after a long searching for the analytic form of finite solenoid field, with the help of from :

- Edmund E. Callaghan, S. H. (1960). The Magnetic Field of a Finite Solenoid (Techical note D-465). Washington, USA: Nation Aeronautics and Space Administration.
- Jackson, J. D. (1998). Classical Electrodynamics. John Wiley & Sons, Inc.
- Milton Abramowitz, I. A. (1965). Handbook of mathematical functions : with formulas, graphs, and mathematical tables. Dover.
- NIST Digital Library of Mathematical Functions. (n.d.). Retrieved from http://dlmf.nist.gov/

we have the analytic form. Finite length Solenoid potential and field

the potential is:

the field vector is:

where

and

those are complete elliptic integral of 1st , 2nd and third kind. i quoted them in here coz the argument is a bit confusing across different references and i settle it down on the context of this page. the below is the plotted field line and intensity for the coil dimension is 1 radius and 1 length.

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elmanuelito

Sep 15, 2013@ 19:13:05I’ve posted a comment for the single coil formula where there seemed to be a factor 2 scaling problem:

https://nukephysik101.wordpress.com/2011/07/17/magnetic-field-of-a-single-coil/#comment-150

It’s possible that the formula you provide also have this problem for the solenoid. Comparison with Callaghan’s formula or formulae for semi-infinite cylinders seems to confirm my statement. But once again, please double check, and I’d love to know if I’m wrong!

jose

Oct 31, 2013@ 00:11:10How did you make this plot?

GoLuckyRyan

Nov 07, 2014@ 12:41:41To elmanuelit: yes, the factor 2 problem in here. You are right.

To Jose: I plot it with Mathematica. They have build in elliptical function and powerful plotting tools.

john compter

Feb 21, 2015@ 01:36:01Who is the author to refer to? Please be aware, fitted for publication in IEEE trans on Magnetics.

GoLuckyRyan

Sep 16, 2015@ 15:16:13If you like to quote, please refer to T.L. Tang. I did not public it in any journals, coz I think it already done for long time. I just re-discover and recalculate it. Of course, as people said, there is a sign error and factor 2 error.

nam

Feb 22, 2015@ 01:17:17You can try this one:

http://onlinelibrary.wiley.com/doi/10.1002/we.1800/abstract

John Compter

Feb 24, 2015@ 15:37:53Besides of the minus sign and somewhere a lost factor of two, world class. I check it by comparing it with an article of H.Haas, Archiv fuer Elektrotechnik, 58, 1976, giving the Br and Bz (a pity, some typo’s in the final expression). This last article gives also the force between two coaxial coils and from the same auther also the mutual induction can be found in Archiv fuer Elektrotechnik, 57, 1975

GoLuckyRyan

Sep 16, 2015@ 15:34:54Thank you! John. I know the result is not new, just not in english.

john compter

Oct 24, 2015@ 17:55:50Dear Sir/Miss,

Below you find the context I included the equation.

4. Tools 14

4.1 Biot Savart and Lorentz 14

4.2 Finite straight wire 15

..1 Field 15

..2 Force 16

..3 Force between round wires 17

4.3 Finite flat plane 18

..1 Field 18

..2 Force 21

..3 Non-cuboidal magnets 23

4.4 Infinite bar 25

..1 Field 25

..2 Force 25

4.5 Finite bar 27

..1 Field 27

..2 Force 28

4.6 Cuboidal magnet 33

..1 Field 33

..2 Force 34

4.7 Round coil or magnet 38

..1 Field 38

..2 Force 42

..3 The mutual inductance, coaxial 43

..4 Self-inductance 44

..5 Mutual inductance between non-coaxial thin wall coils 45

4.8 Self-, mutual inductance and forces of a coils with rectangular shape and cross section 49

..1 Self-inductance 49

..2 Mutual Inductance 55

..3 Force 56

..4 Neumannâs formula 58

4.9 Presence of iron 63

..5 Reluctance force 65

..6 Attraction force on iron 66

4.7 Round coil or magnet

Field

To determine the field of a circular magnet like a disc or cylinder one needs at first the field of a circular loop with

an infinitely thin conductor. One has to apply Biot-Savart in cylinder coordinates as done in [19] and [20].

It is helpful to use the full elliptic integrals of the first and second kind, defined as:

(4.7.1)

The derivation via the vector potential is described in [21]. The vector potential is derived as:

(4.7.2)

with , R as radius and I as current at (r,z) as observation location.

The current has only one component directed tangentially, so in the direction Î¸.

Consequently the same holds for the vector potential. The field components are obtained via:

(4.7.3)

(4.7.4)

As example a picture of a single loop of infinitely thin conductor carrying with 1 Amp:

Fig. 4.7.1, the field lines and flux density of a single loop with its centre as reference

The field lines are drawn in the left-hand figure. The same amount of flux can be found between two field lines.

To get the field lines in an axially symmetric case one should plot r.AÎ¸ .

Prove: from follows with the contour C enclosing surface S,

as the normal unit vector on S and the unit vector along C.

For the flux holds , so .

The vector potential only has a tangential component and this component is not a function

of the tangential position in this case of axial symmetry.

Doing an integration at a fixed value of z over a coaxial circle r leads to

Consequently one should plot lines with a constant value of to get the field lines.

The previous expressions are the base for the field of a coil with an infinitely thin wall.

By integration of the expressions for a single loop over the height of the coil one obtains the field.

The link to a round magnet is that a magnet can be considered as a number of thin coils as e.g. a

round magnet with a hole in its centre.

It is modelled according the following model as two coaxial coils with opposite current direction.

Fig. 4.7.2, a ring magnet as a set of coils

The number of ampere turns is equal to the HcJ times the magnet height.

Representation of a magnet by coils as indicated is fully allowed when:

– the relative permeability Î¼r is near 1 (as holds for NdFeB, SmCo and ferroxdure)

– the point of operation can be found in the linear part of the BH-curve

– the magnetization is uniform and unidirectional.

At first one should define the function J as the complete elliptic integral of the third kind:

(4.7.5)

The full exercise to get the field for a coil is done in [22]. A fast numerical procedure for this integral is given in [23].

(4.7.6)

with: R radius coil,

H half height of the coil,

r, z coordinates,

N number of turns,

I current,

Z=Z(i)

k=k(i) .

The elliptic integral E, K and J are not always available as âin houseâ procedure. Appendix 10 gives a Mathematica implementation.

Fig. 4.7.3 radial field at N.I=-50 amp

Fig. 4.7.4 axial field at N.I=-50 amp

Another recent source for the field of a permanent magnet ring is given in [24].

T.L. Tang published in [25] the following equation for the modified vector potential AÏ(r,z) of a cylindrical coil with N turns, the current I, the radius R and height 2H:

(4.7.7)

with

The combination of the field components and vector potential leads to the following figures.

Fig. 4.7.5 potential lines and field direction of a cylindrical coil

Fig. 4.7.6 potential lines and field direction for two cylindrical coils with opposing current direction

The field of thin wall cylindrical coils and disk and ring magnets can be described well with the previous expressions.

Expressions for the field of thick wall cylindrical coils are still under study in publications,

because a compact formulation is not found yet. Of course, Finite Element Analysis will give correct values,

but a compact analytical expression would be very nice for optimization.

One way to solve this is to cut the thick wall coils in layers and sum the contributions to get the field.

This is demonstrated with the following figures, showing the field of a coil with a circular and rectangular

cross section respectively. The number of ampere-turns equals 1000 and the number of layers equals 40.

Fig. 4.7.7 potential lines, field strength and field direction of a circular loop with a round cross section and uniform current density

Fig. 4.7.8 potential lines, field strength and field direction of a circular loop with a rectangular cross section and uniform current density

Flat coils are applied on printed circuit boards and in wireless energy transfer systems.

Analytical descriptions of the magnetic field of a disk coil have become available in the last 10 years,

but the intended advantages are lost to a high extend because its solution process becomes more

complex and/or time consuming than adding the fields of e.g. 10 concentric rings as given in Fig. 4.7.1.

As example 10 rings with each 100 Amp. with a radius between 0.1 and 0.2 m. leads to the following figure.

Fig. 4.7.9 potential lines, field strength and field direction of a disk coil with 10 wires carrying 100 Amp. each.

Eric Hamel

Mar 18, 2015@ 00:34:08I have check your equation and really don’t understand how you change in your integral the cos(phi’) to -cos(2*theta). You seem to have make a variable change of phi’ = 2*theta. Which means d(phi’) = 2 * d(theta). Which you didn’t put the 2 factor there. Also, I don’t get how you are inverting the minus sign in front of the cos(2*theta)??

Without this minus sign they are a lot of sign that are not right.

For example (a – ro * cos(phi’)) should become

(a – ro * cos( 2 * theta)) = ( a – ro ( 1- 2 * sin^2 (theta) ) )

= ( a – ro + 2 * ro * sin^2 (theta) )

Which is not your case?

How did you do it?

Thank you very much,

Best regards

GoLuckyRyan

Sep 16, 2015@ 15:28:16Hi, sorry for late reply.

I guess you are talking on page 2. Fro the transformation, please check https://nukephysik101.wordpress.com/2011/07/17/magnetic-field-of-a-single-coil/

The attached pdf explained the transformation.Sorry for confusion, I should write equation numbers, references. Hope can help