Examen de Physique : Électromagnétisme cp2 - e

−→
∇·−→ X = ∂Xx
∂x +∂Xy ∂y +∂Xz ∂z
−→
∇×−→ X = ∂Xz
∂y −∂Xy ∂z −→
ux+∂Xx
∂z −∂Xz ∂x −→
uy+∂Xy
∂x −∂Xx ∂y −→
uz
cos p+ cos q= 2 cos p+q
2cos p−q 2
cos p−cos q= 2 sin p+q
2sin q−p 2

•ω
yOz θ −→ uz
−→
ux
−→
Ei= E0cos ωt −−→ ki·−→ r−→ ux
z60
−→
ki−→
kr
z
y
•
x
θ
·−→ ki(−→ ux,−→ uy,−→ uz)λ θ
E0, λ, θ, y, z, ω t
·∂2Ex
∂x2+∂2Ex ∂y2+∂2Ex ∂z2−1 c2
∂2Ex
∂t2= 0
ki= 2π/λ ω c
·−→ BiE0, c, λ, θ, y, z, ω t
−→
ki,−→ Ei,−→ Bi
·
·−→
Er= Er0cos ωt −−→ kr·−→ r−→ ux
Oz
−→
krθ λ
·(y= 0, z = 0) Er0
E0
·−→ BrE0, c, λ, θ, y, z, ω t
−→
kr,−→ Er,−→ Br
·−→ Ez > 0
−→
E = −2E0sin 2π
λcos θzsin ωt −2π sinθ y−→
ux
·
·−→ B
−→
B = 2E0
c
0
−cos(θ) cos 2π
λcos θ zcos ωt −2π λsin θ y
sin(θ) sin 2π
λcos θ zsin ωt −2π λsin θ y
·−→ Rh−→ Ri
·

•γ
−→
E = E0e−z/δ ei(ωt−z/δ)−→ ux
−→
B = √2 E0
δ ω e−z/δ ei(ωt−z/δ−π/4)−→
uy
δ=r2
µ0γω
·
· PJV
δ
PJ=γE02δS
4
·Tf= 933 K γ= 4.106S.m−1cv= 3.106J.m−3.K−1
µ0= 4π.10−7H.m−1S = 1 mm2= 10−6m2ω= 2.1015 rad.s−1E0= 105V.m−1
V= 5 S δ
Q = Vcv(Tf−T0)
300 K
·
•(Oz)a
−→
P||−→ P|| = Cte
·−→ P
·r < a
·−→ ∇·−→ D = ρ`−→ Dρ`
·
−→
∇·−→ P = 1
r
∂(rPr)
∂r +1 r
∂Pθ
∂θ +∂Pz ∂z