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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013, Electromagnetics and Applications
Prof. Markus Zahn, November 8 & 10, 2005
Lecture 16 & 17: Electroquasistatic and Magnetoquasistatic Forces
I. EQS Energy Method of Forces
a) Circuit Point of View
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
q = C (ξ ) v
i=
dC ( ξ )
dq
d
dv
⎡⎣C ( ξ ) v ⎤⎦ = C ( ξ )
=
+v
dt
dt
dt
dt
= C (ξ)
Pin = vi = v
dv
dC dξ
+v
dt
dξ dt
d
dυ
dC dξ
⎡C ( ξ ) v ⎤⎦ = C ( ξ ) v
+ v2
dt ⎣
dt
dξ dt
= C (ξ)
=
=
d
dt
d
dt
1 2 dC dξ
⎡1
2⎤
⎢ 2 C ( ξ ) v ⎥ + 2 v dξ dt
⎣
⎦
dW
dt
W=energy
storage
6.013, Electromagnetics and Applications
Prof. Markus Zahn
⎛1 2⎞
2 dC dξ
⎜ 2 v ⎟ + v dξ dt
⎝
⎠
+
fξ
dξ
dt
mechanical power
(force × velocity)
Lecture 16 & 17
Page 1 of 23
W=
1
1
dC
C ( ξ ) v2 , fξ = v2
2
2
dξ
=
1 q2 dC
1
d ⎛ 1 ⎞
= − q2
⎜
⎟
2
2 C ( ξ) dξ
2
dξ ⎜⎝ C ( ξ ) ⎟⎠
b) Energy Point of View
vi = v
dq dWe
dξ
=
+ fξ
dt
dt
dt
vdq = dWe + fξ dξ ⇒ dWe = vdq − fξ dξ
fξ = −
∂We
∂ξ
;v=
q= cons tan t
∂We
∂q
ξ = cons tan t
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
0
We = −
∫
fξ dξ
q=0
v=
+
∫
vdq ξ = cons tan t
q
C (ξ)
∫
We =
ξ =cons tan t
f=−
∂we
∂ξ
q
1 q2
dq =
2 C (ξ)
C (ξ)
=−
q = cons tan t
1 2 d ⎛ 1 ⎞ 1 q2 dC ( ξ )
q
⎜
⎟=
2
dξ ⎜⎝ C ( ξ ) ⎟⎠ 2 C2 ( ξ ) dξ
=
6.013, Electromagnetics and Applications
Prof. Markus Zahn
1 2 dC ( ξ )
v
2
dξ
Lecture 16 & 17
Page 2 of 23
II. Forces In Capacitors
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
σs = +εEx =
+εv
x
q = σs A = εEx A =
(Lower electrode)
εvA
x
= C ( x ) v
C (x) =
εA
x
From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission.
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17 Page 3 of 23 a) Coulombic force method on upper electrode:
1
1
1 εv2
fx = σsEx A = − εE2x A = −
A
2
2
2 x2
1
because E in electrode=0, E outside electrode = Ex
2
Take average Energy method: C ( x ) =
εA
x
fx =
1 2 dC 1 2
d ⎛ 1
⎞
1 v2εA
v
= v εA
⎜ ⎟=−
2
dx
2
dx ⎝ x ⎠
2 x2
v=
q
qx
1 ε A q2 x2
1 q2
⇒ fx = −
=
=−
2 x2 ε 2A 2
2 εA
εA
C (x)
b)
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
ε A
1
1
1
εA
=
; Ca = 0 , Cb =
+
C ( ξ ) Ca Cb
b
ξ
=
ξ
b
+
ε0 A εA
=
ε ξ + ε0b
ε ε0A
1
− q2
1 2 d ⎛ 1 ⎞
d
1 q2
fξ = − q
ε
ξ
+
ε
b) = −
⎜
⎟= 2
(
0
ε ε0 A dξ
2
dξ ⎜⎝ C ( ξ ) ⎟⎠
2 ε0 A
fξ =
1 2 d
1
d ⎡ ε ε0 A ⎤
1
v
C ( ξ ) = v2
⎢
⎥=−
2
dξ
2
dξ ⎣ ε ξ + ε0b ⎦
2
(
)
6.013, Electromagnetics and Applications
Prof. Markus Zahn
v2ε2ε0 A
( ε ξ + ε0b )
2
Lecture 16 & 17
Page 4 of 23
III. Energy Conversion Cycles
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17 Page 5 of 23 0
∫ vdq = ∫ dw
e
+
∫ f dξ
ξ
∫ vdq = ∫ f dξ
ξ
∫ vdq, ∫ f dξ > 0
ξ
Electric energy in, mechanical energy out.
∫ vdq, ∫ f dξ < 0
ξ
Electric power out, mechanical energy in.
B
D
1
∫ vdq = ∫ vdq + ∫ vdq = 2 C (0) V
A
2
0
C
−
1
C (L ) V 2
2
C ( 0 ) V0 = C (L ) V
∫
vdq =
C (0)
C (L )
∫
⎡
C (L ) C ( 0 ) ⎤ 1
1
⎥ = C ( 0 ) V2
C ( 0 ) V02 ⎢1 −
0
⎢
2
C 2 (L ) ⎥ 2
⎣
⎦
⎛
=
ε A ⎜L + b
⎝
b
⎡
C (0) ⎤
⎢1 −
⎥
C (L ) ⎥⎦
⎢⎣
ε0 ⎞
ε ⎟⎠
(ε A )
0
⎡
⎢
1
2 ⎢
vdq = C ( 0 ) V0 1 −
⎢
2
⎢
⎢⎣
ε0 ⎞ ⎤
⎛
ε
⎜L + b
ε ⎟⎠ ⎥⎥
1
⎝
= − C ( 0 ) V02
⎥
ε0b
2
⎥
⎥⎦
ε L
< 0 (electric energy out)
ε0 b
∫ fdξ = −f L
0
2
2
⎡ εA ⎤
1 q2
1 C ( 0 ) V0
1
f0 = +
=+
= + C ( 0 ) V02 ⎢
⎥
ε0 A
2 ε0 A
2
2
⎣⎢ bε0 A ⎥⎦
1
∫ fdξ = − 2 C (0) V
2
0
εL
=
ε0b
∫ vdq
∫ fdξ < 0 ⇒ mechanical energy out is negative means mechanical energy is put
in
Mechanical energy is converted to electrical energy
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 6 of 23
IV. Force on a Dielectric Material
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
C (ξ) =
fξ =
=
ε0 (b − ξ ) c
a
+
εξc
a
1 2 dC ( ξ )
v
2
dξ
1 2c
v
( ε − ε0 )
2
a
In equilibrium:
Mass density
1 2c
fξ = v
( ε − ε0 ) = ρ g ξ a c 2
a
fluid weight
ξ=
2
1 v ( ε − ε0 )
2
ρga2
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 7 of 23
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
a → αr
2
1 v ( ε − ε0 )
ξ=
2 ρgα2r2
V. Physical Model of Forces on Dielectrics
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
(
) ()
f dipole = q ⎡⎢E r + d − E r ⎤⎥
⎣
⎦
()
() ()
= q ⎡⎢E r + d i ∇ E r − E r ⎤⎥
⎣
⎦
=q
(
(
)
i∇ E
)
= pi∇ E
Kelvin force
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 8 of 23
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17 Page 9 of 23 VI. MQS Energy Method of Forces
A. Circuit Approach
v=
dL ( ξ )
dλ
d
di
⎡L
=
( ξ ) i⎤⎦ = L ( ξ ) + i
⎣
dt
dt
dt
dt
p = vi = L ( ξ ) i
= L (ξ)
vi=
dL ( ξ )
di
+ i2
dt
dt
d ⎛1 2 ⎞ 2 dL ( ξ )
i +i
dt ⎜⎝ 2 ⎟⎠
dt
1 2 dL ( ξ )
⎡1
2⎤
⎢ 2 L (ξ ) i ⎥ + 2 i
dt
⎣
⎦
=
d
dt
=
d ⎡1
⎤ 1 dL ( ξ ) dξ
L ( ξ ) i2 ⎥ + i2
⎢
dt ⎣ 2
dξ dt
⎦ 2
dWm
dξ
1
+ fξ
⇒ Wm = L ( ξ ) i2 ,
dt
dt
2
λ = L ( ξ) i ⇒ fξ =
=
fξ =
1 2 dL ( ξ )
i
2
dξ
1 2 dL ( ξ )
i
2
dξ
1 λ 2 dL ( ξ )
2 L2 ( ξ ) dξ
=−
1 2 d ⎡1
⎤
λ
2
dξ ⎢⎣ L ( ξ ) ⎥⎦
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 10 of 23
B. Energy Method
vi = i
dWm
dλ
dξ
+ fξ
=
⇒ dWm = i d λ − fξ
d ξ
dt
dt
dt
fξ = −
∂Wm
∂ξ
,
i=
∂Wm
∂λ
λ = constant
ξ=constant
0
Wm =
∫
λ =0
fξ =
i dλ
ξ = cons tan t
i=
Wm =
∫
− fξ d ξ +
λ
L (ξ )
λ
λ2
dλ =
L (ξ )
2 L (ξ )
ξ = cons tan t
∫
−∂Wm
∂ξ
=−
⎞
1 2 d ⎛ 1
λ
⎜
⎟
2
dξ ⎝ L ( ξ ) ⎠
=−
1 2⎛
1 ⎞ dL ( ξ )
λ ⎜− 2
⎟⎟
⎜
2
⎝ L ( ξ ) ⎠ dξ
λ = cons tan t
=
6.013, Electromagnetics and Applications
Prof. Markus Zahn
1 2 dL ( ξ )
i
2
dξ
Lecture 16 & 17
Page 11 of 23
VII. Force on a Wire over a Perfectly Conducting Plane
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
L (ξ) =
⎡ξ
µ 0D
ln ⎢ +
2π
⎢R
⎣
2
⎤
⎛ξ⎞ ⎥
⎜ R ⎟ − 1⎥
⎝ ⎠
⎦
[See Haus & Melcher p. 343,
take ½ of Eq. (12) which is the
inductance between 2 cylinders]
A. Energy Method fξ =
µ i2 D
1 2 dL ( ξ)
i
= 0
2
dξ
4 πR
1
( ξR )
2
−1
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
B. Method of Images Approach with Lorentz Force
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 12 of 23
⎛ µ 0i ⎞
µ i2 D
µ 0i2 D
fξ = iD ⎜
= 0
=
⎟
⎜ 2π (2a ) ⎟
4 πa
4π ξ 2 − R 2
⎝
⎠
C. Demonstration: Steady State Magnetic Leviation
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
VIII. One Turn Loop
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 13 of 23
From Electromagnetic Field Theory: A Problem Solving Approach,
by Markus Zahn, 1987. Used with permission.
Hz =
I
,
D
Φ = µ 0H z x l , L ( x ) = Φ
=
I
=
µ0 x l
D
µ0 x l
I
D
A. Energy Method
fx =
1 2 dL ( x )
1 µ l
I
= I2 0
2
dx
2
D
B. Lorentz Force Law
f=
∫ J×B
dV
V
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 14 of 23
Model surface current K y =
Jy =
I
Dδ
∇ ×H = J ⇒
fx =
I
as volume current of small thickness δ
D
∫J
y
∂Hz
I
I
⇒ Hz = −
= − Jy = −
( x − (ξ + δ ))
∂x
Dδ
Dδ
µ 0 Hz
dx dy dx
V
ξ+ δ
⎛ −µ 0 I ⎞
⎜
⎟ ( x − ( ξ + δ ) ) l D dx
⎝ D δ ⎠
=
I
Dδ
x=ξ
=
⎤
−µ 0 I2 l ⎡ x 2
− (ξ + δ ) x ⎥
⎢
2
Dδ ⎣ 2
⎦
∫
ξ+δ
x=ξ
2
⎤
−µ 0 I2 l ⎡ ( ξ + δ )
ξ2
2
⎢
⎥
=
−
−
ξ
+
δ
+
ξ
ξ
+
δ
(
)
(
)
2
2
Dδ 2 ⎢
⎥⎦
⎣
=
⎤
−µ 0 I2 l ⎡ 1
ξ2
2
+ ξδ ⎥
⎢− (ξ + δ ) +
2
2
Dδ ⎣ 2
⎦
=
−µ 0 I2 l ⎡ 1 2 ⎤
− δ
Dδ 2 ⎢⎣ 2 ⎥⎦
=+
1 µ 0 I2 l
⇒ f=
2 D
1
∫ 2 K ×B
dS
S
½ comes from integrating uniform volume
current over small thickness δ
General formula: f = ∫ K × B av dS
S
0
For our case: B av
B
+ B air
1
= metal
= B air
2
2
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 15 of 23
IX. Lifting of Magnetic Fluid
A. Energy Method Approach
H=
Ni
S
Φ = H ⎡µξ
⎣ + µ 0 (l − ξ ) ⎤⎦ d
=
Nd
⎡ µξ + µ 0 (l − ξ ) ⎤⎦ i
s ⎣
λ = NΦ =
L (ξ) =
fξ =
N2 d
⎡ µξ + µ 0 (l − ξ ) ⎤⎦ i
s ⎣
λ
N2 d
⎡µξ + µ 0 (l − ξ ) ⎤⎦
=
i
s ⎣
1 2 dL
1 N2 i2 d
i
=
(µ − µ0 )
2 dξ
2
s
fξ = ρm g h d s =
h=
1 N2 i2 d
2 s2 ρm g d
1 N2 i2 d
(µ − µ0 )
2
s
(µ − µ0 )
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 16 of 23
B. Magnetization force
(
)
FX = µ 0 M i ∇ Hx
∂Hx ⎤
∂Hx
∂Hx
⎡
= µ 0 ⎢Mx
+ My
+ Mz
∂x
∂y
∂ z ⎥⎦
⎣
∂
=0
∂z
∇ ×H = J = 0 ⇒
∂Hy
∂Hx
=
∂y
∂x
∂Hy ⎤
⎡
∂Hx
FX = µ0 ⎢Mx
+ My
⎥
∂x
∂x ⎦
⎣
⎛ µ
⎞
− 1⎟ H
B = µH = µ0 H + M ⇒ M = ⎜
µ
⎝ 0
⎠
(
)
⎡⎛ µ
∂Hy ⎤
⎞
⎞
∂Hx ⎛ µ
Fx = µ0 ⎢⎜
− 1 ⎟ Hx
+⎜
− 1 ⎟ Hy
⎥
∂x ⎝ µ0
∂x ⎥⎦
⎢⎣⎝ µ0
⎠
⎠
⎛ µ
⎞ ∂ ⎡1
2
2 ⎤
= µ0 ⎜
− 1⎟
⎢ 2 Hx + Hy ⎥
µ
∂x
⎣
⎦
⎝ 0
⎠
(
)
fx = ∫ Fx dx dy dz
=
=
=
=
(µ − µ0 )
2
h
d
∫ ∫ ∫
x = −∞ y =0 z =0
( µ − µ 0 ) ds
2
(µ − µ0 ) d s
2
s
(H
2
x
∂
Hx 2 + Hy 2 dx dy dz
∂x
(
+ Hy 2
)
)
h
x = −∞
N2i2
s2
1
N2i2
(µ − µ0 ) d
2
s
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 17 of 23
X. Magnetic Actuator
∫ H i ds = H ( x + a ) + H ( a − x ) = N i
1
2
11
+ N2i2 C
µ0H1 A1 = µ0H2 A2 ⇒ H1 =
H2 A2
A1
⎡
A ⎤
H2 ⎢( a − x ) + ( a + x ) 2 ⎥ = N1i1 + N2i2
A1 ⎦
⎣
H2
=
(N1i1 + N2i2 ) A
1
A1 ( a − x ) + ( a + x ) A2
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17 Page 18 of 23 H1 =
(N1i1 + N2i2 ) A2
A1 ( a − x ) + ( a + x ) A2
λ1 = N1 µ0 H1 A1 =
µ0 N1 A1 A2 (N1i1 + N2i2 )
λ2 = N2 µ0 H2 A2 =
A1 ( a − x ) + ( a + x ) A2
µ0 N2 A1 A2 (N1i1 + N2i2 )
A1 ( a − x ) + ( a + x ) A2
λ1 = L1 ( x ) i1 + M ( x ) i2
λ2 = M ( x ) i1 + L 2 ( x ) i2
L1 ( x ) =
µ0 A1 A2 N12
µ0 A1 A2 N22
µ0 A1 A2 N1 N2
; L2 ( x ) =
; M (x) =
A1 ( a − x ) + ( a + x ) A2
A1 ( a − x ) + ( a + x ) A2
A1 ( a − x ) + ( a + x ) A2
= L1 ( x ) L 2 ( x )
dw = i1 dλ1 + i2 dλ2 − f dx
d (i1λ1 + i2 λ2 − w ) = λ1 di1 + λ2 di2 + f dx
w’ (co−energy)
dw ' = λ1 di1 + λ2 di2 + f dx
f=+
∂w '
∂x
i1 ,i2 cons tan t
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 19 of 23
dw ' =
∫
f dx +
1
i1 =i2 = 0
∫
dw ' =
f = +
∫
λ1 di1 +
L1 ( x ) i1 di1 +
i2 =0
x=constant
=
∫
2
i2 = 0
x=cons tan t
λ2 di2
3
i1
=cons tan t
x=cons tan t
∫
(M ( x ) i
1
+ L 2 ( x ) i2 ) di2
i1 =constant x = cons tan t
1
1
L1 ( x ) i12 + M ( x ) i1i2 + L 2 ( x ) i22 2
2
∂w '
∂x
=
i1 ,i2
1 2 dL1 1 2 dL 2
dM
+ i2
+ i1i2
i1
2
dx 2
dx
dx
XI. Synchronous Machine
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17 Page 20 of 23 λ as = L s ias + Mir cos θ
λ bs = L s ibs + Mir sin θ
λ r = L r ir + M (ias sin θ + ibs sin θ )
dw = ias d λ as + ibs d λbs + ir d λr − T e dθ
d ( w − ias λ as − ibs λbs − ir λr ) = −dw '
w' = ias λ as + ibs λbs + ir λr − w
co−energy
dw' = λ as dias + λbs dibs + λr dir + T e dθ
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 21 of 23
w'=
∫
T dθ +
ias =0
ibs =0
ir = 0
∫
1
∫
λ as dias +
θ = cons tan t
ibs =0
ir = 0
2
3
∫
ias =0
ibs =0
ir = 0
T d θ + ∫ L s ias dias + ∫ L s ibs dibs +
1
w' =
2
3
λ r dir
θ = cons tan t
ias = cons tan t
ibs = cons tan t
4
0
w'=
∫
λ bs dibs +
θ = cons tan t
ias = cons tan t
ir =0
∫
θ =cons tan t
ias = cons tan t
ibs = cons tan t
⎡L
⎣ r ir + M (ias cos θ + ibs sin θ ) ⎤⎦ dir
4
1
1
1
L s ias2 + L s ibs2 + L r ir2 + Mir (ias cos θ + ibs sin θ )
2
2
2
Te = +
∂w '
∂θ
ias ,ibs ,ir
= Mir ( −ias sin θ + ibs cos θ )
Balanced 2 phase currents
ias = Is cos ωt , ibs = Is sin ωt , ir = Ir , θ = ωmt + γ
T e = M Ir Is ( − cos ωt sin θ + sin ωt cos θ ) = M Ir Is sin ( ωt − θ )
= M Ir Is sin ( ( ω − ωm ) t − γ )
< T e > ≠ 0 ⇒ ω = ωm
T e = −M Ir Is sin γ
J
d2 θ
dθ
= Te − β dt
dt2
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
Page 22 of 23
θ = ωm t + γ 0 + γ ' ( t ) ,
γ ' ( t ) << γ 0
−M Ir Is sin γ 0 − βω = 0
sin γ 0 = −
βω
M Ir Is
Pullout when sin γ 0 = 1 ⇒ βω = M Ir Is
Hunting transients: sin ( γ 0 + γ ') ≈ sin γ 0 cos γ '+ cos γ 0 sin γ ' ≈ sin γ 0 + γ ' cos γ 0
J
d2 γ '
= −M Ir Is cos γ 0 γ '− βγ ' = − (M Ir Is cos γ 0 + β ) γ '
dt
2
d2 γ '
+ ω02 γ' = 0 ;
dt2
ω02 = ⎡M
⎣ Ir Is cos γ 0 + β⎤⎦ J
γ ' = A1 sin ω0 t + A2 cos ω0 t
Stable if ω02 > 0
( ω0 real)
Unstable if ω02 < 0
( ω0 imaginary)
6.013, Electromagnetics and Applications
Prof. Markus Zahn
Lecture 16 & 17
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