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6.641 Electromagnetic Fields, Forces, and Motion, Spring 2005
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6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3: Electroquasistatic and Magnetoquasistatic Fields and Boundary
Conditions
I. Conditions for Electroquasistatic Fields
A. Order of Magnitude Estimate [Characteristic Length L, Characteristic
time τ ]
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
∇ iE = ρ ε ⇒
∇×H =
ε
∂E
H εE
εEL L2ρ
⇒
=
⇒H=
=
L
τ
∂t
τ
τ
∇ × E = −µ
Eerror
E
Eerror
E
=
E
ρL
=ρ ε⇒E=
L
ε
E
∂H
µH µρL2
µρL3
⇒ error =
=
⇒E
=
error
∂t
L
τ
τ
τ2
µρL3
τρL
ε=
1⇒
L
cτ
µεL2
2
τ
=
L2
( cτ )
2
;
c =
1
ε µ
1
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 1 of 12
B. Estimate of Error introduced by EQS approximation
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
E=
_
V_
i z = E0 i z
d
⎪⎧−εE0
σsu = ⎨
⎪+εE0
⎩
Kr 2πb + πb2
z=d
z=0
dσsu
b dσsu
b dE
= 0 ⇒ Kr = −
= − ε 0
dt
2 dt
2 dt
∫ H i ds = ∫ ∂t ( εE) i da ⇒ H 2πr = πr ε
∂
C
φ
S
2
dE0
r dE0
⇒ Hφ = ε
dt
2 dt
∂H
∫ E i ds = −∫ µ ∂t i da
C
S
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 2 of 12
b
d2E0
µε
⎡⎣Ez (b ) − Ez (r ) ⎦⎤ d = +
r ' dr ' d
2
dt2
∫
r
=
Ez (r ) =
E0 +
d2E0
µεd 2
b − r2
4
dt2
(
εµ d2
E0
(r
4 dt2
2
)
− b2
)
If E0 ( t ) = A cos ωt
Eerror
E0
=
Eerror
εµ d2
E0
4E0 dt2
1⇒
E0
fλ = c =
(b
2
ω2 εµb2
4
)
− r2 =
(
1 2
ω εµ b2 − r2
4
1
1
εµ
2πc
ω
ω2εµb2
π2
λ=c⇒ω=
⇒
= 2 b2
2π
4
λ
λ
f=1 MHz in free space ⇒ λ =
If b
)
100 m
3 × 108
106
1⇒b
λ
π
= 300 m
EQS approximation is valid.
II. Conditions for Magnetoquasistatic Fields
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 3 of 12
∇×H = J ⇒
∇ × E = −µ
H
= J ⇒ H = JL
L
E µH
∂H
µHL µJL2
⇒ =
⇒E=
=
L
∂t
τ
τ
τ
∇ × Herror =
ε
H
εE
εEL εµJL3
∂E
⇒ error =
⇒ Herror = 2 =
L
∂t
τ
τ
τ2
Herror
εµJL3 εµL2
L2
= 2
= 2 =
H
τ JL
τ
( cτ )2
1⇒L
cτ
Courtesy of Hermann A. Haus and James R. Melcher. Used with permission.
τem =
L
=L
c
εµ
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 4 of 12
III. Boundary Conditions
1. Gauss’ Continuity Condition
Courtesy of Krieger Publishing. Used with permission.
∫ ε E i da = ∫ σ dS ⇒ ε (E
0
s
S
S
0
2n
- E1n ) dS = σ sdS
ε0 (E2n - E1n ) = σ s ⇒ n i ⎡⎣ε0 (E2 - E1 ) ⎤⎦ = σ s
2. Continuity of Tangential E
Courtesy of Krieger Publishing. Used with permission.
∫ E i ds = (E
1t
- E2t ) dl = 0 ⇒ E1t - E2t = 0
C
(
)
n× E1 - E2 = 0
Equivalent to Φ1 = Φ2 along boundary
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 5 of 12
3. Normal H
∫µ
0
H i da = 0
S
µ 0 (H an - Hbn ) A = 0
H an = Hbn
n i ⎡⎣H a - H b ⎤⎦ = 0
4. Tangential H
d
∫ H i ds = ∫ J i da + dt ∫ ε E i da
0
C
S
S
Hbt ds - H at ds = Kds
Hbt - H at = K
n × ⎡⎣H a - H b ⎤⎦ = K
5. Conservation of Charge Boundary Condition
d
∫ J i da + dt ∫ ρdV = 0
S
n i ⎡⎣ J a - J b ⎤⎦+
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
V
∂
σs = 0
∂t
Lecture 3
Page 6 of 12
6. Electric Field from a Sheet of Surface Charge
a. Electric Field from a Line Charge
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 7 of 12 dEr =
dq
(
4πε0 r2 + z2
+∞
Er =
∫
dEr =
z = −∞
)
λ0r
4πε0
λ0rdz
cos θ =
(
4πε0 r2 + z2
+∞
3
)
2
dz
∫
z = −∞
(r
2
+ z2
3
)
2
+∞
=
=
λ0r
4πε0
z
(
r2 z2 + r2
)
1
2
z = −∞
λ0
2πε0r
Another way: Gauss’ Law
∫ ε E i da = ε E 2πrL = λ
0
0 r
0
L
S
Er =
λ0
2πε0r
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 8 of 12
b. Electric Field from a Sheet Charge
dEy =
dλ
(
2πε0 x2 + y2
+∞
Ey =
∫
dEy =
x = −∞
=
)
σ0 y
2πε0
1
cos θ =
2
+∞
∫
x = −∞
(
2πε0 x2 + y2
)
dx
2
x + y2
σ0 y 1
x
tan−1
2πε0 y
y
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
σ0 ydx
+∞
−∞
Lecture 3
Page 9 of 12
⎧ σ0
⎪
⎪ 2ε0
=⎨
⎪− σ0
⎪ 2ε0
⎩
y>0
y<0
Checking Boundary condition at y=0
Ey ( y = 0+ ) − Ey ( y = 0−
) =
⎛ σ
σ0
− ⎜− 0
2ε0 ⎜⎝ 2ε0
σ0
ε0
⎞ σ0
⎟⎟ =
⎠ ε0
c. Two sheets of Surface Charge (Capacitor)
⎧ σ0 _
iy
⎪
⎪ 2ε0
E1 = ⎨
_
⎪− σ0 i y
⎪ 2ε
0
⎩
E = E1 + E
2 =
y > −a
y < −a
σ0
ε0
0
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
_
i y
, E2
⎧ σ0 _
iy
⎪−
⎪ 2ε0
= ⎨
_
⎪ σ0
i y
⎪ 2ε
⎩ 0
y>a
y<a
y <a
y >a
Lecture 3
Page 10 of 12
7. Magnetic Field from a Sheet of Surface Current
From a line current I
Hφ =
I
2πr
_
_
_
i φ = − sin φ i x + cos φ i y
Thus from 2 symmetrically located line currents
dHx =
dI
(
2
2
2π x + y
)
1
2
( − sin φ )
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 11 of 12
K dx
y
= − 0
2π x2 + y2
Hx = −
K0
y
2π
+∞
∫
x = −∞
dx
2
x + y2
+∞
=−
K0 y 1
x
tan−1
2π y
y
x = −∞
⎧ K0
⎪⎪− 2
= ⎨
⎪ + K0
⎪⎩ 2
y>0
y < 0
Check boundary condition at y=0:
Hx ( y = 0 + ) − Hx ( y = 0 − ) = −K 0
−
K 0 ⎛ K 0 ⎞
−⎜
⎟ = −K 0
2
⎝ 2 ⎠
6.641, Electromagnetic Fields, Forces, and Motion
Prof. Markus Zahn
Lecture 3
Page 12 of 12