Forces électromagnétiques Les points du cours à connaître

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−m~v
ττ
~
E
~
B
~
E~ B
~
E
•−m~v
ττ
• −e~ E−e~v ∧~ B
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•~
j=ne(−e)~ve
q m O
(O, x, y, z)~v0=v0x.~ux+v0y.~uy
~
E=E0.~uy
y=q.E0
2.m.v2
0x
x2+v0y
v0xx
q m O
(O, x, y, z)~v0=v0x.~ux+v0y.~uy
~
B=B0.~ux
q m O
(O, x, y, z)~v0=v0x.~ux+v0y.~uy
~
E=E0.~uy~ B=B0.~ux
R= 14cm e = 1,6.10−19C m = 1,67.10−27kg
Ec= 1,2MeV ∆V= 4000V
vmax = 1,52.107m.s−1
U= 1,2MV
f= 17,3MHz
n= 150tours
B= 1,13T
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~v =v0.~uzxOy O
Z a Oz D Z
∆t Z
~
E=−E0.~uyZ
y
∆vy~uyZ
y D v0
∆vye
mE0∆t
Z~ B=B0.~ux
B0O
B0E0v0
e
m
y a D B0E0
Z
∆t=a
v0
~
E=−E0.~uy
∆vy=y.v0
D
∆vy=e
mE0.∆t
~
B=B0.~ux~ E=−E0.~uy
B0=E0
v0
e
m=y.E0 D.a.B2
0
m q
U= 10kV ~v =v0.~uz
v0
O z > 0~ B=B0.~uy
B0= 0,10T
R
xOy
x
39K+ 41K+
v0=q2.q.U
m
R=m.v0
q.B0
x= 2q2.U
q.B2
0√m2−√m1
x= 4,6cm
e
m
m−e u =
2,5kV R = 3,3cm
B= 5,1mT
v0m e u
R v0m e B
e
mu R B
e= 1,6.10−19C m = 9,1.10−31kg
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v0=q2.e.u
m
R=m.v0
e.B e
m=2.u R2.B2= 1,77.1011C.kg−1
e
m= 1,76.1011C.kg−1
m−e v0O
y > 0~ E=E.~ey
xOy α ∈0; π
2~ex
x(t)
y(t)
xsS y > 0
∆α α ∈α0−∆α
2;α0+∆α 2
α0S
x(t) = v0.cos(α).t
y(t) = v0.sin(α).t −e.E.t2
2.m
xs=m.v2
0
e.E sin α 2
α0=π
4
m−e v0O
y > 0~ B=−B.~ez
xOy α ∈]0; π[~ex
R
C
xsS y > 0
∆α α ∈α0−∆α
2;α0+∆α 2
α0S
R=m.v0
e.B
xC=R. sin(α) = m.v0
e.B sin(α)yC=−R. cos(α) = −m.v0 e.B cos(α)
xs=2.m.v0
e.B sin(α)
α0=π
2
~v ~ E0
~
fi→e=−λ.~v
~v
ne
γ
γ=n.e2
λ
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