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The magnetic potential of axially symmetric fields
NUCLEAR INSTRUMENTS
33 (I965) 68-70;
AND METHODS
© NORTH-HOLLAND
PUBLISHING
CO.
THE MAGNETIC POTENTIAL OF AXIALLY SYMMETRIC FIELDS G. BOSI Istituto di FiMca "Guglielmo Marconi", UniversiM degli Studi di Roma
Received 18 August 1964 A general expression for the magnetic potential of axially symmetric fields is carried out: the results are rigorously correct, if no currents are present. The problem is solved by a harmonical function, whose axial derivative fits the given field form in the median plane: this coincidence is imposed up to second order
terms. The given solution succeeds in the range 0 < r < 2ro, r being the distance from the symmetry axis and r0 the central orbit radius. Formulae are checked by comparison with an analytical method proposed by Gersch, which yields approximate results.
1. Introduction I f the influence of exciting coils is neglected, the magnetic field within the gap o f an electromagnet is defined by
2. The magnetic potential Gersch 1) carried out an approximate expression for V: it succeeds at large distances from the symmetry axis. Thus Gersch's potential is determined by (6), (7) and
curl H = 0,
(1)
div H = O.
(2)
~2 V Or 2
Due to (1) it is possible to introduce the magnetic potential V in such a way that H = grad V. Combination o f (2) and (3) yields (4)
where A is the Laplace's operator. We will consider axially symmetric fields falling off radially. I f cylindrical co-ordinates r, 0, z are used, the field 0-component is zero everywhere, while the radial one vanishes in the median plane (z = 0). Here the field form is given: H(r)=
Ho
]
r-,'o
l+at\
ro /
+ .2 \
r-'ol ro /
]
+ ....
(7)
@2V @z2
(ll)
+
1
~V
~2V
~--
= 0,
/-/('1) = /4o[1 + a , , + ,,2,7 ~ + . . . ] , (aV/@{)¢=0
=
roll(q),
(@V/@q)¢= o :
0.
(12) (13) (14) (15)
2".1. FIRST CASE: 2a2 ,-}-at -----0 Examples of such fields are given by homogeneous field (al = a2 . . . . . 0) and double-focusing field with al = - ½ , a2 = ¼. As
I f the magnetic susceptivity is supposed to be infinite, the magnet pole surfaces are coincident with a pair o f equipotential surfaces being symmetric to the median plane. That is why it should be profitable to integrate eq. (4) according to (6) and (7). As @V/@O = O, eq. (4) may be written down as @2V 1 @V --+---+-@r2 r @r
(10)
Then (8), (5), (6), (7) become respectively:
where ro is the central orbit radius. Then V should be a harmonical function and its derivatives should satisfy the following conditions:
(@Vl@r)z=o -- 0.
(9)
= z / r o.
~,2V
(6)
= O.
rl = (r - ro)/r o,
(5)
(aW/aZ)z=O = H(r),
~z 2
Our purpose consists o f finding an exact solution o f eq. (8), which should succeed in a suitably large range a r o u n d ro, for this is the most important region. Our solution shall fit the field (5) up to second order terms : in practice it is quite difficult to design such a magnet as to reach a higher accuracy. We shall distinguish three cases: 2a 2 + a t = O, 2a2 + al > 0, 2a 2 + a t < 0. Let us introduce adimensional co-ordinates
(3)
a v = 0,
~2 V
- - + -
In (1 + q) = t / - ½r/2 + . . . .
(16)
we will try such a kind o f solution as = 0.
(8)
vQ/,{) = f ( { ) [ l + 2 In (1 + q)] 68
(17)
THE
MAGNETIC
POTENTIAL
OF
with 2 = constant. F r o m (12), (13), (14), (15) we obtain :
2 = a 1, f(4) = Horo4,
SYMMETRIC
69
FIELDS
2 , . . . ) being given by (25a) and (25b), converges uniformly in the range - 1 < v/ < 1. 2.3. THIRD CASE: 2a 2 -}-a 1 < 0
We will try
whence, for vt > - 1 (say r > 0), V(q,4) -- Horo4[1 + al In (1 + q)].
V(q, 0 = Horoh(q ) sinh re,
(18)
We remark that, according to (18), equipotential surfaces of a homogeneous field (al = 0) are planes parallel to the median plane (4 = const.), as is well known otherwise. Further the tangent in (0, 40) to the meridian cross-section of an equipotential surface is given by 4 = 40( 1 -- alq). For a I =
AXIALLY
(19)
1. -- ~.
(27)
where v is a positive constant. By substituting into (12) we obtain h"(v/) + ~-~--~ h'(q) + v2h(r/) = 0,
(28)
which is equivalent to (1 + q)h"(q) + h'(v/) + v2(1 + v/)h(r/) = 0
(29)
for v/ > - 1. We take 4 = ½40(2 + r/),
(20) h(t/) = v k=o -
which is usually taken 2) as the shape of pole pieces producing al = - ½, a 2 = 1.
ckrt k.
(30)
We find : 2c2 + c I + veco = 0,
2.2. SECOND CASE: 2az @at > 0 We will try
ck+z(k+ 1)(k+2)+Ck+ (21)
where # is a positive constant. By substituting into (12) we obtain 1
1) 2
+v2(ck+Ck-1)
= 0,
(k = 1,2,3 . . . . ).
V(q,4) = Horog(q) sin #4,
g"(v/) + ~
l(k+
(31a) (318)
Again : Co = 1, e I = a1, c2 = ax, v = ( - 2 a z - a l )
~-.
Hence: g'(,) -/~2g(q) = 0.
(22)
V(vl,¢) = H o r o ( - 2az - al) -*~ Ck~k sinh (4(-- 2a2 -- a,)~).
For q > - 1, eq. (22) is equivalent to
(32)
k
(I + q)g"(q) + g'(,) -/~2(1 + q)g(q) = 0.
(23)
We take
l ~ bkrlk. g(") = 7
(24)
It is easy to demonstrate: 2b2 + bl - / * 2 b o = 0,
(25a)
bk +2(k + 1)(k + 2) + bk +l(k + 1) 2 - Iz2(bk + bk- 1) = 0. (k = 1,2,3 . . . . ).
(25b)
(13), (14), (15), (25a) are satisfied by bo = 1, bl = al, b2 = a2,/* = ( 2 a 2 + a a ) ~. Hence : oo
V(q,4)=Horo(2a2+al)-÷(k~=obkVlk)sin(4(2a2+at)}). (26) It is easily seen that the series ~°=o bkq k, bk(k= O, 1,
The series ~ = o Ckqk is endowed with the same convergence radius as ~ ° = o bk~ k. 3. C o n c l u s i o n s
For sake of synthesis we remark that both (26) and (32) just approach (18) as 2a2 + at approaches zero. Also it is easily seen that (26) is just equivalent to (32) for 2az + a I < 0. Thus we may infer that there is but one solution of eq. (12) [-under conditions (13), (14), (15)] in the range - 1 < v/ < 1: it is given by (26) independently of (2a 2 + al)-values. Finally we will calculate the a m o u n t of discrepancies between our formulae and Gersch's 1) results. According to this author the magnetic potential is roughly given by V(q,¢) = (ro/i) Im[-iH(o))dco ],
(33)
where ~o = q + i4 and Im [ ~ H(w) d o ] is the imaginary part of ~ H(~o)&o. In the third approximation (33) yields :
70
G. BOSI v(~,¢)
(34)
Hz(rl,~) = H o [ l + a l r l + a z r l 2 - ( a 2 + 1 a l ) ~ 2 ]. (37)
1 ~V _ Ho[1 + a,rl + a2(r/2 _ ~2)j. (35)
The held (35) differs from (37) by second order terms: however (37) turns out to be the correct field form, see ref. 3).
= ro/-/o¢[1 + a l . + a 2 ( , 2 - ~g 1.2 )j,
whence
In the same a p p r o x i m a t i o n (18), (26) and (32) give: V(t/,~) = roHo~[1 + a , r / + a 2 r / 2 - ½ ( a 2 + ½ a l ) ~ 2] and
References
(36)
1) H. Gersch, Nucl. Instr. and Meth. 13 (1961) 297. 2) N. Svartholrn, Ark. Fys. 2 (1950) 115. 3) E. Persico and C. Geoffrion, Rev. Sci. Instr. 21 (1950) 959.
33 (I965) 68-70;
AND METHODS
© NORTH-HOLLAND
PUBLISHING
CO.
THE MAGNETIC POTENTIAL OF AXIALLY SYMMETRIC FIELDS G. BOSI Istituto di FiMca "Guglielmo Marconi", UniversiM degli Studi di Roma
Received 18 August 1964 A general expression for the magnetic potential of axially symmetric fields is carried out: the results are rigorously correct, if no currents are present. The problem is solved by a harmonical function, whose axial derivative fits the given field form in the median plane: this coincidence is imposed up to second order
terms. The given solution succeeds in the range 0 < r < 2ro, r being the distance from the symmetry axis and r0 the central orbit radius. Formulae are checked by comparison with an analytical method proposed by Gersch, which yields approximate results.
1. Introduction I f the influence of exciting coils is neglected, the magnetic field within the gap o f an electromagnet is defined by
2. The magnetic potential Gersch 1) carried out an approximate expression for V: it succeeds at large distances from the symmetry axis. Thus Gersch's potential is determined by (6), (7) and
curl H = 0,
(1)
div H = O.
(2)
~2 V Or 2
Due to (1) it is possible to introduce the magnetic potential V in such a way that H = grad V. Combination o f (2) and (3) yields (4)
where A is the Laplace's operator. We will consider axially symmetric fields falling off radially. I f cylindrical co-ordinates r, 0, z are used, the field 0-component is zero everywhere, while the radial one vanishes in the median plane (z = 0). Here the field form is given: H(r)=
Ho
]
r-,'o
l+at\
ro /
+ .2 \
r-'ol ro /
]
+ ....
(7)
@2V @z2
(ll)
+
1
~V
~2V
~--
= 0,
/-/('1) = /4o[1 + a , , + ,,2,7 ~ + . . . ] , (aV/@{)¢=0
=
roll(q),
(@V/@q)¢= o :
0.
(12) (13) (14) (15)
2".1. FIRST CASE: 2a2 ,-}-at -----0 Examples of such fields are given by homogeneous field (al = a2 . . . . . 0) and double-focusing field with al = - ½ , a2 = ¼. As
I f the magnetic susceptivity is supposed to be infinite, the magnet pole surfaces are coincident with a pair o f equipotential surfaces being symmetric to the median plane. That is why it should be profitable to integrate eq. (4) according to (6) and (7). As @V/@O = O, eq. (4) may be written down as @2V 1 @V --+---+-@r2 r @r
(10)
Then (8), (5), (6), (7) become respectively:
where ro is the central orbit radius. Then V should be a harmonical function and its derivatives should satisfy the following conditions:
(@Vl@r)z=o -- 0.
(9)
= z / r o.
~,2V
(6)
= O.
rl = (r - ro)/r o,
(5)
(aW/aZ)z=O = H(r),
~z 2
Our purpose consists o f finding an exact solution o f eq. (8), which should succeed in a suitably large range a r o u n d ro, for this is the most important region. Our solution shall fit the field (5) up to second order terms : in practice it is quite difficult to design such a magnet as to reach a higher accuracy. We shall distinguish three cases: 2a 2 + a t = O, 2a2 + al > 0, 2a 2 + a t < 0. Let us introduce adimensional co-ordinates
(3)
a v = 0,
~2 V
- - + -
In (1 + q) = t / - ½r/2 + . . . .
(16)
we will try such a kind o f solution as = 0.
(8)
vQ/,{) = f ( { ) [ l + 2 In (1 + q)] 68
(17)
THE
MAGNETIC
POTENTIAL
OF
with 2 = constant. F r o m (12), (13), (14), (15) we obtain :
2 = a 1, f(4) = Horo4,
SYMMETRIC
69
FIELDS
2 , . . . ) being given by (25a) and (25b), converges uniformly in the range - 1 < v/ < 1. 2.3. THIRD CASE: 2a 2 -}-a 1 < 0
We will try
whence, for vt > - 1 (say r > 0), V(q,4) -- Horo4[1 + al In (1 + q)].
V(q, 0 = Horoh(q ) sinh re,
(18)
We remark that, according to (18), equipotential surfaces of a homogeneous field (al = 0) are planes parallel to the median plane (4 = const.), as is well known otherwise. Further the tangent in (0, 40) to the meridian cross-section of an equipotential surface is given by 4 = 40( 1 -- alq). For a I =
AXIALLY
(19)
1. -- ~.
(27)
where v is a positive constant. By substituting into (12) we obtain h"(v/) + ~-~--~ h'(q) + v2h(r/) = 0,
(28)
which is equivalent to (1 + q)h"(q) + h'(v/) + v2(1 + v/)h(r/) = 0
(29)
for v/ > - 1. We take 4 = ½40(2 + r/),
(20) h(t/) = v k=o -
which is usually taken 2) as the shape of pole pieces producing al = - ½, a 2 = 1.
ckrt k.
(30)
We find : 2c2 + c I + veco = 0,
2.2. SECOND CASE: 2az @at > 0 We will try
ck+z(k+ 1)(k+2)+Ck+ (21)
where # is a positive constant. By substituting into (12) we obtain 1
1) 2
+v2(ck+Ck-1)
= 0,
(k = 1,2,3 . . . . ).
V(q,4) = Horog(q) sin #4,
g"(v/) + ~
l(k+
(31a) (318)
Again : Co = 1, e I = a1, c2 = ax, v = ( - 2 a z - a l )
~-.
Hence: g'(,) -/~2g(q) = 0.
(22)
V(vl,¢) = H o r o ( - 2az - al) -*~ Ck~k sinh (4(-- 2a2 -- a,)~).
For q > - 1, eq. (22) is equivalent to
(32)
k
(I + q)g"(q) + g'(,) -/~2(1 + q)g(q) = 0.
(23)
We take
l ~ bkrlk. g(") = 7
(24)
It is easy to demonstrate: 2b2 + bl - / * 2 b o = 0,
(25a)
bk +2(k + 1)(k + 2) + bk +l(k + 1) 2 - Iz2(bk + bk- 1) = 0. (k = 1,2,3 . . . . ).
(25b)
(13), (14), (15), (25a) are satisfied by bo = 1, bl = al, b2 = a2,/* = ( 2 a 2 + a a ) ~. Hence : oo
V(q,4)=Horo(2a2+al)-÷(k~=obkVlk)sin(4(2a2+at)}). (26) It is easily seen that the series ~°=o bkq k, bk(k= O, 1,
The series ~ = o Ckqk is endowed with the same convergence radius as ~ ° = o bk~ k. 3. C o n c l u s i o n s
For sake of synthesis we remark that both (26) and (32) just approach (18) as 2a2 + at approaches zero. Also it is easily seen that (26) is just equivalent to (32) for 2az + a I < 0. Thus we may infer that there is but one solution of eq. (12) [-under conditions (13), (14), (15)] in the range - 1 < v/ < 1: it is given by (26) independently of (2a 2 + al)-values. Finally we will calculate the a m o u n t of discrepancies between our formulae and Gersch's 1) results. According to this author the magnetic potential is roughly given by V(q,¢) = (ro/i) Im[-iH(o))dco ],
(33)
where ~o = q + i4 and Im [ ~ H(w) d o ] is the imaginary part of ~ H(~o)&o. In the third approximation (33) yields :
70
G. BOSI v(~,¢)
(34)
Hz(rl,~) = H o [ l + a l r l + a z r l 2 - ( a 2 + 1 a l ) ~ 2 ]. (37)
1 ~V _ Ho[1 + a,rl + a2(r/2 _ ~2)j. (35)
The held (35) differs from (37) by second order terms: however (37) turns out to be the correct field form, see ref. 3).
= ro/-/o¢[1 + a l . + a 2 ( , 2 - ~g 1.2 )j,
whence
In the same a p p r o x i m a t i o n (18), (26) and (32) give: V(t/,~) = roHo~[1 + a , r / + a 2 r / 2 - ½ ( a 2 + ½ a l ) ~ 2] and
References
(36)
1) H. Gersch, Nucl. Instr. and Meth. 13 (1961) 297. 2) N. Svartholrn, Ark. Fys. 2 (1950) 115. 3) E. Persico and C. Geoffrion, Rev. Sci. Instr. 21 (1950) 959.