Lecture 3 - Magnets

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Lecture 3 - Magnets

ACCELERATOR PHYSICS

MT 2004

E. J. N. Wilson

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ACINST3 - E. Wilson – 3-Oct--04 - Slide 2

Lecture 3 – Magnets - Contents

Magnet types

Multipole field expansion

Taylor series expansion

Dipole bending magnet

Diamond quadrupole

Various coil and yoke designs

Power consumption of a magnet

Magnet cost v. field

Coil design geometry

Field quality

Shims extend the good field

Flux density in the yoke

Magnet ends

Superconducting magnets

Magnetic rigidity

Bending Magnet

Fields and force in a quadrupole

(3)

Components of a synchrotron

(4)

Dipoles bend the beam

Quadrupoles focus it

Sextupoles correct chromaticity

Magnet types

x By

x B

y

0 0

(5)

Multipole field expansion

φ (r ,θ)

Scalar potential obeys Laplace

whose solution is

Example of an octupole whose potential oscillates like sin 4θ around the circle

∂

²

φ

∂

^x² ⁺

∂

²

φ

∂

^y² ^{= 0 or}

1 r²

∂

²

φ

∂θ

² ⁺

1 r

∂

^r ^r

∂φ

∂

^r

⎛

⎝ ⎜ ⎞

⎠ ⎟ = 0

φ

=

φ

_nrⁿ sin n

θ

n=1

∑

∞

(6)

Taylor series expansion

φ = φ_n

n=1

∑

∞ ^r ⁿ^{sin n}^θ

Field in polar coordinates:

B_r = − ∂φ

∂r , B_θ = 1 r

∂φ

∂θ

B^r = φ_nnr ⁿ⁻¹sin nθ , B_θ = φ_nnr ⁿ⁻¹ cos nθ

B_z = B_r sin θ + B_θ cosθ

= −φ_nnrⁿ⁻¹[cosθ cos nθ + sinθsin nθ]

= φ_nnr ⁿ⁻¹ cos n( − 1)θ = φ_nnxⁿ⁻¹ (when y = 0)

Taylor series of multipoles To get vertical field

Fig. cas 1.2c

Octupole

Sext

Quad

Dip.

! ...

3 2 1

! 1

1

...

4 . 3

. 2

.

3 3 3 2

2 2 0

3 4

2 3

2 0

∂ + + ∂

∂ + ∂

=

+ +

=

x x x B

x x B

x B B

x x

x B

z z

z

φ φ φ φ

(7)

Dipole bending magnet

(8)

Diamond dipole

(9)

Diamond quadrupole

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Various coil and yoke designs

''C' Core:

Easy access Less rigid

‘H core’:

Symmetric;

More rigid;

Access problems.

''Window Frame'

High quality field;

Major access problems Insulation thickness

μ >>

g λ

1

NI/2

(11)

Power consumption of a magnet

= σ A R N

2l 2

σ μ

A

B R g

I

o 2

2 2

Power = = l

μ >>

g λ

1

NI/2

B_air = μ₀ NI / (g + λ/μ);

g, and λ/μ are the 'reluctance' of the gap and iron. A is the coil area, l is the length σ is

conductivity

Approximation ignoring iron reluctance (λ/μ << g ):

NI = B g /μ₀

(12)

Magnet cost v. field

Variation of cost with field B

0 1 2 3 4 5 6 7 8 9

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1

Field B (T)

Coil costs Running costs Pole costs Yoke costs Total

Pressure from need to save real estate

Constraint from saturation or critical current, synchrotron radiation

(13)

0.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0

Current density j

Lifetime cost

running

Coil design geometry

Standard design is rectangular copper (or aluminium) conductor, with cooling water tube. Insulation is glass cloth and epoxy resin.

Amp-turns (NI) are determined, but total copper area (Acopper) and number of turns (N) are two degrees of freedom and need to be decided.

Current density:

j = NI/A_copper Optimum j

determined from economic criteria.

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Field must be flat to 1 part per 10000

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Shims extend the good field

To compensate for the non-infinite pole, shims are added at the pole edges. The area and shape of the shims determine the amplitude of error harmonics which will be present.

Shim

A Quadrupole

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Flux density in the yoke

(17)

Magnet ends

The 'Rogowski' roll-off:

y = g/2 +(g/π) exp ((πx/g)-1);

Three dimensional computer calculation

1 0

-1 -2

0 1 2

(18)

Superconducting magnets

(19)

dp

dt = p dθ

dt = p dθ ds

ds

dt = p ρ

ds dt

= ev × B = eds dt B

Magnetic rigidity

1

ρ ^{= d} θ ds

Bρ

( )

^[^T.m] = pc eV[ ]

c m.s

[

⁻¹

]

= 3.3356 pc

( )

^[^GeV^]

Bρ

( )

= p

e = pc

ec = βE

ec = βγE₀

ec = m⁰c

e

( )

βγ

d

θ

(20)

Bending Magnet

F Effect of a uniform bending (dipole) field

F If then

F Sagitta

θ << π / 2

( )

=

( )

ρ

= ρ

θ B

B 2 2 2

sin l l

( )

ρ

≈

θ B

B 2

l

( )

16 2 16

cos 2 1

2 = θ

≈ ρθ θ

ρ −

± l

(21)

Fields and force in a quadrupole

No field on the axis Field strongest here

B ∝ x

(hence is linear) Force restores Gradient

Normalised:

POWER OF LENS Defocuses in

vertical plane ( )

^. ^y ¹

k B

B

ρ

x f

= − ∂ =

∂ l l

( )

¹ ^. ^y

k B

B

ρ

x

= − ∂

∂

x B_y

∂

(22)

Summary