# Magnetic Field inside Rectangular Toroid

Discussion in 'Homework Help' started by x1222, Apr 2, 2012.

1. ### x1222 Thread Starter Member

Oct 22, 2011
31
0
Consider a toroidal structure with a rectangular cross-section. If the toroid is defined by
the surfaces r = 1cm and r = 4cm and the planes z = 0 cm and z = 2 cm, and the surface
current density on the surface defined by r = 4cm is given by -60 A/m in the z direction.
(a)
Specify the current densities on the surfaces at r = 1, z = 0, and z = 2.
(b)
Find the expression for H inside the toroid (i.e. in the region 1 < r < 4 cm and 0 <
z < 2cm).

I have the solution, but I'm having trouble understanding it. The current density, K, for r=1 was found to be 240 A/m.

To find the field inside the toroid, ampere's law is used with ∮H dl = 2∏r = ∫(K at r=1)dθ from θ=0 to 2∏. Giving the final answer of H=15/(2∏r)

So in this question K is different at r=4 and r=1. Is it changing due to distance from the sources of the field, or is it uniform and I'm missing some simple math ratio? Since K is different at different radii, how come we integrate only integrate with Kr=1 and not other any other K's? I mean it is changing with distance and I'm assuming since the cross section of the toroid is a square loop, all four sides would contribute to the field inside the toroid, not just K at r=1.

How come I don't need to calculate for Kr=4 and Kz=0 =Kz=2 and sum them up? Hope that made sense, thanks in advance.

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
When you apply Ampere's law in a case like this, you choose a loop to use for the calculation. This loop needs to be one that has symmetry in order to do the calculation. For this reason you must choose a circle with a radius that is inside the toroid.

Now Ampere's law says the current that pierces a surface that is bounded by the circular path equals the line integral of the field around the boundary of that surface. So, which surface do you want to assign to the circular boundary? The simplest choice is a flat plane circle, right? This circular area will be pierced by the current on the interior surface r=1. You are free to choose any other surface, - for example one which is pierced by the z=0, or z=2 surface, or both. You can even use a crazy surface that is pierced by the r=4 surface, or some crazy combination of all surfaces. However, if you do, you just made the problem more difficult to visualize. Still, you'll get the same answer by any of these methods, and the same current pierces any proper surface assigned to the circular boundary.

EDIT: By the way, on your question about the K changing with radius, that is correct. The reason why this does not matter is because K gets integrated over a surface area to give total current. A large K integrated over the smaller area of the R=1 surface equals the smaller K integrated over the larger area of the R=4 surface. In fact, that is the property which allowed you to know that the inner surface had K=240 A/m. The radius is 4 times less (hence the area is 4 times less) and so the K must be 4 times more than that at r=4 cm.

EDIT: By the way, the answer of H=15/(2 pi r) is a little unusual. Numerically, it is nearly the correct answer, but the value of pi should not really be in there. It should be H=2.4/r exactly. When you include pi in the answer, it gives the impression that the value is exact, which would not be true in this case.

Last edited: Apr 3, 2012
x1222 likes this.
3. ### x1222 Thread Starter Member

Oct 22, 2011
31
0
Thanks for taking the time to write that, very in-depth. I think I was getting confused what K really was. It is the same as Gauss law really then, only inside the imaginary surface matters.

Ah yes, the final answer is actually H=2.4/r. I just thought /2pi*r might have been more clear.

4. ### steveb Senior Member

Jul 3, 2008
2,433
469
Looking back at what I wrote late last night, I see I miss worded the description of getting current from K. Since K is a linear surface current, the integration is over length not area. I misspoke because K is current density integrated over the thin surface thickness. So, there is an area involved, but thickness is ignored, which only leaves the surface length..

There is nothing wrong with leaving the 2 PI r in your formula. As you say, it gives insight into how you arrived at the value. However I would include a PI in the 4.8 PI in the numerator to let the answer be exact. This probably seems strange because approximate answers are allowed in engineering, but somehow when you include PI in the answer it gives the false impression that the answer is exact. Hence I tend to make sure any answer that has PI in it is an exact answer.