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# Re: [femm] Re: inductance - revised axisymmetric model - fem files attached

- To: femm@yahoogroups.com
- Subject: Re: [femm] Re: inductance - revised axisymmetric model - fem files attached
- From: "Jim in_Dallas" <jim271@hotmail.com>
- Date: Wed, 14 Nov 2001 13:45:41 -0600
- Bcc:

>     From: "Jim in_Dallas" <jim271@hotmail.com>
>     Reply-To: femm@yahoogroups.com
>     To: femm@yahoogroups.com
>     Subject: Re: [femm] Re: inductance - revised axisymmetric model
>     Date: Wed, 14 Nov 2001 13:29:55 -0600

> `
> You are right, I'm sorry I didn't model the multi-turn coil correctly. I
> assumed as long as the turns are much closer together than the radius of the
> coil, the actual turn spacing was not critical. This is incorrect. For a
> multi-turn coil, the turn spacing is very important and the radius of the
> wire used is less important.`

>     The equation for a single turn coil can be applied to multi-turns only if
>     the turns are closely packed. This would be the case if the coil is made
>     with several turns of small wire that occupy the same volume as a single
>     turn of larger wire. My FEMM model didn't take this into account, so it was
>     a coincidence that the results appeared to be in agreement.

> `Today I ran FEMM for several single-turn coils with the following results:`
> `Let F1=formula 1: L=R*u0*Ln(R/r) , R=coil radius, r=wire radius`

>     For all the single turn coils:
>     Coil radius=0.4cm
>     J=1 MA/m^2

>     (a) wire radius=0.025 cm, L(femm)=15.4nH, L(F1)=13.9nH,
>     ratio[L(femm)/L(f1)]=1.11
>     (b) wire radius=0.05 cm , L(femm)=12.1nH, L(F1)=10.4nH,
>     ratio[L(femm)/L(f1)]=1.16
>     (c) wire radius=0.075 cm, L(femm)=10.1nH, L(F1)= 8.4nH,
>     ratio[L(femm)/L(f1)]=1.21

> `The last coil (c) is shown in the attached file: singleturn1.fem.`

>     It appears that as the wire radius becomes smaller, the ratio of the FEMM
>     calculation to the inductance calculated by formula 1 gets closer to unity.
>     I repeated the femm simulations with different mesh sizes and that has only
>     a small effect on the results. (I expected the results of these two methods
>     to agree more closely. Does anyone have an idea why the difference is this
>     much?)

>     I made another coil using 4 turns of wire with a diameter of 0.06cm and
>     configured it so the 4 turns fit in the same space as the single wire in
>     coil (c) above which has a wire diameter of 0.15cm. This is file:
>     four_turns.fem
>     The inductance of this coil calculated by FEMM is 167.8nH. This is close to
>     16X the inductance of the single turn coil (c) above (16*10.1=161.6nH).

> `Therefore, if you have a multi-turn coil with the turns closely packed, then
> formula 1 gives fairly accurate results is you use the outer radius of the
> wire bundle as the "wire_radius" term in the equation and then multiply by
> N^2.`

>     During the past year I've compared the FEMM inductance value for a variety
>     of long solenoids with the value calculated using Wheeler's formula and the
>     difference is typically less than two or three percent.

> `-- Jim`
> ``

>     >From: tborzic@yahoo.com
>     >Reply-To: femm@yahoogroups.com
>     >To: femm@yahoogroups.com
>     >Subject: [femm] Re: inductance - axisymmetric model
>     >Date: Tue, 13 Nov 2001 22:50:40 -0000
>     >
>     >--- In femm@y..., "Jim in_Dallas" <jim271@h...> wrote:
>     >
>     > > >
>     > > If we use the axisymmetric model for the coil, the FEMM results is
>     >very
>     > > close to the value calculated using the formula for a current loop.
>     > >
>     > > Example:
>     > >
>     > > current loop: L=N^2*R*mu0*Ln(R/r)
>     > > where R=radius of loop
>     > > r=radius of wire
>     >
>     >Jim,in Your smallcoil femm file r is not 0.0005cm it is d,(d=2r)
>     >so r=0.000025
>     >so this formula works only for ....Ln(R/d)?
>     >because you draw d=0.0005 not r (r is a half of diameter)
>     >
>     > >
>     > > Using your dimensions:
>     > > R=4cm
>     > > r=0.0005cm
>     > > 4 turns
>     > > then L = 167.2nH
>     > >
>     > > An alternative formula found in some references is: L=N^2*r*mu0*[Ln
>     >(8*R/r)
>     > > -2]
>     > >
>     > > Since the Ln(8) approx = 2, these formulae give similar results.
>     > >
>     > > Using FEMM:
>     > > Let J=1.0 then
>     > > integral(A.J)=6.236e-9
>     > > I = 0.1954 Amps/turn
>     > >
>     > > L=163.4nH which differs by less than 3 percent. I've attached my
>     >fem file.
>     > >
>     >
>     >when i changed gap between wires to something smaller result is 167.x
>     >nH. Only thing that bothers me is that radius in formula..
>     >So, now i will bite my nails all night (it is night in Croatia,
>     >actually midnight) hoping that r in formula is actually d.
>     >Or I made mistake when mesuring dimensions in smaillcoil file?
>     >Hope You will soon read this:)))
>     >Thanks again
>     >tomislav
>     >
>     >
>     >

>     _________________________________________________________________
>     Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp

``

    _________________________________________________________________
    Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp

Attachment: `singleturn1.FEM`
Description: Binary data

Attachment: `four_turns.FEM`
Description: Binary data

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