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