3 gabarito prova 3 { eletromagnetismo i { noturno · 3a gabarito prova 3 { eletromagnetismo i {...
TRANSCRIPT
3a Gabarito Prova 3 – Eletromagnetismo I – Noturno
Q1 [4.0]
(a) [0.5]
~B(t, r, θ) ' µ0k2c
4π
eikr
rr × ~p =
µ0 k2 c
4π
eikr
rp0e−iωtr ×
(p0ze
−iωt)=µ0k
2cp04π
ei(kr−ωt)
r(z cos θ + ρ sen θ)× z
= −µ0k2cp0
4π
ei(kr−ωt)
rsen θϕ
(b) [1.5]
~∇× ~B =1
r
[1
sen θ
∂ (Bϕ sen θ)
∂θr − ∂ (rBϕ)
∂rθ
]= −µ0k
2cp04π
1
r
[ei(kr−ωt)
r2sen θ
∂(sen2 θ
)∂θ
r − sen θ∂(ei(kr−ωt)
)∂r
θ
]
≈ iµ0k3cp0
4π
ei(kr−ωt)
rsen θθ
Entao
~E = c2∫
(~∇× ~B)dt
~E = c2µ0k
3p04πω
ei(kr−ωt)
rsen θθ
~E =µ0k
3c3p04πω
ei(kr−ωt)
rsen θθ
(c) [1.0]
~Sp =1
µ0~E × ~B
= − 1
µ0
µ0k3c3p0
4πω
ei(kr−ωt)
rsen θ
µ0k2cp0
4π
ei(kr−ωt)
rsen θ(θ × ϕ)
=µ0k
5c4p2016π2ω
e2i(kr−ωt)
r2sen2 θ r =
ε0ε0
µ0ω5c4p20
16π2c5ω
e2i(kr−ωt)
r2sen2 θ r
=ω4p20
16π2ε0c3e2i(kr−ωt)
r2sen2 θ r
(d) [1.0]
〈~Sp〉t =1
T
∫ T
0
1
µ0Re[ ~E]×Re[ ~B]dt
=ω4p20
16π2ε0c3sen2 θ
r2rω
2π
∫ 2π/ω
0cos2(kr − ωt)dt
=ω4p20
32π2ε0c3sen2 θ
r2r
Q2 [3.0]
1
a) [1,0] Condicoes de contorno
~Etang1 = ~Etang2 → (E1p − E′1p) cos θ1 = E2p cos θ2
~Htang1 = ~Htang2 →n1µ1c
(E1p + E′1p) =n2µ2c
E2p
Segue que
E1p − E′1p =cos θ2cos θ1
E2p; E1p + E′1p =n2µ1n1µ2
E2p
entao
E′1p =
n2µ1n1µ2
− cos θ2cos θ1
n2µ1n1µ2
+ cos θ2cos θ1
E1p; E2p =2
cos θ2cos θ1
+ n2µ1n1µ2
E1p
Daı
rp =E′1E1
=
n2µ1n1µ2
− cos θ2cos θ1
n2µ1n1µ2
+ cos θ2cos θ1
=
n2µ2
cos θ1 − n1µ1
cos θ2n2µ2
cos θ1 + n1µ1
cos θ2
e
tp =E2
E1=
2cos θ2cos θ1
+ n2µ1n1µ2
=2n1µ1
cos θ1n1µ1
cos θ2 + n2µ2
cos θ1
b) [0,5]
rp = 0→ n2µ2
cos θ1 −n1µ1
cos θ2 = 0
(n2µ1n1µ2
)√1− sin2 θB −
√1− n21
n22sin2 θB = 0(
n2µ1n1µ2
)2
−(n1µ2n2µ2
)2
sin2 θB − 1 +n21n22
sin2 θB = 0
sin2 θB =
1−(n2µ1n1µ2
)2
n21n22−(n1µ2n2µ1
)2
c) [0,5]
tp =2
cos θ2cos θ1
+ n2µ1n1µ2
=2
µ1n2
µ2n1+ n2µ1
n1µ2
=µ2n1µ1n2
Os coeficientes associados a conservacao de energia sao o R e T (e nao r e p).
d) [0,5]
T =〈~S2 · k〉t〈~S1 · k〉t
=(ε2/v2)E
2p2 cos θ2
(ε1/v1)E2p1 cos θ1
=ε2v1ε1v2
t2pcos θ2cos θ1
=ε2v1ε1v2
(µ2n1µ1n2
)2 µ1n2µ2n1
=ε2µ2v1n1µ1ε1v2n2
n1n2
=v2v1
v1/c
v2/c= 1
2
e) [0,5]
rs = 0→ n1µ1
cos θ1 −n2µ2
cos θ2 = 0(n1µ2n2µ1
)√1− sin2 θ1 −
√1− n21
n22sin2 θ1 = 0(
n2µ1n1µ2
)2
−(n1µ2n2µ2
)2
sin2 θ1 − 1 +n21n22
sin2 θ1 = 0
sin2 θ1 =
1−(n2µ1n1µ2
)2
n21n22−(n2µ1n1µ2
)2
Q3 [3.0]
a) [0,5]
∇× ~E = −∂~B
∂t→ ey
∂
∂y×[Eei(ky−ωt)ez
]= ikEei(ky−ωt)ex = −∂
~B
∂t
∂ ~B
∂t= −ikEei(ky−ωt)ex → ~B =
k
ωEei(ky−ωt)ex
b) [0,5] Para que a onda possa se propagar, o vetor de onda precisa ser real; como
k =
√ω2 − ω2
p
c2
entao e preciso que ω > ωp.
c) [0,5]
∂
∂ω(k2) =
∂
∂ω
(ω2 − ω2
p
c2
)
2∂k
∂ωk = 2
ω
c2→ ∂ω
∂k
ω
k= vgvf = c2
d)
~S =1
4µ0
[~E × ~B∗ + ~E∗ × ~B
]=
1
4µ0
[Eei(ky−ωt)z × k
ωE∗e−i(ky−ωt)x+ E∗e−i(ky−ωt)z × k
ωEei(ky−ωt)x
]=
1
2µ0
k
ω|E|2y
e) [0,5] Se ω < ωp, entao,
k = i
√ω2p − ω2
c2=i
λ; λ ≡ c√
ω2p − ω2
e~E = Eei(ky−ωt)z = Ee−iωe−y/λz
3
f) [0,5]
~E = Ee−y/λe−iω z ~B = i|k|ωEe−y/λze−iωtex
~S =1
4µ0
[~E × ~B∗ + ~E∗ × ~B
]=
1
4µ0
[Ee−iωte−y/λez × (−i) |k|
ωE∗eiωte−y/λex + E∗eiωte−y/λez × i
|k|ωEe−iωte−y/λex
]= ey
1
2µ0
|k|ω|E|2e−2y/λ[−i+ i] = 0
4