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Re: [femm] inductance/flux calculations in 2D/3D



corneliuspaul@xxxxxx wrote:
OK Dave, many thanks for the advice.
I tried out some configurations and it really seems to work
in practice and I am ready to accept it as the "basic rationale"

Although I still have difficulties understanding WHY it is like that.
A very simple hypothetical experiment:

- take a core with a quadratic cross section

- wind a one turn coil around it. that is, this coil consists of
  4 wires, each on every side of the core

- now every wire produces a H-Field around it  H = I/(2*pi*r)
  where r is the distance from the wire surface, I is current

- if the core is of linear material then B = mu * H

- flux = integrate B over cross section

THIS IS WHY I "FEEL" EVERY ONE OF THE 4 WIRES PRODUCES ONE FOURTH
OF THE INDUCTION OR FLUX IN THE CORE.

WHERE DOES "the only parts of the wire that actually "drive" flux are those

that get encircled by the flux path" COME FROM???

what am I missing ?

Why, this is just a statement of Ampere's Loop Law:
ó
(ç)
õ
(B/m) ·dl =  ó
õ
ó
õ
J ·dA
The left-hand side is the line integral of magnetomotive force drop on a path around the core. The right-hand side is then the current that is enclosed by the path. The end-turns aren't enclosed by this path (i.e., each turn in the coil contributes only once to the integral on the right-hand side, not once for each side of the coil). You can draw paths that loop around the end turns, but they typically traverse a lot of air and don't add much flux (they are high-reluctance flux paths and the left-hand side integral is dominated by the parts of the flux path in air because of it low m).

I think what is playing games with your intuition is that the "H = I/(2*pi*r)" formula only works for a wire in air, not for problems that have iron.

Dave

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