Example- Magnetic field profile of a cylindrical wire
As another example to the applications of Ampere’s Law, let’s consider the magnetic field profile of the current-carrying wire. B field inside and outside of a current-carrying wire. Let’s consider a cylindrical wire. Something like this carrying a current i, let’s say, in the outward direction. And let’s give some dimensions, say, the radius of the wire is big R. And we’d like to determine the magnetic field that this current generates inside and outside regions of this wire. And in order to do this, first let’s consider the inside region and then look at the wire from the top view, or from the cross-sectional point of view. And it will look like something like this.
OK, the radius of the wire is big R, and we’d like to determine the magnetic field that the current i generates, say, it is coming out of plane throughout this wire. I is flowing out of plane throughout the whole cross-sectional area of this wire. Let’s say our point of interest is somewhere over here, point p, and that is little r distance away from the center. At this location, by looking at the current flow direction using right hand rule, we’ll see that such a current is going to generate concentric circular magnetic field lines rotating in counter-clockwise direction. We’re going to choose, again, our Amperian loop such that it coincides with the field line passing through the point of interest. That field line is going to be circling in counter-clockwise direction. Therefore, the loop c that we will choose to apply Gauss’s Law, is going to be coinciding with the magnetic field line passing through point p.
The Ampere’s Law says that b dot dl over this loop c is equal to Mu 0 times in closed. The left hand side, at such a loop, will satisfy the conditions to apply Ampere’s Law as in the previous examples. And the left hand side of Ampere’s Law will give us b times dl, b magnitude times dl magnitude times cosine of the angle between b and dl. And again, the magnetic field is going to be tangent to the field line at every point along this loop. Something like this. And dl is going to be along the loop.
Therefore, the angle between b and dl will always be 0 degrees along this loop. So, we’ll have cosine of 0 here, and that will be equal to Mu 0 times in closed. An integral is taken over this loop c. Cosine of 0 is just 1, and here, again, the magnitude of the magnetic field is going to be always the same as long as we’re along the same field line. So, this will be constant. Then, we can take it outside of the integral. Constant over that loop. So, the left hand side will be equal to b times integral of dl over this loop c, and that will be equal to Mu 0 times in closed. Integral of dl along this loop c, again, means that we’re adding the magnitude of all these displacement vectors, one another, along this loop. And if you do that, we will end up with the circumference of the circle, which is equal to 2 Pi times little r. Therefore, the left hand side of Ampere’s Law will give us b times 2 Pi r. And on the right hand side, we have Mu 0 times in closed.
In closed is the net current flowing through the area, or the surface, surrounded by the Amperian loop, or the closed loop c that we choose. And that is this region. So, we’re interested with the net current flowing through that region. And we know that the net current flowing through the whole cross-sectional area is i. I, current i, is flowing through this wire. To be able to express the amount of current flowing through only the shaded region, we need to express, first, the current density. And as we recall, the current density was the total current, which is i divided by the total surface area through which this current is flowing. And that is the total cross-sectional area of this wire, which will be Pi r squared.
Now, that expression is going to give us the magnitude of the current density, i over Pi r big R squared. Current per unit area. If I multiply this density by the area of the region that I’m interested with, which is this shaded area, then I’m going to end up with the net current passing through that area. Current per unit area times area. Areas will cancel, and I’m going to end up with the total current flowing through that region. Therefore, in closed is going to be equal to current density, j, times the area of the region that I’m interested with. And that is the area of that shaded region, in other words, the area surrounded by loop c. And that area is equal to Pi times little r squared. Or, in explicit form, j is i over Pi r squared times Pi little r squared. And here, Pis will cancel. Therefore, in closed is going to be equal to i times little r squared over big R squared.
OK, so the left hand side of the equation is b times 2 Pi r. And now, the right hand side is Mu 0 times in closed, which is i times little r squared over big R squared. We can cancel r squared, little r squared, with this r over here. And leaving b alone on one side of the equation, we will end up with magnetic field magnitude inside of this wire is equal to Mu 0 times i divided by 2 Pi big R squared times little r. So, b inside becomes linearly proportional to the radial distance. In other words, inside of the magnetic field profile of the current-carrying wire is linearly dependent to the radial distance. Well, we have already done the outside earlier.
Let’s do it one more time for the outside region. And again, if you consider the top view of our wire carrying the current outside of the surface. And let’s say the radius is big R. Current is coming out of plane everywhere. And now our point of interest is outside of this wire, say, some r distance away from the center of the wire. In order to apply Ampere’s Law, again we’re going to choose a closed loop which will coincide with the magnetic field line passing through this point. Therefore, it’s going to be a circular loop, something like this, with radius little r. And again, the magnetic field is tangent to this loop. Therefore, the magnetic field line at every point. So, it will have the same magnetic field along this loop wherever we go. And dl is incremental displaced vector along this loop everywhere. Therefore, the angle between b and dl will be 0 degrees. So, b dot dl over this loop c, let’s call this one a c2 now, will be equal to Mu 0 in closed from Ampere’s Law. Writing down the left hand side in explicit form, we will have b magnitude dl magnitude times cosine of 0 integrated over loop c2 is equal to Mu 0 times in closed. Cosine of 0 is 1. Again, b magnitude is constant along this loop.
We can take it outside of the integral, leaving us b times integral of dl over loop c2 is equal to Mu 0 times in closed. And the integral of dl over loop c2 will give us the length of that loop, which is the circumference of this circle. And that’s going to give us just 2 Pi r. So, b times 2 Pi r on the left hand side will be equal to Mu 0 times in closed. In closed, again, is the net current passing through the area surrounded by the Amperian loop. In other words, this loop c2 now. And that area is this area. When we look at that area, we see that the whole current that the wire is carrying is passing through that surface. Therefore, in closed is equal to i. And solving for b, we end up with b is equal to Mu 0 i over 2 Pi r. And that is, of course, for the regions for the points that they’re outside of the wire. In other words, little r is larger than br. Whereas, the previous expression that we found is for the regions inside of the wire. And that is the case that the little r is less than big R.
So, in the inside region, magnetic field is linearly proportional to the radial distance, whereas for the outside region, the magnetic field is inversely proportional to the distance. In other words, b is proportional to 1 over r. Now, if we plug the magnetic field profile as a function of the radial distance, and let’s say here’s our wire over here carrying some certain current of i with radius big r. Inside of the wire, the dependence of magnetic field to the radial distance is a linear one. In other words, the magnetic field increases with radial distance r. And when r is equal to 0, in other words, along the axis of the wire, you see that the magnetic field is 0. And it is understandable because at that location, there will not be any current enclosed, no i enclosed.
And then, the magnetic field increases with the radial distance. Therefore, it increases linearly with the radial distance inside of the wire. And it takes its maximum value when the radial distance becomes equal to the radius. So when little r becomes equal to big R, then that r and this r squared will cancel. And we’re going to end up with Mu 0 i over 2 Pi r. That’s the value at the surface. Mu 0 i over 2 Pi big R. Once we leave the wire, then the current decreases with 1 over r. I’m sorry, the magnetic field decreases with 1 over r. And it goes to 0 as r goes to infinity. And, this is a typical magnetic field profile of a current-carrying wire.