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**Example- Magnetic field of a coaxial cable**

Now let’s calculate the magnetic fields of a coaxial cable in different regions.

B field of a coaxial cable. A coaxial cable consists of two concentric cylindrical regions, an inner core, an outer cylindrical shell, something like this. These conducting cylindrical regions are separated by an insulating medium from one another, and as one of these cylinders carry the current in one direction, that’s called the current flowing the inner core as i sub a. The outer cylindrical shell carries the current i sub b in opposite direction.

If we give some dimensions to this cable, let’s say this radius is a, the inner radius of the outer cylindrical shell is b, and outer radius of the other cylindrical shell is c.

Therefore, current is flowing through these cylinders in opposite directions, and we’d like to determine the magnetic field of such a cable in different regions. Let’s start with the region such that our point of interest, distance to the center, is less than the radius a. In other words, inside of the inner cylinder.

And let’s look at this case from the top view and so here we have, let’s say, the inner cylinder from cross sectional point of view, and the outer cylindrical shell, something like this, and the inner cylinder is carrying the current i sub a out of plane, and the outer cylinder is carrying the current i sub b into the plane, everywhere throughout these regions.

Again, the inner cylinder’s radius is a, and this radius is b and the radius of the outer region is c. Well, we have done a very similar example earlier. Our first region of interest is that our point of point a is inside of the inner cylinder. Let’s say somewhere around here, and in order to find the magnetic field at this location, which is little r distance away from the center, we place an empirical loop in the form of a circle which coincides with the magnetic field line passing through that point, and let’s call this loop as c1 for the first region.

And the empire/s law says that B of dl integrated over this loop, c1, will be equal to neu 0 times the net current passing through the region, or the surface, surrounded by this loop c1.

As we did in the earlier examples, such a loop will satisfy the conditions to apply empire’s law, and the magnetic field will be tangent to the field line, and that field line coincides with the loop that we’re choosing and dl is an incremental displacement element along this loop, therefore the angle between b and dl will always be 0 degrees for this case.

So, the left hand side will give us b magnitude, dl magnitude times cosin of 0, integrated over loop c1, will be equal to neu 0 times i enclosed.

Cosin of 0 is 1 and b is constant over this loop because the loop coincides with the magnetic field line passing through that point, and as long as we are on that field line we will see the same magnetic field magnitude. Therefore, since the magnitude is constant, we can take it outside of the integral, therefore, the left hand side we end up with b times integral of dl over loop c1 is equal to neu 0 times i enclosed.

Integral of c1, integral of dl, over loop c1 will give us the length of that loop, which is the circumference of that circle, and that’ll be equal to 2pi times the radius of that circle, which is little r times b will be equal to neu 0 times i enclosed.

I enclosed is the net current passing through the region surrounded by this loop c, so that is the surface. The loop c surrounds this green shaded region, and we know that through the whole inner surface, the current flowing is i sub a, which basically covers this whole region over here, and in order to get the net current flowing through this green shaded region we will define the current density, which is current per unit cross sectional area, and if we multiply that current density by the area surrounded by the loop c, we will get the amount of current passing through that surface.

Therefore, if we move on, we will have b times 2pir, this is the left hand side, which is equal to neu0 times i enclosed, where in this case i enclosed will be equal to J times the area of that region, which is pir squared, and here the current density is total current i divided by the total cross sectional area of this wire, and that is pi times a square.

So, b times 2pir is going to be neu0 times, where i enclosed we will have i over pia square, and this is the current density for the current flowing through the inner cylinder, and I should use the subscript a over here because we defined the amount of current flowing through the inner cylinder as i sub a. I sub a over pia square is going to give us the current density, and if we multiply this current per unit area by the area of the region that we’re interested, which is pir squared, then we end up with the total current passing through that surface.

Here, this pi and that pi will cancel, and we can cancel one of these r squares with the r on the left hand side, and leaving b alone we will end up with magnetic field inside of the inner cylinder as neu0 i sub a divided by 2pia square times r.

And, of course, this is identical result with the example that we did earlier to get the magnetic field profile of a current carrying cylindrical wire.

Now, as a second region let’s consider the magnetic field for the region that our point of interest is between the two cylinders. In other words, r is less than b and greater than a region.

If we look at that region we’re talking about this part, and in this part let’s say our point of interest is now located somewhere over here. Again, we choose a closed loop. In this case, let’s call this one as c2, which coincides with the magnetic field line passing through the point of interest p. Now it is located in this region.

And for that region, this is our outer cylindrical shell region which is carrying the current i sub b into the plane. Now, for this region, again, when we choose this loop which coincides with the field line passing through that point it will satisfy the conditions in order to apply the ampere’s law, and therefore, the left hand side of the ampere’s law will be identical to the previous part, and its going to give us b note dl integrated over now loop c2, which is equal to neu0 i enclosed. The left hand side is going to give us, again, b times 2pir. Of course, now, the distance, little r, is the distance from center to this point for this region.

And the right hand side, for this case, now we’re going to look at the net current passing through the region surrounded by loop c2, in other words, the area surrounded by loop c2, and that is this yellow shaded area, and when we look at that surface we see that the whole current flowing through the inner cylinder is passing through this surface, and of course anything outside of this surface is of interest, and therefore, in this case, i enclosed is going to be equal to simply the current flowing through the inner cylinder, which is i sub a. Therefore, on the right hand side, we will have neu0 times i sub a, and solving for the magnetic field we will have neu0 i sub a over 2pir for this region.

So, this is the case, that r is between b and a and for the previous part we calculated the magnetic field for the region such that r is less than a.

Now let’s move forward and let’s calculate the magnetic field inside of the other cylindrical shell. So, in this case, we’re talking about b at the region where r is between c and b.

In other words, now we’re interested with the interior region of this other cylindrical shell. Let’s assume that in this case our point of interest is somewhere over here.

Now again we choose our empirical loop such that it coincides with the field line passing through that point, therefore, it is going to be, again, in the form of a circle, and its radius, r, now is measured from the center, pointing this [inaudible 16:45].

Now let’s call this loop as c3. Again, the left hand side calculations will be similar to the previous parts. This loop will satisfy the conditions in order to apply Ampere’s law. The magnitude of the magnetic field will be constant everywhere along this loop, and the angle between b and dl will be 0.

So, Ampere’s law, which is b dot dl, integrated over loop c3 equal to neu0 i enclosed is going to eventually give us, for the left hand side, same as above, will give us d times dpir, and on the right hand side we will have neu0 times i enclosed.

Now here we’re talking about the net current passing through the area surrounded by loop c3. If we look at that area we will see that, first of all, we’re talking about this area now here, this blue shaded area, in that region we see that the whole inner cylinder, or the current flowing through the inner cylinder will be passing through that area, and for the other cylindrical shell we see that only this much section of the cylinder will contribute to the magnetic field, because the current flowing through the region which is our side of this specific surface is of interest.

Therefore, since i sub a is flowing out of plane, and i sub b is flowing into the plane, the net current is going to be basically the difference between these two currents. So we can express i enclosed as i sub a, let’s choose this direction, our plane direction as positive, and that is moving out of plane, that is positive, and the other one is the fraction of the current which is moving into the plane and in order to express that one we need to now express the current density associated with the outer shell, which is total current flowing through that shell, and that is i sub b, divided by total cross sectional area of the conductor, we’re talking about the outer shell, and the total cross sectional area of that outer cylindrical shell is the area of this big cylinder minus the area of this small cylinder.

So, in other words, that will be equal to pic squared minus pib square, and that part, this expression, is going to be equal to current density of the outer cylinder.

And this density times the area of interest will give us the net current flowing through that area. So, in other words, if we take the product of current density with this blue shaded region, shaded region’s area I should say, then we will get the net current flowing through that surface, and that is basically pir squared minus pib squared.

Okay. We can simplify this expression by writing it as i enclosed is equal to i sub a minus i sub b over pi parenthesis c squared minus b squared times pi times r squared minus b squared.

Here the pis will cancel, and therefore i enclosed will be equal to this quantity. Then b times 2pir will be equal to neu0 times i enclosed and that is i sub a minus r square minus b square, i sub b divided by c square minus b square.

In order to get the magnetic field we leave that quantity alone on the left hand side of the equation, therefore, b will be equal to neu0 over 2pir times i sub a minus i sub b times r square minus b square, divided by c square minus b square equals parenthesis.

So inside of the outer cylindrical shell, the magnetic field magnitude is going to be equal to this quantity. Of course, the direction, the net direction of magnetic field, whether this is clockwise or counterclockwise, depends on the magnitude of these currents, and this is for the region that r is between c and b.

The last region is the outside region of this coaxial cable. So we go back to our diagram, then we’re talking about that our point of interest is located somewhere over here, and again, by choosing an empirical loop, which is passing through the point of interest and coinciding with the field line passing through that point, point p, and that is r distance away from the center.

The left hand side of the Ampere’s law, let’s call this loop as c4, the Ampere’s law for this case will be the b note dl integrated over loop c4, which will be called neu0 times i enclosed, and the left hand side, again, will be similar to the previous parts, which will give us b times 2pir, and that’ll be equal to, for the i enclosed now, we’ll look at our diagram, we’re talking about the net current passing through the area surrounded by now, this whole region, and it’s surrounded by loop c4, which we’re talking about this whole region, and we can easily see that the whole current passing through the coaxial cable is passing through this point, passing through this surface, and that is i sub a is coming out of plane and i sub b is going into the plane.

As a result of this, the net current passing through the area surrounded by empirical loop c4 will be equal to i sub a minus i sub b since they’re flowing in opposite directions, therefore, on the right hand side we will have neu0 times i sub a minus i sub b, and solving for magnetic field we’re going to end up with the final expression of neu0 2pir times i sub a minus i sub b.

And this is the magnetic field generated outside of this coaxial cable. That is for the region that r is greater than c.

Okay. Well, if i sub a is equal to i sub b, if these two currents, that they’re equal in magnitude since they’re flowing in opposite directions, then i enclosed is going to be equal to 0. It means that magnetic field outside of the coaxial cable will be 0 for r greater than c region.