东南大学《电机内的电磁场》课件第一章

× A
And
A
Uniqueness of a Vector
Considering
× f ≡ 0 B = × A = × ( A + f ) = × A'
Uniqueness of a Vector
Gauge
A = Arbitrary value
For simplicity ( Coulomb gauge)
– 无限薄电流片
H1t H 2 t = lim Jh = K
h→ 0
That is
H1t H 2 t = K
Interface Conditions (3)
For electric field
E1t = E 2 t
And
J1n = J 2 n
D1n D2 n = ρ s
Interface Conditions: A?
Interface Conditions (1)
Flux continuity law
B1n = B2 n A1t = A2 t
Interface Conditions (2)
Ampere law
H1t t H2t dl
J h
Interface Conditions (2)
Ampere law
( H1t H 2 t )dl = dl lim Jh
h→ 0
Interface Conditions (2)
If J is finite
H1t H 2 t = lim Jh = 0
h→ 0
That is
H1t = H 2 t
Interface Conditions (2)
If J is infinite and Jh=K
ψ φ ψdv + ∫ φψdv = ∫ φ ds ∫ n v v s
2
Uniqueness of the Scalar
In ular, φ=ψ and 2ψ=0
ψ | ψ | dv = ∫ψ ds ∫ n v s
2
Two Theorems (1)
Theorem 1
–
ψ = 0 in V on S ψ = 0
Considering
B = 0
Note that
× A ≡ 0
Vector Potential
A magnetic vector potential is introduced by
B = × A
And, we define
A = 0
Helmholtz 亥姆霍兹 Theorem
The conditions to determine a vector uniquely
Considering
B = 0
We have
ψ = 0
Governing Equation of Scalar Potential
If media is homogeneous
– =Constant – satisfies Laplace (拉普拉斯） equation
ψ = ψ = 0
Vector Potential And Magnetic Energy
On the infinitely far boundary
H =0
Hence
1 A × H ds = 0 ∫ 2s
Magnetic Vector Potential In 2D Problems
Only one component
Vector Potential And Flux
(2) 等A线
– In 2D magnetic field, flux is just 等A线
A = kAz ( x , y ) Az Az i j B = × A = y x
– In 3D, no this conclusion
Vector Potential And J
– Further
– 第二类齐次边界条件
Boundary Conditions (3)
Third kind boundary…Robin
u [ α ( x , y , z )u + β ( x , y , z ) ] |Γ3 n = g( x , y , z )
Practical Boundary Conditions
We have
ψ =C =0
Two Theorems (2)
Theorem 2
ψ = 0 in V ψ on S =0 n
2
ψ = C in V
Uniqueness of the Magnetic Scalar Potential
The following Boundary value problem (BVP) has unique solution
u |Γ1 = g ( x , y , z )
– Further
g( x , y , z ) = 0
– 第一类齐次边界条件
Boundary Conditions (2)
Second kind boundary…Neumann
u |Γ2 = g ( x , y , z ) n g( x , y , z ) = 0
2
ψ = 0 in V
Two Theorems (1)
Test of Theorem 1
– By Green theorem
2
ψ | ψ | dv = ∫ψ ds = 0 ∫ n v s
ψ = 0
ψ =C
Two Theorems (1)
Considering
– ψ is continuous in V
全区
源区
Integral is limited in the source zone
Vector Potential And Magnetic Energy
1 1 Wm = ∫ H Bdv = ∫ H × Adv 2v 2v 1 1 = ∫ ( A × H )dv + ∫ A Jdv 2v 2v 1 1 = ∫ A × H ds + ∫ A J dv 2s 2v
Uniqueness of the Magnetic Scalar Potential
Let φ= 1 - 2
2φ = 0 in V on S1 φ = 0 φ =0 on S 2 n
φ =0
ψ1 = ψ 2
Homework 1. Review Chapter 1
–Pp. 3~15
A = 0
Governing Equation of Magnetic Vector Potential
Considering
× H = J
We have
×
1
× A = J
Governing Equation of Magnetic Vector Potential
When media is homogeneous
Ht=0
ψ =C
Bn=0
1 A =0 n
ψ =0 n
A=C
Boundary Conditions: Examples (1)
Field in air-gap and slot
ψ =C1
ψ =0 n
ψ =0 n
ψ =C2
Boundary Conditions: Examples (2)
Field in synchronous motor
– To simplify analysis
Vector Potential And Flux
(1) A的环量
Φ = ∫ B ds
s
= ∫ × A ds = ∫ A dl
s l
Vector Potential And Flux
Φ ds dl
∫ B ds = Φ
s
A dl = Φ ∫
l
Chapter 1
Basic Theory of Magnetostatic Field
Basic Equations
× H = J
B = 0
B = H
Scalar Potential
For the magnetic field in zerocurrent region
× H = 0 无旋场
2ψ = 0 in V ψ = ψ 0 on S1 ψ =0 on S 2 n
Uniqueness of the Magnetic Scalar Potential
Test: assume both ψ1 and ψ 2 are the solution of the BVP
2ψ = 0 in V 1 ψ 1 = ψ 0 on S1 ψ 1 on S 2 =0 n 2ψ = 0 in V 2 ψ 2 = ψ 0 on S1 ψ 2 on S 2 =0 n
2
Laplace Operator
ψ = ψ
2
ψ ψ ψ = 2 + 2 + 2 y z x
2 2 2
Scalar Potential: Meaning or Means
Has no physical meaning Just a mathematical means Purpose to use potential
– =Constant – A satisfies Poisson equation
×× A = ( A) (A) = A = J
2
Vector Laplace Operator
A = Ax i + Ay j + Az k
2 2 2 2
Vector Decomposition
Ax = J x
Note that