Does Ampere's Law Hold for Magnetic Circuits with Non-Symmetric Geometry?

[TFH]

Background: In the textbooks I’ve read the simple magnetic circuit is developed by first looking at a ferrous torus symmetrically wrapped with a condutor. Symmetry arguments indicate that the H field is radial through the torus and 0 on the inside and outside of the torus. Then from Ampere's law H is obtained.

H = NI/L, N = number of turns, I = current, L = length around the torus (Pi R).

I'm okay with this but next the shape of the torus changes to a square and the turns are bunched together on 1 leg of the square, say l1. And two assumptions are discussed: (1) flux is constand around the square core and (2) no leakage field. With these assumptions Ampere's law is used again to obtains the same equation for H except the Lenght "L" changes to match the square core.

With a square core: L = l1 + l2 + l3 + l4. (l's are lengths of the square edges)

My Problem: If I use Ampere's law for other paths I get "VERY" different answers.

I can say integrate through l1 (that contains all the conductors) and would obtain: H = NI/l1. Or I could integrate say the half of the square that does not have a conductor and obtain:
H = NI/(l2 +l3+l4), but here I is zero. So H=0 under the assumptions.

So does the assumptions of no leakage field and constant flux for this geometry immediately cause a violation of Ampere's law?

Thanks for any comments.

 

[Charles Link]

You have it correct, but the $H$ is just a mathematical construction. The real field is $B=\mu_o \mu_r H$ or, in vacuum, $B=\mu_o H$. This does allow you to estimate the leakage flux out of the transformer, but it won't be exact.
For a thorough discussion of transformers, see also: https://www.physicsforums.com/threa...ly-the-biot-savart.927681/page-3#post-5996404