2
\$\begingroup\$

I need help understanding Ampere's circuital law. Consider a square shaped iron core with an N turn coil on one side. Let each side of the core has the length a an has equal cross-section. The textbook approach is to take a closed contour passing through the middle of the core. All magnetic flux is inside the core, so we have (given the coil current I)

NI = aH + aH + aH + aH

But what if I take a contour that pass inside the coil and then goes outside? Outside the core there is no (or very little) magnetic field, so

NI = aH + 0

The two equations do not match. What is wrong?

Two contours

Update

After reading comments and thinking on the problem I have advice for future readers.

  1. Magnetic flux concentrates inside the core. This is true enough to assume that flux is constant along the core, but not enough to assume that field in the air is zero.

  2. From Ampere's circuital law we may conclude that H field must be stronger in the surrounding air that in the core to satisfy the law for different contours. B field on the other hand is much stronger inside the core than outside because it is proportional to the flux.

I have modeled the problem in the FEMM and attach supporting figures below. If someone wants to model it himself, be careful not to saturate the core; otherwise you'll get disturbed field distribution.

enter image description here

enter image description here

\$\endgroup\$
11
  • \$\begingroup\$ I'd say there are two strong assumptions here: first, you assume the field is uniform inside the core (I can hardly imagine the field taking sharp turns), second what about this tiny pieces of path that you need to get out of the core? \$\endgroup\$ – Sredni Vashtar Jan 31 at 16:57
  • \$\begingroup\$ @SredniVashtar I think field uniformity is a common assumption. You can imagine core with curved corners; this won't change much. As for the second point, the flux is perpenicular to the contour there, so Hdl integrates to zero; it is the same thing that is done when deducing a formula for infinite solenoid. \$\endgroup\$ – AlexVB Jan 31 at 17:27
  • \$\begingroup\$ Yes, the contribution along those tiny pieces is not the cause, I was too quick in judging. But you can't confine the field in the core - even in the case of a linear solenoid you need to make it infinite to make the outside contribution tend to zero. I mean, when you have the solenoid alone the field lines are cylindricall symmetric, but when you have the core in place, the lines are 'crammed into the core' but they still have to go around the solenoid. Have you tried to solve the problem in terms of B field \$\endgroup\$ – Sredni Vashtar Jan 31 at 19:01
  • 1
    \$\begingroup\$ Alex, what I meant is that the lines of B that are in the opposite side of the coil wrt the core won't just disappear because you put a high my core inside. If you considered a high but finite my you will avoid the 0/0 paradox. You have most B field inside the core and a little outside it in a ratio that depends on mu in the materials. That little B field outside is related to H by the inverse ratio of permeability. And yes, I guess if you solve for B you need to take into account the magnetization currents around the core. It's not an easy problem but imo the paradox is linked to infinite mu \$\endgroup\$ – Sredni Vashtar Feb 1 at 20:16
  • 1
    \$\begingroup\$ "my" means "mu". ***** Autocorrect. And since I have some more space, what I believe is that some B field is present in the space enclosed by the core. It's very difficult to shield a magnetic field (jackson has a section on that), what the core does is to 'refract' the field lines but cannot trap them all. Three equations are needed, circulation of H, flux of B, and constitutive equation in the two materials. All this IMO. \$\endgroup\$ – Sredni Vashtar Feb 1 at 20:28
0
\$\begingroup\$

Its a very good question. If the core are ideal, so relative permmeativity is very high (infinite) and H is equal to zero.

I know its not answering your question but maybe its some direction to the solution.

\$\endgroup\$
1
  • \$\begingroup\$ I think it can't be equal to zero exactly because of the Ampere's law. What you shall get in case of infinite permeability is infinite flux for any given current. \$\endgroup\$ – AlexVB Jan 31 at 19:14
0
\$\begingroup\$

Now I manage to refresh my memory with a little help from my notes...

As I mentioned before - in the ideal core the H field is zero because of an infinite relative permittivity.

so in both cases, the right side of the equation is equal to zero.

now, let's figure out the left side of the equation - NI... In an ideal core, the current needed to excite a finite magnetic flux is zero! this is because the magnetization inductance is infinite... So both of the equations are correct.

\$\endgroup\$
2
  • \$\begingroup\$ Well, alright. In fact H=0 and I=0 is a solution for the equations regardless of the core permeability. Now the problem is what to do with a real core of finite permeability. Well, I'm almost convinced that H field is present in the air and is even stronger than in the core. \$\endgroup\$ – AlexVB Feb 1 at 13:49
  • 1
    \$\begingroup\$ Daniel, this isn't a forum for discussing things. If you have additional information, you can add it to your existing answer. You can merge the two and delete one of them, or you can leave it for someone else to clean up (and make them cranky.) \$\endgroup\$ – JRE Feb 1 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.