Design of Magnets for FFAGs; with a practical example.

1. Design of Magnets for FFAGs;with a practical example. Neil Marks, STFC- ASTeC / U. of Liverpool, Daresbury Laboratory, Warrington WA4 4AD, U.K. Tel: (44) (0)1925 603191 Fax: (44) (0)1925 603192 n.marks@stfc.ac.uk

2. Objectives • present a short overview of the design process to determine FFAG magnet pole shapes; • present, as an example, the pole design of 2 magnets for an FFAG lattice (Grahame Rees: ‘PUMPLET’) • introduce a (new?) concept for reducing both capital and running costs for certain types of FFAG magnet, where the beam is off-centre.

3. Are FFAG magnets ‘complex’? • Sometimes are complex – but not difficult! • Consideration in pole design: • what is the lattice specification – either variation on axis of By vs x, or , better still, the harmonic components of the By? • then (for the magnet designer) – what basic type of magnet is it (dipole, quadrupole, sextupole, etc). • finally, what procedure should be used to establish the pole profile? • See next slide.

4. Detailed Procedure • To determine the ‘perfect’ pole shape to generate the required field: • i) establish coefficients of Taylor series (harmonic amplitudes) up to 3rd or 4th order (or higher) to ‘fit’ the specified By(x) curve; • ii) develop the equation for scalar potential f (x,y) by summing the scalar potentials for all harmonics (equations for scalar potential given in next four slides); • iii) generate a table of (x,y) values corresponding to a fixed value of f – this is (one of) the ‘perfect’ pole shapes (not taking account of pole edge or end effects).

5. Scalar potential. • In the absence of currents and steel (ie, between poles): • H = 0; • So can define: B = -  • As: .B = 0 • Then: 2 = 0 (Laplace’s equation) • In two dimensions, in cylindrical coordinates (r,θ): •  = n (Jn r ncos n +Knr n sin n), • with n integral (1 → ∞) and Jn,Kn are funs. of geometry. Note components of flux density: Br = - n (n Jn r n-1 cos n +nKn r n-1 sin n) B = - n (-n Jn r n-1 sin n +nKn r n-1 cos n)

6. Scalar potential for a Dipole (n=1): • Cylindrical:Cartesian: •  =J1 r cos  +K1 r sin . 1 =J1 x +K1 y • Br = J1 cos  + K1 sin ; Bx = -J1 • B = -J1 sin  + K1 cos ; By = -K1 So, K1 (≠ 0) is the coefficient for vertical dipole field; (J1 = 0)

7. Scalar potential for n = 2 (a Quadrupole) • Cylindrical: Cartesian: •  = J2 r 2 cos 2 +K2 r 2 sin 2; 2 = J2 (x2 - y2)+2K2 xy • Br = 2 J2 r cos 2 +2K2 r sin 2; Bx = -2 (J2 x +K2 y) • B = -2J2 r sin 2 +2K2 r cos 2; By = -2 (-J2 y +K2 x) So K2 (≠ 0) is the coefficient for 'normal' or ‘upright’ quadrupole field; (J2 = 0) Line of constant scalar potential Lines of flux density

8. -C +C +C -C -C +C Sextupole field n=3: • Cylindrical; Cartesian: •  = J3 r3 cos 3 +K3 r3 sin 3; 3 = J3 (x3-3y2x)+K3(3yx2-y3) • Br = 3 J3r2 cos 3 +3K3r2 sin 3; Bx = -3J3 (x2-y2)+2K3yx • B= -3J3 r2 sin 3+3K3 r2 cos 3; By = -3-2 J3 xy +K3(x2-y2) -C K3 (≠ 0) is the coefficient for 'normal' or ‘right’ sextupole field; (J3 = 0) . +C +C Line of constant scalar potential -C -C Lines of flux density +C

9. Higher order fields (eg octupole) • Obtain scalar potential equation for the appropriate n in Cartesian coordinates. • eg – for octupole (n = 4) • 4 = J4(x4-6y2x2+y4) + 4K4(yx3-y3x) K4 (≠ 0) is the coefficient for 'normal' or ‘right’ octupole field; (J4 = 0) .

10. Design Exercise - FFAG Magnets for ‘Pumplet’ (*) (*) as specified by Grahame Rees.

11. By(T) and x (T) specifications

12. Magnet bd – By curve fitting • Fit: • Series: b0 + b1x + b2 x2 + b3 x3; • Coefficients: b0 = 0.04693; b1 = 2.9562 E-4; b2 = -2.9366 E-6; b3 = -1.6920 E-7; • RMS fitting error: 3.67 E-5; 8:104 of mean (need to be better for actual project).

13. bd - lines of iso-scalar potential Pairs of (x,y) and (x,-y) to give f = ± 0.300000 T mm; this gives the poles shapes.

14. bd poles and vac vessel. • What By distribution does this give? Model using OPERA 2D.

15. OPERA 2D model

16. By vs x at y = 0. How does this compare with the specified data?

17. bd -comparison of OPERA 2D with defined By • RMS error (fitting + determining potentials + OPERA FEA) : 3.75 E-5

18. Magnet BF curve fit • What sort of magnet is this? dipole/quadrupole/sextupole? See next slide.

19. Magnet BF curve fit – extended. • Extrapolated down to x = – 55 mm. • It’s a sextupole (with dipole, quadrupole and octupole components) • magnetic centre at  - 40 mm !!!

20. Magnet BF pole arrangement with vac vessel. • Poles have: • f = ± 0.35 T mm; • Note all that wasted space at -70 < x < -25? See next slide but one!

21. OPERA Model of magnet BF

22. Comparison between OPERA BY prediction and defined data; top pole at full potential

23. Solution to wasted space problem. • Side poles have: • f = ± 0.35 T mm; • BUT: • Central poles have: • = ± 0.01 T mm; • Central poles: • i) are much closer to • median line; • ii) now require only • 1/35 of the coil • excitation current. • STFC have now applied for a patent for this arrangement. f = 0.35 f = - 0.01

24. OPERA model of BF;(low potential central poles).

25. Comparison of By: between OPERA & defined data; reduced f on top pole.

26. Conclusion • We have demonstrated a combination of analytical and numeric methods that can define the ‘ideal’(*) pole geometry for any complexity of an FFAG lattice. • (*) ie for: • infinite permeability steel; • infinitely wide poles; • infinitely long magnets. } We always have these problems!