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( 2n
+ 1)
2 2
∞
∞
⎛ k ⎞ 4π 2 * ⎛ k ⎞ 4 π 2 n( n+
2)
* *
4 ⎜ ⎟ 2 0 ∑ Re
n n ⎜ ⎟ 2 0 ∑ Re
n n+ 1 n n+
1
µω k n= 1 n( n+ 1) µω k n=
1 ( n+
1)
I = E ⎡⎣ a b ⎤⎦+ E ⎡⎣b b + a a ⎤⎦⇒
⎝ ⎠ ⎝ ⎠
( 2n
+ 1)
4π 4 π nn ( + 2)
I E ⎡⎣ a b ⎤⎦ E ⎡⎣b b a a ⎤⎦
∞
∞
2 * 2
* *
4
=
2 2 0 ∑ Re
n n
+
2 2 0 ∑ Re
n n+ 1+
n n+
1
µω n= 1 nn ( + 1) µω n=
1 ( n+
1)
π
2
2
∞
k 2π
E0
5 2 2 2 ∑
µω k n=
1
{ [ ]}
I =− Re (2n+ 1) an + bn
⇒
2
∞
2π
E0
∑
5 2 2
µω n=
1
{ [ ]}
I =− Re (2n+ 1) an + bn
2
( µω)
[ bn n n′ an n n′ bn n n′ bn n n′ bn n n′ an n n′ an n n′ an n n′
]}
e
π k
* n′+ 1 n+ 1 2iρ
I6 = ∫ dθsenθcos
θ ∑ E ′ 2 Re −( − ) ( − ) ×
2 2 nEn
i i e
2k
0
× τπ + ππ − ππ − ττ + πτ − τπ − πτ + ττ ⇒
π
π
* n′+ 1 n+ 1 2iρ
I6 = ∫ dθsenθcos
θ ∑ E ′ 2 Re −( − ) ( − ) ×
2 nEn
i i e
2
0
( µω)
[ bn n n′ an n n′ bn n n′ bn n n′ bn n n′ an n n′ an n n′ an n n′
]}
× τπ + ππ − ππ − ττ + πτ − τπ − πτ + ττ
{
{
Combinando I1 com I4 = I7, I2 com I5 = I8
e I3 com I6 = I9, e de acordo com a
1
{ i}
* * * *
expressão de força
i
= ∫ − ε ( θ θ
+
ϕ ϕ) +µ ( θ θ
+
ϕ ϕ)
F Re [ E E E E H H H H ] n da, ficamos com:
2
( 2n
+ 1)
∞
∞
⎛4π 2 4 π 2 ⎞⎛
* nn ( + 2)
* *
⎞
I7 = ⎜ E
2 0
ε+µ E
2 2 0 ⎟⎜
∑ Re ⎡anbn Re bnbn+ 1
anan+
1
k n= 1 n( n 1)
⎣ ⎤⎦+ ∑ ⎡ + ⎤⎟
µω + n=
1 ( n+
1)
⎣ ⎦
⎝ ⎠⎝ ⎠
2
usando a relação k⋅ k =ω εµ :
( 2n
+ 1)
∞
∞
8 π 2 ⎛
* nn ( + 2)
* *
⎞
I7 =
⎛
E
⎞
⎜ 2 0
ε ⎟⎜
∑ Re ⎡anbn Re bnbn+ 1
anan+
1
k n= 1 n( n 1)
⎣ ⎤⎦+ ∑ ⎡ + ⎤⎟
⎝ + n=
1 ( n+
1)
⎣ ⎦
⎠⎝ ⎠
⎛ ε µ ⎞
I =−2π E + Re (2n+ 1) an + bn
=
∞
2
8 0 ⎜ 2 2 2 ⎟ ∑
⎝k
µω ⎠ n=
1
{ [ ]}
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