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**Example- Variable current density**

Now, let’s consider a cylindrical wire with a variable current density. Consider a cylindrical wire with radius r and variable current density given by j is equal to j zero times 1 minus r over big R. And we’d like to determine the magnetic field of such a current inside and outside of this cylindrical wire. So we have a cylindrical wire, let’s draw this in an exaggerated way, with radius r and carrying the current i, let’s say in upward direction. Current density is not constant, but it is is varying with the radial distance, little r, according to this function. In other words, if we look at this function over here, we see that the current density is j zero along the axis of the wire. In other words, when little r is zero, then the current density is constant and it is equal to j zero through the cross sectional area of this wire. And whenever little r becomes equal to big R, in other words, at the surface of the wire, then the current density becomes zero because in that case 1 minus 1 will be zero. So along the surface of this wire, current density is zero and we have a maximum current density along the axis of the wire and it is changing with the radial distance.

We’d like to calculate the magnetic field first for a region such that our point of interest is inside of the cylinder. In other words, b is question mark for points such that their location is inside of the wire. In other words, little r is smaller than the big R. To be able to calculate this, first let’s consider again, the top view of this wire. When we look at the wire from the top view, we will see that the current i is coming out of plane and and if you choose a point over here, it’s location, relative to the center, is given with little r and the radius is big R. Like in the similar type of geometries earlier, we’re going to choose an emperial loop in order to calculate the magnetic field at this point such that the loop coincides with the field line passing through this point. And if we apply right hand rule, holding the thumb in the direction of the flow of current, which is coming out of plane, and the corresponding magnetic field lines will be in the form of concentric circles, and circling right hand fingers about the thumb, we will see that field lines will be circling in the counterclockwise direction.

So at the point of interest, we’re going to have a magnetic field line in the form of a circle. Something like this. Going in counterclockwise direction. And we’re going to choose an emperial loop which coincides with this field line. Such a choice will make the angle between the magnetic field line, which will be tangent to this. A magnetic field which will be tangent to this field line and every point along the loop. And d l, which is also going to be in the same direction for this case, incremental displacment vector, along the loop, and the angle between them will always be zero degree. And also furthermore, since the magnetic field is tangent to the field line and we are always along the same field line, the magnetic field magnitude will be constant always along this loop c. And let’s call this loop as c one for the interior region.

Ampere’s law says that b dot dl, over this closed loop c that we choose, should be mMu zero times i enclosed. If we write down the left hand side in explicit form, that will be b magnitude, dl magnitude times cosine of the angle between these two vectors which is zero degree, integrated over loop c one will be equal to Mu zero times i enclosed. Since cosine of zero is one and the magnitude of the magnetic field is constant over this loop, we can take it outside of the integral. Then we end up with b times integral of dl over loop c one is equal to Mu zero times i enclosed. Integral of dl over loop c one means that the magnitude of these displacement vectors are added to one another along this whole loop and if you do that, of course, eventually we’re gonna end up with the length of this loop, in other words, the circumference of this circle. And that’s going to give us 2 Pi r, so b times 2 Pi r will be equal to Mu zero times i enclosed.

Well, if the current density were constant, to be able to calculate the i enclosed, which is the net current flowing through the area surrounded by this loop, we are going to just take the product of the current density with the area of the region that we’re interested with. That direct product will have given us the net current flowing through this shaded region, but since the current density is changing, we cannot do that.

Current density is changing as a function of radial distance little r. In other words, as we go from the center, current density takes different values. Since the change is as a function of this radia distance little r, we can assume that the whole surface consists of incremental rings with very small thicknesses. In other words, we can choose an incremental ring, something like this, with very small thickness. Let’s say the radius of this ring is s, therefore its thickness is ds and that thickness is so small, that as we go along this radial distance, along the thickness of this ring, the change in current density can be taken as negligible. In other words, this r change is so small such that the whole function for such a small change can be taken as constant. In that case, we can calculate the net current flowing through the area surrounded by the incremental ring surface. And if we call that current as d i, once we calculate that current, then we can go ahead and calculate the current flowing through the surface of the next incremental ring. And then do the same procedure for the next one. And so on and so forth. Once we get all those incremental current values, if we add them, for this region, then we can get the total current flowing through this region of interest.

Okay then. We can say that d i is going to be equal to current density j totaled with the area vector of this incremental ring surface and let’s called that one as d a. Well, since current is flowing out of plane, therefore the current density vector is also pointing out of plane. In the meantime, the area vector of this ring is perpendicular to the surface area of the ring. That too will be pointing out of plane there. Then the angle between these two vectors will be just zero degree. So that product will give us j times d a times cosine of zero. Cosine of zero is just one and the explicit for of j is j zero times 1 minus the radial distance divided by the radius of this disk. Now here, we will change r variable to s, therefore the current density function is going to be as j zero times 1 minus s over R.

For the d a, in other words, the surface area of this incremental ring, if we just cut that ring open, it will look like a rectangular strip. Something like this. The length of this strip will be equal to the circumference of that ring and that is 2 Pi s. The thickness is going to be equal to d s. For such a rectangular strip, we can easily express the area, d a, which is going to be equal to length times 2 Pi s times the thickness, which is the s. Therefore, the explicit form of d i, the incremental current is going to be equal to j zero times 1 minus s over R times 2 Pi s d s. So, this is going to give us the incremental current flowing through the surface of an incremental strip or the incremental ring and applying the same procedure, we can calculate the next d i and so on and so forth. If we add all these d i’s to one another, this addition process is integration, then we’re going to end up with the enclosed current flowing through the area surrounded by this closed loop c one.

All right then, moving on. I enclosed is going to be equal to, here 2 Pi and j zero are constant, we can take it outside of the integral. 2 Pi j zero. Writing this integral in explicit form, we will have integral, the first term is going to give us s d s and then for the second one, we will have integral of s squared over R d s. If we look at the boundaries of the integral, we’re going to be adding these incremental rings up to the region of interest. In other words, the corresponding radii of these rings will start from the innermost ring with a radius of zero and will go all the way up to the outermost ring in this region, therefore up to little r. So s is going to vary from zero to little r in both of these integrals. i enclosed therefore will be equal to 2 Pi j zero. The first integral is going to give us s squared over 2 evaluated at zero and r. Here big R is constant, we can take it outside of the integral and the integral of s squared will give us s cubed over 3, so from there we will have s cubed over 3 r, which also be evaluated at zero and little r.

Substituting the boundaries, i enclosed will be equal to 2 Pi j zero times, if you substitute r for s squared, we will have r squared in the numerator, and divided by 2, zero will give us just zero, minus, now we will substitute r for s here, so we will end up with r cubed divided by 3 r. Again, when we substitute zero for s, that’s going to give us just zero. Here, we can express the quantities inside of the bracket in r squared common parantheses, then i enclosed becomes equal to 2 Pi r squared times j zero and inside of the bracket we will have one half minus the little r divided by 3 big R.

Okay. Since we calculated the i enclosed, going back to the Ampere’s law on the left hand side, we had b times 2 Pi r and on the right hand side, we will have Mu zero times i enclosed. Mu zero times, and the explicit form of i enclosed is 2 Pi r squared times j zero and multiplied by one half minus r over 3 r. Here we can divide both sides by little r, therefore eliminating this r and r squared on the right hand side. Solving for b, we can cancel 2 Pi on both sides also, we end up with magnetic field magnitude is equal to Mu zero times j zero times one half minus little r over 3 R times little r. Therefore, the magnitude of the magnetic field, for this current carrying cylindrical wire, r distance away from the center, is going to be equal to this quantity.

Well, if we look at the second region, which is the outside of the wire, with this variable current density, and if we re-draw the picture over here from the top view, here’s the radius of the wire. Now our point of interest is outside of the wire. A point somewhere around here, let us say. Applying the same procedure, since the current is coming out of plane, it will generate a magnetic field line in the form of a circle rotating in counterclockwise direction about this wire. The magnetic field will be tangent to this field line everywhere along this field line. So we choose a hypothetical closed loop, which coincides with the field line passing through our point of interest. Let’s call this loop as c two.

The Ampere’s law says that b dot d l, integrated over this loop c two, should be equal to Mu zero times i enclosed. The left hand side of the Ampere’s law will be exactly similar to the previous part, and it will eventually give us b times 2 Pi r, which will be equal to Mu zero times i enclosed. Here, now we’re interested with the net current passing through the surface surrounded by loop c two, which is a shaded region. When we look at that region, we see that the whol wire is passing through that region, therefore whatever net current carried by the wire is going to be flowing through that region. Again we have a variable density which is variable in the radial direction and we will choose our incremental ring region with an incremental thickness at an arbitrary location and calculate the current flowing through the surface of this ring assuming that the thickness of the ring is so thin, so small, such that when we go from s to s plus d s, the current density remains constant. Or te change in radial distance for j, which was j zero times 1 minus s over R, so for such a small increment in s, is negligible, therefore one can take the change in current density for such a small radial distance change as negligible, so we treat the current density for that thickness as constant.

Then, again, d i becomes equal to j dot d a, which is going to be equal to j d a cosine of zero as in the previous part. Again, exactly like in the previous part, j zero 1 minus s over R and for d a we will have 2 Pi s d s. By integrating this quantity throughout the region of interest, then we will get the i enclosed. In this case, our region of interest is the whole cross sectional are of the wire and the corresponding s therefore will vary from zero to big R to be able to get the total current flowing through the cross sectional area of this wire. s is going to vary from zero to big R in this case.

Then i enclosed will be equal to, again take 2 Pi and j zero outside of the integral since they are constant. The first term is going to give us integral of s d s and then the second one is going to give us integral of s squared over R d s. The boundaries are going to go from zero to big R, and from zero to big R also for the second integral. By taking this integral, we will have 2 Pi j zero times s square over 2 evaluated at zero and big R and minus s cubed over 3 r, evaluated again at zero and big R. Substituting the boundaries, we will have 2 Pi j zero times r square over 2 from the first one. Zero will give us again zero minus r cubed over 3 r from the second integral and here, we can cancel r cubed with the r in the denominator, therefore we will end up only with r squared in the numerator.

Here we have r squared over 2 minus r squared over 3. We can have Â common denominator in order to express i enclosed as 3 r squared minus 2 r squared divided by 6. Numerator is going to gives us just one r squared, therefore i enclosed is going to be equal to 2 Pi j zero times r squared over 6. That is the explicit form of enclosed current, which is also the net current flowing through the wire. We can also express this quantity in terms of the cross sectional area of the wire since Pi times r square is equal to the cross sectional area of the wire. Then, this can also be expressed as j zero times a, then we can make one more cancellation over here between 2 and the 6, we will end up with 3 in the denominator. So we can express this as 1 over 3 times j zero a in terms of the cross sectional area of the wire.

Now going back to the Ampere’s law, we have found that the left hand side was b time 2 Pi r. Right hand side will be Mu zero times i enclosed, which is, in terms of the radius, 1 over 3 j zero times Pi big R squared and in this form, we can cancel the Pi’s on both sides and leaving b alone, we end up with Mu zero, j zero, 2 will go to the other side as dividend, 2 times 3 will make 6, and big R squared. Of course we will also have little r in the denominator. Or we can also write this in terms of the cross sectional area of the wire as Mu zero j zero a divided by 2 Pi. 2 times 3 is 6 Pi r. So, both of these quantities will give us the magnitude of the magnetic field outside of this wire carrying variable current density.