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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013/ESD.013J — Electromagnetics and Applications
Fall 2005
Problem Set 5 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 5.1
A
�
�
∂
∂
∂
x̂
+ ŷ
+ ẑ
· εE0 (x̂x + ŷy) sin ωt = 2εE0 sin ωt
∂x
∂y
∂z
�
� �
�
� x̂
�
�
ŷ
ẑ �� �� x̂
ŷ ẑ ��
�∂
∂y
∂
∂x
∂
∂ �
∂
∂
�
�
�
E0 sin ωt = 0
− µH = � × E = �∂x ∂y ∂z � = � ∂x ∂y 0� E0 sin ωt = ẑ
−
∂t
∂x ∂y
� Ex Ey Ez � � x
y 0�
ρ = � · D = � · εE =
⇔ H = C(r) = 0
Time-independent magnetic field that could be set to zero, since it is not generated by the time dependent
electric field.
J=�×H−
∂
∂
εE = 0 − εE0 (x̂x + ŷy) sin ωt = −ωεE0 (x̂x + ŷy) cos ωt
∂t
∂t
B
�
�
�
�
∂
∂
∂
∂y
∂x
ρ = � · εE = x̂
+ ŷ
+ ẑ
· εE0 (x̂y − ŷx) cos ωt =
−
εE0 cos ωt = 0
∂x
∂y
∂z
∂x ∂y
�
� �
�
� x̂
ŷ
ẑ �� �� x̂
ŷ ẑ��
�
∂
∂
∂
∂ �
∂
�∂
0�� E0 cos ωt = ẑ (−2E0 cos ωt)
− µH = � × E = �� ∂x
∂y
∂z � = � ∂x
∂y
∂t
�Ex Ey Ez � � y −x
0 �
⇔ H = ẑ
0 2E0
2E0
�=
sin ωt + �
C(�
r)
ẑ
sin ωt
µω
ωµ
∂
εE = 0 − [−ωεE0 (x̂y − ŷx) sin ωt] = ωεE0 (x̂y − ŷx) sin ωt,
∂t
where the first term is 0 because H does not depend on position.
J=�×H−
Problem 5.2
A
rad
ω
=⇒ f =
= 2 · 106 Hz = 2 MHz
sec
2π
1
2π
1
k = 4π · 10−2
=⇒ λ =
= ·
102 m = 50 m
m
k
2
ω = 4π · 106
cn =
4π · 106 m
m
ω
=
= 108
−2
k
4π · 10
sec
sec
1
Problem Set 5
6.013, Fall 2005
B
m
3 · 108 sec
c
c
⇔n=
=
=3
m
n
cn
108 sec
√
√
n = εr µr = εr · 1 ⇔ εr = n2 = 9
�
�
�
� �
µ
µr µ0
µr µ0
1
1
η=
=
=
=
· 120π Ω = · 120π Ω = 40π Ω
ε
εr ε0
εr ε0
9
3
�
n = index of refraction
Note:
η = impedance
�
�
� x̂ ŷ ẑ �
�
�
∂Ex
∂
∂
∂
�
− µ0 H = �× E = �� 0 0 ∂z
� = +ŷ ∂z = ŷ ∂z E0 cos(ωt − kz)
∂t
�Ex 0 0 �
cn =
=⇒ −
∂
E0 k
�0
µ0 H = ŷ E0 k sin(ωt − kz) =⇒ H = ŷ
cos(ωt − kz) + �
C(�
r)
∂t
ωµ0
=⇒ H = ŷ
E0
E0
cos(ωt − kz) = ŷ
cos(ωt − kz)
cn µ0
η
=⇒ H = ŷ
1
cos(4π · 106 t − 4π · 10−2 z) Ampères/m
4π
C
�
�
� x̂
S = E × H = �� Ex
�Hx
=⇒ S = ẑ E0
=⇒ S = ẑ
ŷ
Ey
Hy
� �
ẑ �� �� x̂
Ez �� = ��Ex
Hz � � 0
ŷ
0
Hy
�
ẑ��
0�� = ẑ Ex Hy
0�
E0
1
W
cos2 (ωt − kz) = ẑ 10 ·
cos2 (4π · 106 t − 4π · 10−2 z) 2
η
4π
m
W
2.5
cos2 (4π · 106 t − 4π · 10−2 z) 2
π
m
Problem 5.3
A
∂
∂
J = ωp2 εE =⇒
Re[Ĵejωt ] = ωp2 εRe[Êejωt ]
∂t
∂t
=⇒ Re[jωĴejωt ] = Re[ωp2 εÊejωt ] =⇒ jωĴ = ωp2 εÊ
=⇒ σ(ω) =
Ĵ
Ê
=
ωp2
ωp2 ε
= −jε
jω
ω
B
jωε(ω)Ê = Ĵ + jωεÊ = σ(ω)Ê + jωεÊ =⇒ jωε(ω) = σ(ω) + jωε
�
�
ωp2
ωp2
jωε(ω) = −jε
+ jωε =⇒ ε(ω) = ε 1 − 2
ω
ω
2
Problem Set 5
6.013, Fall 2005
C
k=ω
D
�
ε(ω)µ0 =⇒
√
• k0 = ω ε0 µ0
⎧
�
� �
�
�
√
ω2
ωp2
ωp2 ⎨ ω εµ0 1 − ωp2 , ω ≥ ωp
√
�
• kp = ω ε 1 − 2 µ0 = ω εµ0 1 − 2 =
ωp2
⎩ jω √εµ
ω
ω
0
ω 2 − 1, ω < ωp
�
� x̂
1 ��
ˆ
ˆ
ˆ
0
� × E = −jωµ0 H =⇒ H = −
jωµ0 �� ˆ
Ex
ŷ
0
0
�
ẑ ��
1 ∂Êx
∂ �
y
∂z � = −ˆ
jωµ
0 ∂z
0�
ˆ0 ejk0 z = −ŷ k0 Eˆ0 ejk0 z , z < 0
ˆ 0 = −ŷ jk0 E
•H
jωµ0
ωµ0
−jkp ˆ −jkp z
kp ˆ −jkp z
ˆ 0 = −ˆ
•H
y
Ep e
= ŷ
Ep e
, z>0
jωµ0
ωµ0
Note: The hatˆabove x, y, z denotes a unit vector, but above field components denotes a complex phasor.
E
• ẑ × (Êp − Ê0 ) = 0 =⇒ ŷ(Êp − Ê0 ) = 0 =⇒ Êp = Ê0 ≡ Ê
ˆ 0 ) = K0 x̂ =⇒ Ĥp − H
ˆ 0 = −K0
ˆp − H
ˆ 0) = K
ˆ =⇒ −x̂(Ĥp − H
• ẑ × (H
F
k0 ˆ
kp ˆ
D,E
ˆ0 − H
ˆ p = K0 =⇒
ˆ = − ωµ0 K0
H
−
E−
E = K0 =⇒ E
ωµ0
ωµ0
k0 + kp
Therefore:
Ê(z) =
�
−x̂
−ˆ
x
ˆ
H(z)
=
�
−ŷ k0 +pkp K0 e−jkp z , z > 0
0
ŷ k0k+k
K0 ejk0 z ,
z<0
p
ωµ0
−jkp z
,
k0 +kp K0 e
ωµ0
jk0 z
,
k0 +kp K0 e
z > 0
z < 0
and
k
G
In the field expressions above µ0 , ω, k0 , K0 are real, therefore:
�S� =
1
ˆ ∗ ] = 1 Re[Êx Ĥy∗ ]ẑ =⇒
Re[Ê × H
2
2
�
�
k0
ωµ0
1
−jk0 z
K0 ejk0 z ·
K
e
• �S0 � = −ˆ
z Re
0
2
k0 + kp
k0 + kp∗
�
�
1
ωµ0 k0
ωµ0 k0
= −ẑ Re
K 2 = −ẑ
K2
2
|k0 + kp |2 0
2|k0 + kp |2 0
3
Problem Set 5
6.013, Fall 2005
�
�
kp∗
∗
1
ωµ0
+jkp
z
Re
K0 e−jkp z ·
K
e
0
2
k0 + kp
k0 + kp∗
�
�
ωµ0 kp∗
ωµ0
1
= ẑ Re
K 2 e2Im{kp }z = ẑ
K 2 Re{kp }e2Im{kp }z
2
|k0 + kp |2 0
2|k0 + kp |2 0
�
�
ω2
ωµ0
2 √
K
ω
εµ
1 − ωp2 , ω ≥ ωp
ẑ
2
0
0
2|k0 +kp |
=⇒ �Ŝp � =
0,
ω < ωp (because Re{kp } = 0)
• �Sp � = ẑ
When the wavevector is purely imaginary inside a medium, the fields decay exponentially (they are called
“evanescent”) and no power is carried by them.
Problem 5.4
A
Figure 1: Diagram for Problem 5.4 Part A. (Image by MIT OpenCourseWare.)
For no power to be transmitted across the prism hypotenuse, the angle of incidence must be above the critical
angle. In general, from Snell’s Law:
n1 sin θ1 = n2 sin θ2
and for total internal reflection:
√
n1
θ =45◦
sin θ1 ≥ 1 1=⇒ n1 ≥ n2 2
n2
√
So for free space (n2 = 1) : n1,min = 2 ≈
√ 1.414
and for water (n2 = 1.33) : n1,min = 1.33 2 ≈ 1.88
sin θ2 ≥ 1 =⇒
B
The reflection coefficient at the input surface (1) is
r(1) =
n1 − n2
= r
n1 + n2
4
Problem Set 5
6.013, Fall 2005
and at the output surface (2) is
r(2) =
n2 − n1
= −r.
n2 + n1
Therefore the reflectivity is
�
�2
�
�
� n1 − n2 �2 �� nn21 − 1 ��
�
�
= � n1
R = |r | = �
� .
� n2 + 1 �
n1 + n2 �
2
√
For n1 ≥ n2 2 no power is lost at the hypotenuse, so the power transmitted over the input is:
S(2) = (1 − R)(1 − R)S(1)
=⇒
S(2)
S(1)
⎡
�n
�2 ⎤2
1
−
1
S(2)
⎦
=⇒
= (1 − R)2 = ⎣1 − nn21
S(1)
n2 + 1
� �� �
� ⎫2 ⎡
⎧ �� �2
⎤2
2
⎪
n1
⎪
n1
n1
n1
⎪
n1
⎨ n2 + 2 n2 + 1 − n2 − 2 n2 + 1 ⎪
⎬
4
⎢
⎥
= ⎣ � n2 �2 ⎦
=
�2
�
⎪
⎪
n1
n1
⎪
⎪
⎭
⎩
n2 + 1
n2 + 1
√
For the values calculated in part A n1 = n2 2 for both cases, therefore:
�
�2
√
S(2)
4 2
= √
≈ 0.943
S(1)
( 2 + 1)2
for both cases.
5