TD B3 - Correction 4 - Analyse qualitative du phénomène d

UAB = 0
−→
v A B UAB
−→
B
φ
t6= 0
−→
A−→ uz
−→
Em=−∂−→
A
∂t −→
uz−→ B
−→
A−→ B=−→−→ A

Q=Z∞
0
I(t)t=aire sous la courbe '30 carreaux
100.10−6×20.103= 2 Q= 60
Imoy =Q
T=60 10−3= 6.104
U=Eh = 5.106Eeclair =
1
2Cu2=1 2QU = 1,5×1081 = 3,6×106Eeclair = 41
D'1
c'
3.108−1
τ'10−3
T¿τ
T'D
c'3,3µs ¿τ= 1000 µ
−→
B=µ0I
2πr −→
uθ
−→
BCS
Φ = RR−→ B·−→ S
e
e=−Φ
t=−ZZ ∂−→
B
∂t ·−→
S
e
I
t
I(t) 0 100 µ
700 µ
1/r

−→
FLaplace =Ztige i−→ `∧−→ B=−I`(x)B0cos α−→ ux−I`(x)B0sin α−→ uz
`(x)=2xtan θ
x−→ uz
m−→
v
dt =−→
P+−→ RN+−→ FLaplace
−→
RN
x
m¨x=mg sin α−2Ix tan θB0cos α
¨x+ω2
0x=gsinα
ω02=2Itan θB0cos α
m
x=gsin α
ω02(1 −cos ω0t)
ω0
x
FLaplace =I`B0= 1 ×5.10−2×0,1 = 5.10−3
∆m=FLaplace
g= 5.10−4

−→
B−→ B=µ0n i(t)−→ uz
−→
B
θ
z−→ E θ
z
M r
−→
uz−→ ur−→
E−→ uθ
−→
E=Eθ(r, t)−→ uθ
−→
E r r < a
S
IC
−→
E·−→ `=ZZS
−→−→
E·−→ S=−ZZS
∂−→ B
∂t ·−→
S=−Φ
t
C
IC
−→
E·−→ `= 2πEθ(r)−Φ
t=−πr2B t=πr2µ0ni0sin ωt
−→
E=rµ0ni0
2sin(ωt)−→
uθ
−→
j=γ−→ E=γrµ0ni0
2sin(ωt)−→
uθ
hPLori=ZZZcylindre
−→
j·−→ E τ =ZZZ γE2τ=ZZZ γÃωµ0ni0r
2!2
hsin2ωtir r θ z

hPLori=π
16 (µ0ni0)2γ ω2b4L