Piece of Wire Through a Toroid

Why not be more complicated, and try calculating results when wire is against inner surface of a (say) 10mm toroid compared to same toroid, wire and current when wire is centerd?

Turns count on a toroid is the number of passes through the center.

If it is through the center, it represents one turn, and will behave as such.

=3D=3D

No. When you place a ferrite or other flux concentrator in the system, all the in-air calculations and considerations are invalid. All the flux is captured, and distributed evenly throughout the ferrite's area. The flux density is the flux divided by ferrite cross-section area. Use the standard inductor and transformer equations to evaluate issues like core saturation.

=3D=3D=3D

I think D from BC has a point, at least for the simple case of a straight wire - the equation B =3D (uo*ur*I)/2*pi*r applies. In a similar analysis of a current transformer (Chai, H. 1998. _Electromechanical_Motion_Devices_, p.119), the equivalent expression is integrated from the inner to the outer radii of the toroid (thickness d, inner and outer radii ri and ro) to obtain the total flux p:

p =3D uo*ur*I*d*ln(ro/ri)/2*pi

That ri in the denominator foretells trouble in tiny toroids.

-- Joe

This law does not apply here, see other postings.

And:

(4*pi*10e-7 * 4000 * 1) / (2 * pi * 1E-3) = 0,8 Gauss ?

P.

Yes. In cgs units: H = .4(pi)NI/l = .4(pi)N I/ (2(pi)r). For N = 1, I =

1, r = 0.1, H = .4/(2(.1)) = 2 Oe

B = uH, So, B = 4000 X 2 = 8,000 Gauss!

Keep in mind that this flux density is at the radius of 1mm and no where else. As you progress outward through the thickness of the core, the flux density decreases so an accurate answer has to integrate through the total core thickness. Another poster posted an equation that takes this into account.

The relative permeability is a function of the flux density, B as illustrated by the B-H curve. Permeability is the slope of the B-H curve. Long before saturation occurs, the permeability drops off controlling the flux density for a given magnetization, H. This complicates the above analysis because "u" is changing and probably keeps the core from actually saturating

Also, when a magnetic material saturates, its relative permeability drops to unity. This has the effect of magnetically increasing the torroid's inner radius as though it were air. Since the dimensions in these examples are all very small, this phenomena is not particularly an issue in most cases.

cut....

Well, you're only off by 10,000, what the hell?? It's 0.8 Tesla not 0.8 Gauss. A Tesla is 10^4 Gauss. 0.8 Tesla = 8,000 Gauss

Like this?

• = wire () = core

Side view:

I ^ * ri ro * / / * / / ( (*) ) ]

=3D=3D=3D=3D

of

re

.

st

The equation gives the total flux (Webers), not the flux density (Tesla or Webers/m^2). You are using SI units so if you want to talk about flux density in gauss instead of Tesla you must multiply by

10^4. To convert Webers to Tesla (averaged over the core), divide by the cross-sectional area.

For your example:

p =3D uo*ur*I*d*ln(ro/ri)/2*pi =3D 4pi*10^-7*4000*1 amp*.01m*ln(.0013/.0003)/2pi =3D 2*10^-7*4000*1*.01*ln(.0013/.0003) =3D 0.0000117 Wb

divide by cross-section: =3D .01m*.001m =3D 10^-5 m^2

=3D 1.17 T =3D 11,700 gauss

-- Joe

Yikes.. 11.7kG I suppose all the ferrite material will saturate in the above example. The small signal inductance will be no different than an air core.

ok... How about I try to solve what toroid ID and OD is barely ok.

Say B = 4000 Gauss

0.4T = Webers/10^-5m^2

Webers = 4E-6

4E-6 = uo*ur*I*d*ln(ro/ri)/2*pi

ln(ro/ri) = 4E-6*pi*2/uo*ur*I*d

Say ur = 4000

ln(ro/ri) = 0.5

e^0.5 = 1.65

So if ri = 0.3mm then ro = 0.3mm*1.65=~ 0.5mm

The toroid got thinner??? I suppose the less core volume enveloping the wire.. the less flux density in the core.

Spray on ferrite...

D from BC myrealaddress(at)comic(dot)com British Columbia Canada

No, you forgot that you set the core's cross-sectional area at 10^-5 m^2, which constrains ro and ri.

Solving for ri:

A =3D d (ro-ri) =3D d (1.65ri - ri) =3D d*0.65ri ri =3D A/0.65d =3D 10^-5 m^2 / 0.65*10^-2m =3D 1.54mm ro=3D 1.65*1.54mm =3D 2.54 mm

As you suspected, you need a bigger core. The fact that you managed to need one with an O.D. of exactly 0.1" must be some kind of cosmic proof that you're on the right track

-- Joe

Correction: O.D. of 0.2"

So much for Karma.

-- Joe

oops.. That's right. No wonder my electronics is so crappy, my math sucks.. Grrr... :(

ok...trying again...Perhaps I'll get your result...

(uo*ur*I*d*ln(ro/ri)*(1/(2*pi) B= _____________________________ Tesla d(ro-ri)

d cancels out. ur=4000, B=0.4T

ro-ri uo*ur*I*(1/(2*pi)) 4E-7*4000*1*0.5

Is pi part of the denominator, i.e. should there be parentheses around 2*pi?

Shouldn't that be:

B = ur*uo*H

H is in terms of Amps per length B is a density

ur is unitless..A factor above free space. (ferrite)

B =urH is missing a dimension.

Instead:

B = uo*ur*H

amps B= ----- * uo * ur length

uo = free space 4*pi*E-7 amperes/meter^2

amps amps B= ----- * ------ * dimensionless length area

B = amps/length per area or amps/volume

It's B for flux density...I guess that's why it's called density. Like density of matter is weight/volume.

But B is in terms of webers/area=tesla

What I'm working on is: What are the benefits of a straight wire in ferrite. NO turns. (From a power conversion point of view.) Say 80cm of wire..

Note: No turns = the wire is not looped on the ferrite. However the wire does return to the current source and that's a loop so N=1.

D from BC myrealaddress(at)comic(dot)com British Columbia Canada

Here's a long link..

's+law+H&source=web&ots=MG4FYgt_9D&sig=yL8X8-E4xa0zHrSMK3nvLbIZIvw&hl=en&sa=X&oi=book_result&resnum=4&ct=result#PPA119,M1

Handbook of Engineering Electronics By Rajeev Bansal Page 119 is viewable in Googele books.

Flux = (uNIh/2*pi) * ln(b/a)

So yes... parentheses around 2*pi.

D from BC myrealaddress(at)comic(dot)com British Columbia Canada

Pardon the pun, but donut answer the question.

Shhh...do not give away the secret!