9.6 Inductance of a Toroid
Let’s consider a toroid with a rectangular vertical cross-section. Therefore, we are going to observe the toroid from its vertical cross-sectional point view, something like this. And, let’s give some dimensions, the inner radius is a, outer radius is b, and, let’s say, the height is h, and the width is d. Actually, that will be b minus a, so we don’t have to give another dimension for that part.
Let’s assume that the turns are in clockwise direction. Therefore, the current is going to be flowing through these turns in clockwise direction. And, if we just plot several of these turns, they’re going to look like, something like this. So, here’s the direction of the current flow along these turns.
If we look at the magnetic field that these current loops will generate, by looking at, first, this cross-sectional region and the current is here, going in clockwise direction, keeping the right-hand fingers in the directional flow of current, and keeping the thumb in up, or open position, we will see that the magnetic field generated by this loop is going to be pointing into the plane. So it will be pointing like this.
If we look at the right hand, the left hand side up here, again, current is clockwise direction holding the right hand fingers curling in clockwise direction and keeping the thumb up. For this case, we will see that the magnetic field lines are going to be coming out of plane.
As a matter of fact, in a three dimensional sense, we can say that, okay, they’re going to be coming out and going into the plane all around this toroidal direction, and closing up upon themselves by generating these closed magnetic field lines. Therefore, the magnetic field at this point is going to be pointing into the plane.
First, let’s try to calculate that magnetic field, and following the steps of the procedures to determine the inductance of a toroid, something like this. First we assume that the current i is flowing through the toroid, and second, we will calculate the magnetic field by applying Ampere’s law, which is B dot dl integrated over closed contour C is equal to μ0 times i enclosed.
Well, here, we’re going to choose, as we did earlier, again, a loop which coincides with the field line passing through the point of interest. Let’s assume that our point of interest is at this location. If we choose our loop coinciding with the field line passing through that point, we can see that along this point, magnetic field is always tangent to this loop, and along this loop, since we’re always on the same field line, then the magnitude of the magnetic field will be constant. Furthermore, dl, is the incremental displacement vector. At every point here, B field is into the plane and so is the dl. Therefore, the angle between them will be 0 degree everywhere along this slope.
So, in explicit form, B magnitude dl magnitude times cosine of the angle between these two vectors, which is, in this case, 0 degrees integrated over all this closed loop C will be equal to μ0 times i enclosed. Cosine of 0 is 1, and again, since is the magnitude of the magnetic field is constant everywhere along this loop, we can take it outside of the integral, leaving us B times integral of dl along this loop C will be equal to μ0 times i enclosed.
Let’s say our point of interest is r distance away from the center of the toroid. Then the integral of dl is going to give us the length of this loop. And, since it is a circle, this will be equal to the circumference of that circle, and that is 2πr. B times 2πr will be equal to μ0 times i enclosed. And, solving for B, that will be equal to μ0 times i enclosed over 2πr.
To be able to complete the calculation, we need to determine i enclosed. And, i enclosed is the net current passing through the area surrounded by this loop. In other words, we’re looking at this surface and we’re looking at how much current is passing through that surface. As you can see that surface will enclose all the turns of this inner loop. In other words, all the turns are going to be coming out of plane of that specific area.
So, here, all these points are going to give us the amount of current passing through this surface. All this inner branch will be included in that region. Therefore, as we did earlier, i enclosed is going to be equal to number of turns passing through that surface, and that is the number of turns of the toroid times the current carried by each turn, which is i. So, here, N is the number of turns of the toroid.
Therefore, the magnetic field becomes equal to μ0Ni over 2πr. As major difference from the magnetic field of a solenoid or perfect solenoid, we see that this quantity is not constant. It varies with the radial distance, r. In other words, as we go along this radial distance, from inner wall towards outer wall, the strength of magnetic field will decrease with 1 over r.
Okay, once we calculate the magnetic field, our third step is to calculate the magnetic flux. Well, if magnetic field were constant, then that magnetic field directed, dotted with the area vector, in this case we are talking about this cross-sectional area of the toroid, vertical cross-section. Then, by taking that dot product we could have easily calculated the flux.
But, since the magnetic field is changing with the radial distance r, and it is inversely proportional to that distance, then we’re going to use our usual procedure such that we will choose an incremental strip, rectangular strip, in this region and we will assume that the thickness of the strip is very, very small and we will call it as dr. That dr is so small as we go from inner region to outer region along this rectangular strip, the change in magnetic field magnitude can be taken as constant. Therefore, one can calculate the flux through this strip as an incremental flux simply by taking the direct dot product of magnetic field vector and the area of this strip.
So, dΦB, incremental flux, through such a strip is going to be equal to B dot the area of that incremental strip that we choose. So, in other words, the area of this rectangle.
Well, if we look at the area vector of this, again, following the direction of the current loop throughout this whole surface and holding the right hand fingers in the direction of current flow, the thumb will point into the plane. Therefore the area vector of that strip and the magnetic field will be in the same direction.
So, dΦB is going to be equal to B magnitude dA magnitude times cosine of the angle between them, and that will be cosine of 0. Since cosine of 0 is just 1, and if we express dA in explicit form, the area of this strip, by going back to our diagram, the width of this strip is dr, and it’s height is h. So, h dr is going to give us the area of that rectangular strip.
Now, once we calculate the incremental flux through that incremental rectangular strip over here, then we can go to the next strip and do the same process, and then the next one, and so on, so forth, we do this throughout the whole region that we’re interested with. And, finally, we add all those incremental fluxes to one another to be able to get the total flux through that region.
Well, here, the summation is integration ΦB will be equal to integral of dΦB, and that will be equal to integral of B dA. And, writing this in explicit form, it’s going to be equal to B magnitude — we just calculated that — and it was equal to μ0 Ni over 2πrr, and dA is equal to h dr. So, this is the magnitude of the magnetic field, and this is the area of the rectangular strip with an incremental width.
Here, our variable is r and we’re going to calculate the flux through this vertical cross-sectional area of the toroid. Therefore we will add these fluxes throughout the surface of this vertical cross-sectional. And, therefore, r is going to vary from inner radius a to outer radius b. So, the boundaries are going to go from a to b.
Okay, here μ0 Ni and 2π, these are all constant, we can take it outside of the integral. And, leaving us, flux is equal to μ0 Ni over 2π times, actually, we can take h, also outside, that is also constant. So, μ0 Nih times integral of dr over r integrated from a to b. Now, integral of dr over r is going to give us ln of r, so ΦB will be equal to μ0 Nih divided by 2π times ln of r evaluated at a and b. Substituting the boundaries, flux is going to be equal to μ0 Nih over 2π times ln of b over a.
Now, once we calculated the flux, we can move to the last step, and calculate the inductance from it’s definition, which is number of flux linkages NΦB divided by the current. That will be equal to μ0 times this quantity, will give us N2 ih divided by 2π ln of b over a divided by i.
Currents, again, will cancel, and we will end up with inductance of a toroid equal to μ0 times N2h divided by 2π times ln of b over a. Again, here we see that inductance is dependent to the geometrical properties of the inductor and, for a toroid, to its inner radius, outer radius, and its height, and, as well as its number of turns.