentric, to this circle with a very, very thin, thickness. And, so the radius of this ring, is little r, and the thickness is DR. And, this ring is so thin, it means that, when we go along the thickness of this ring, we can assume that the current density remains constant.
So, since J is equal to magnitude of the current density, is J0 times 1 minus little r over big R. The DR is so small, such that R plus DR, which is the change that the current density will experience will be so small and neglect that change, and assume that it remains constant through the thickness of this ring, this incremental ring. And, therefore, we can calculate the current passing through the thickness of this incremental current. We can say that, current through the thickness of incremental ring, and let’s call this one as DI, that DI is going to be equal to current density J dotted with the area of this ring, where DA is area of the ring, or surface area vector. That we say that surface area vector of the ring. And we know that the surface area vector is perpendicular to the surface. It we assume that the current is going into the plane, for example, and this is the direction of I, and as well as DI, of course, the surface area vector, DA, is going to be, also, into the plane because it is going to be perpendicular to the surface. Then the angle between these two vectors will going to be 0 degree, so DI is going to be equal to J magnitude DA magnitude times cosine of 0. And, cosine of 0 is just 1.
Okay, now, we will write down these quantities in explicit form, we know that the J is given as J0 time 1 minus little r over big R, and that is constant through the thickness of this ring. And, DA, can be obtained as, by cutting this ring open, as we have seen this earlier in a different example. We will end up with a rectangular strip, the length of this rectangular strip will be equal to the circumference of this ring, which is, 2 Pi R. And, the thickness of this strip will be the thickness of this ring, and that is DR. Therefore, DA is going to be equal to 2 Pi R times DR. So, DI, is going to be equal to, for J, we have J0 times 1 minus little r over big R, and times DA and that is 2 Pi little r times DR. This will be the incremental current passing through this incremental ring surface.
Now, we can apply this same procedure in order to get the incremental current passing through next incremental ring. And, then the next incremental ring, which is concentric to that ring, and then the next one, and so on, so forth. And, we do this throughout the surface of the cross-sectional area of this conductor. So, we calculate all these little DI’s through these incremental rings and then we add them. Addition process is the integration, therefore, if you take the integral of both sides, then we will get the total current passing through the cross-sectional area of this cylindrical conductor.
The boundaries of the integration will be, we will start from the inner-most ring, which will have the radius of 0, and we will go to the outer-most ring, which will have the radius of big R. Okay, let’s go ahead and take this integral, the first term will give us integral from 0 to R of J0 times 2 Pi R DR. And, then the second term will give us minus integral from 0 to big R of J0 times 2 Pi R times R is R squared divided by big R times DR. We can easily take these integrals here, J0, and 2 Pi, constant, we can take it outside of the integral, and as well as here, J0, 2 Pi, and big R. These are constant, we can take it outside of the integral. And, then, we will have J0 time 2 Pi R square over 2 from the first integral, which will be evaluated at 0 and big R minus J0 times 2 Pi over big R times integral of R square is R cube over 3 evaluated at 0 and big R.
Moving on, these 2’s will cancel and if we substitute the boundaries, we will substitute big R for the little r, the first one will give us G0 times Pi times big R square, and 0 will give us just 0 minus, here now we will substitute big R for R cube. Therefore, we will have J0 time 2 Pi over big R times R cube over 3. Again, if we substitute 0 for the little r, we’re going to end up with 0. Here, we can cancel this R with the R cube, leaving R square in the numerator. And, therefore, we will have, since A, the total cross-sectional area of the cylindrical conductor is Pi times the radius square, and then this quantity is nothing but A, and as well as here, we will have Pi R square, and that is also equal to A. Therefore, first term will be J0 times A minus the second term will give us 2 over 3 J0 times A.
So, finally, I will be equal to having common denominator here, 3 J0 A minus 2 J0 A divided by 3 is going to give us 1 over 3 times J0 A. Which is the current flowing through this cylindrical conductor in terms of it’s cross-sectional area and the current density J0.