PHYS-F-205 - Electricité et magnétisme Correction séance 4 et 5

I= 20kA
B=µ0I/2πr r
B=12,57 ×10−7×20 ×103
2π×0,1= 4mT
I B =
µ0I/2R R
B=12,57 ×10−7×0,1×10−3
2×0,1= 0,63nT
n B =µ0nI
n= 100/cm = 10 000/m
I=U
R=12 60 = 0,2A
⇒B= 12,57 ×10−7×10000 ×0,2 = 2,5mT

R B =
µ0I/2πR
Imax =2πRBmax
µ0
=2π×10−3×0,1
12,57 ×10−7= 500 A
~p d
~
Bfil ~ Bterrestre
~
Btotal =µ0I
2πd , Bterrestre=3×10−5,5×10−5
tan θ=By
Bx
=5×10−5
3×10−5⇒θ= 59 degrs

Oz ρ Ri<ρ<
RoN I
N
I~
B·~ dl =µN I µ
~
B
~
dl
~
B·~ dl =B dl dl =ρdθ .
I~
B·~ dl =Z2π
0
Bρdθ =µN I ⇒B=µN I
2πρ
Oz
NI B
µ/µ0B µ
B
µ
(µ, B)

d~ B P
Id~ l
d~ B=µ0I
4π
d~ l×r~ 1r
r3
·~ 1rdl P
·~ 1r⊥d~ l⇒I|d~ l×~ 1r|=Idl sin(90) = Idl
·r=√R2+x2
dB =µ0
4π
Idl
R2+x2
d~ B x Ox
dB⊥dB Ox
dBx= cos φdB
B=Z2π
0
µ0
4π
Idl
R2+x2cosφ =Z2π
0
µ0
4π
IRdθ
R2+x2
R
r
dl =Rdθ cosφ =R/r
B=µ0I
2
R2
(R2+x2)3/2

~
FCoulomb =q~ E~ FLorentz =q~v ×~ B
~
FCoulomb +~ FLorentz =q(~ E+~v ×~ B) = 0
~
E ~v×~ B⇒E−vB = 0
⇒v=E/B
dl d ~ F=Id~ l×~ B
dF =IdlB sin θ
Ftot =IlB sin θ= 6 ×0,01 ×0,01 ×sin(31,2) = 0,31mN
N
I B
τ=NISB sin φ
S φ ~ B
τ= 20 ×1,5×1,3×10−3×0,9×sin 32 = 0,0186Nm