# Homework Help: Toroid of Circular Cross-section

1. Oct 5, 2011

### runnergirl

1. The problem statement, all variables and given/known data
A toroid of circular cross-section of radius a and mean radius r_{m}. Show that the inductance of this coil is given by L = μ_{r}μ_{o}N^{2}[r_{m}-(r_{m}^2-a^{2})^{1/2}].

2. Relevant equations
The B-field of a toroid: B = μNI/2πr (phi-direction)

3. The attempt at a solution
Total flux: N∫Bds where ds in the phi direction in spherical coordinates is: r_{t}dr_{t}dθ. In this case we need to solve for r which can not be assumed to be approximated by the mean radius r_{m}. Given that the vectors can vary as a function of θ within the toroid, I used the law of cosines to solve for r = √r_{m}^2+r_{t}^2 - 2r_{m}r_{t}cosθ and then integrated from 0 to a for the radial component and 0 to 2π for the θ component. The issue I'm having is that the integral isn't coming out and I'm not sure where I went wrong. Any help would be much appreciated, thank you.

2. Oct 6, 2011

### rude man

I think the trick here is to realize that what is needed is the magnetic field distribution inside a single-turn coil with radius a-R carrying current i, where R is the inner diameter of the toroid, found from r(m) = (a + R)/2. Then, work with a and R. L is then the integral of the found B field over the coil area divided by i. i of course disappears explicitly in the expression for L.

Then, account for the fact that there are N turns to the toroid.

3. Oct 6, 2011

### runnergirl

So in this case, do I still use the standard B-field of a toroid of B = μNI/2πr where now instead r = (a+R)/2 and integrate R? and what would I integrate R from? Would it be from r(m)-a to r(m)+a? And of course still integrate phi over 0 to 2π. Thanks again for your help.

4. Oct 6, 2011

### rude man

The radius of your circular loop is going to be a-rm. I think I misled you by saying you need to solve for R, the distance from the toroid center to the inside diameter of the toroid. You don't.

So you're really given the radius of a winding as = a-rm. Now, you have to integrate the B field over an infinite number of annular sections of one winding. Can you visualize the annuli I'm talking about? It should be obvious from the symmetry of the winding that B is a function of the radial coordinate only. You will now be integrating from x = 0, the center of a winding, to x = a-rm (I don't want to use r since r is the distance form the center of the toroid. This distance has no relevance to your integration.) You're integrating annular sections of area 2πx*dx from the center of the coil winding to its periphery. And you know how to use Ampere's law to get B(x), right? Hope I'm making sense here.

Last edited: Oct 6, 2011
5. Oct 6, 2011

### runnergirl

I just want to make sure I'm understanding this correctly. So I am now disregarding the standard formula for a toroid of B = μNI/2πr and instead visualizing this as a loop of radius a - rm. And if I use Ampere's law it should be B = μI/2πx still. However if I integrate from 0 to a - rm with ds = 2πx*dx then I would get, with N turns, μIN(a - rm)...So I guess I'm not seeing perhaps what Ampere's Law should be for what I'm visualizing as a disk...

6. Oct 6, 2011

### rude man

dφ = BdA where dA is an annular section as I described and B = μi/2π(a-rm-x) integrated from x=0 to x = a-rm.

You're dealing with 1 winding for the moment so N=1. When you've found the total flux in 1 winding you can the multiply by N^2 to get the entire flux in the toroid.

7. Oct 6, 2011

### runnergirl

When I integrate B=ui/2(pi) (a-rm_x) *2(pi) x*dx I get a logarithmic term which even applying the limits does not work out. Thanks again for all your help.

8. Oct 7, 2011

### rude man

runnergirl - I owe you an apology. I have not been looking at this right.

I will post my solution tomorrow. Again, I apologize for misleading you. It doesn't happen very often but unfortunately you were the victim in this instance.

One thing - your use of Latex doesn't seem to translate properly. I never could make out the expression for L and am not familiar with the Latex compiler language.

9. Oct 7, 2011

### rude man

OK, good news & bad news.

The good news is I have the correct expression:

φ = (μiN/π)∫[√(a^2 - x^2)/(x+ rm)]dx from -a to a. rm = toroid mean radius.

Then, L = Nφ/i

If you look at a cross-section of the toroid and set up an x-y coordinate system with the origin at the center of the cross-section, then B(r) = μiN/2πr or, since r = rm + x,
B(x) = μiN/2π(rm + x) with x ranging from -a to a.

Slicing up the cross-section into vertical (parallel to the toroid principal axis) slices, an element of area is dA = 2√(a^2 - x^2)dx.
So dφ = B(x)dA and the above integral expression follows.

The bad news is my table of integrals doesn't include the above integrand. I also tried to integrate wrt to r instead of x and ran into an even worse situation.

This is one of the most difficult e-m problems I've run across, mathematically.

Last edited: Oct 7, 2011
10. Oct 7, 2011

### runnergirl

Ok, I see how B is being set up with the r = rm+x. However, I don't see how you're getting dA. Is the cross-section as if you're slicing up the toroid and it's a disk like dimension? and are you essentially solving a right triangle? Otherwise, I'm just having a rough time visualizing it.

Sorry about the LaTeX, this is my first time posting on here and I don't know why the code didn't compile correctly. I may be using the wrong format for here.

11. Oct 7, 2011

### rude man

Thanks for not giving up on me.

I scanned an illustration for you, how do I get it to you? I don't have your e-mail & I don't see how I can do it using personal message facility on this board. I don't know how to do it the way others do here.

Try here. Never used this before ....

Last edited: Oct 7, 2011
12. Oct 7, 2011

### runnergirl

13. Oct 7, 2011

### rude man

Hope you got my e-mail. The shaded area is dA = 2√(a^2 - x^2)dx.
r = r(mean) + x.
r extends from [r(mean) - a] to [r(mean) + a].

Maybe you'll let me know if anybody solved this. I'd bet nobody will unless they get a really comprehensive table of integrals like Jahnke & Emde, or unless someone figures out how to do the integration more simply by a substitution of variables, shift of coordinates, etc.

14. Oct 16, 2011

### runnergirl

So since the integral is essentially the ∫(√a^2-x^2)/x dx you have to use the integral of the form X = a + bx + cx^2 which in this case a = a^2 and c = -1... so the integral table for this is ∫√X/x dx = √X + b/2∫dx/√X + a∫dx/x√X...it's not a very nice integral but if you follow through it will give you the right answer.

15. Oct 16, 2011

### rude man

Thanks for getting back to me. Remember when I said I tried it two ways and the second was even worse? Your solution was the second one! I looked at those (what looked like) recursion integrals and lost interest.

I think whoever gave you that problem is a sadist! :)

16. Oct 16, 2011

### runnergirl

Indeed it was a difficult problem... I've posted another one that's similar that involves resistance rather than inductance. Thanks again for all your help :)

17. Oct 16, 2011

### rude man

I'll look for your new post.

The second way I tried it was to put the coordinate origin at the center of the toroid instead of at the center of the cross-section. I'm pretty sure that's how your solution proceeded from also.