Une page de Wikiversité, la communauté pédagogique libre.
En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Champ magnétique d'un aimant d'un courantChamps magnétique et notions électromagnétiques/Exercices/Champ magnétique d'un aimant d'un courant », n'a pu être restituée correctement ci-dessus.
Une bobine
B
{\displaystyle B}
parcourue par un courant d'intensité
i
{\displaystyle i}
créé en
M
{\displaystyle M}
un champ magnétique
B
→
1
{\displaystyle {\vec {B}}_{1}}
. Un aimant
A
{\displaystyle A}
créé en
M
{\displaystyle M}
un champ magnétique
B
→
2
{\displaystyle {\vec {B}}_{2}}
.
Tracer
B
→
2
{\displaystyle {\vec {B}}_{2}}
sachant que
B
2
=
B
1
⋅
(
1
+
3
)
{\displaystyle B_{2}=B_{1}\cdot (1+{\sqrt {3}})}
.
Le champ résultant de l'action simultanée en
A
{\displaystyle A}
et de
B
{\displaystyle B}
est
B
→
=
B
→
1
+
B
→
2
{\displaystyle {\vec {B}}={\vec {B}}_{1}+{\vec {B}}_{2}}
. Tracer
B
→
{\displaystyle {\vec {B}}}
.
Calculer le module de
B
{\displaystyle B}
en fonction de
B
1
{\displaystyle B_{1}}
.
Quelle est la force subie par une particule de charge
q
{\displaystyle q}
animée d'une vitesse
v
→
{\displaystyle {\vec {v}}}
qui se trouve en
M
{\displaystyle M}
à l'instant
t
{\displaystyle t}
?
Données
α
=
π
3
{\displaystyle \alpha ={\frac {\pi }{3}}}
B 1 = 1 mT
Vitesse d'un électron dans un conducteur de cuivre v 1 = 60 cm/h
v 2 = 10 m/s
Masse d'un électron m e = 9,109 382 15 × 10−31 kg
Charge électrique fondamentale négative q = −1,602 176 53 × 10−19 C
Solution
Voir la question suivante.
Soit
B
x
=
B
1
x
+
B
2
x
{\displaystyle B_{x}=B_{1_{x}}+B_{2_{x}}}
et
B
y
=
B
1
y
+
B
2
y
{\displaystyle B_{y}=B_{1_{y}}+B_{2_{y}}}
.
On sait que
B
1
x
=
B
1
{\displaystyle B_{1_{x}}=B_{1}}
et
B
1
y
=
0
{\displaystyle B_{1_{y}}=0}
.
On détermine l'intensité du vecteur
B
→
2
{\displaystyle {\vec {B}}_{2}}
.
B
2
x
=
B
2
⋅
cos
(
α
)
{\displaystyle B_{2_{x}}=B_{2}\cdot \cos(\alpha )}
avec
α
=
π
3
{\displaystyle \alpha ={\frac {\pi }{3}}}
B
2
x
=
B
2
⋅
cos
(
π
3
)
=
1
2
⋅
B
2
{\displaystyle B_{2_{x}}=B_{2}\cdot \cos({\frac {\pi }{3}})={\frac {1}{2}}\cdot B_{2}}
Or
B
2
=
B
1
⋅
(
1
+
3
)
{\displaystyle B_{2}=B_{1}\cdot (1+{\sqrt {3}})}
donc
B
2
x
=
1
2
⋅
B
1
⋅
(
1
+
3
)
{\displaystyle B_{2_{x}}={\frac {1}{2}}\cdot B_{1}\cdot (1+{\sqrt {3}})}
De même
B
2
y
=
B
2
⋅
sin
(
α
)
=
B
2
⋅
sin
(
π
3
)
=
3
2
⋅
B
2
{\displaystyle B_{2_{y}}=B_{2}\cdot \sin(\alpha )=B_{2}\cdot \sin({\frac {\pi }{3}})={\frac {\sqrt {3}}{2}}\cdot B_{2}}
Donc
B
2
y
=
3
2
⋅
B
1
⋅
(
1
+
3
)
{\displaystyle B_{2_{y}}={\frac {\sqrt {3}}{2}}\cdot B_{1}\cdot (1+{\sqrt {3}})}
Enfin, on détermine l'intensité du vecteur
B
→
{\displaystyle {\vec {B}}}
B
x
=
B
1
x
+
B
2
x
=
B
1
+
1
2
⋅
B
1
⋅
(
1
+
3
)
=
B
1
⋅
(
1
+
1
2
+
3
2
)
=
B
1
⋅
(
3
2
+
3
2
)
=
B
1
⋅
(
3
(
1
+
3
)
2
)
{\displaystyle {\begin{aligned}B_{x}&=B_{1_{x}}+B_{2_{x}}\\&=B_{1}+{\frac {1}{2}}\cdot B_{1}\cdot (1+{\sqrt {3}})\\&=B_{1}\cdot (1+{\frac {1}{2}}+{\frac {\sqrt {3}}{2}})\\&=B_{1}\cdot ({\frac {3}{2}}+{\frac {\sqrt {3}}{2}})\\&=B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}})\end{aligned}}}
De même
B
y
=
B
1
y
+
B
2
y
B
y
=
0
+
3
2
⋅
B
1
⋅
(
1
+
3
)
B
y
=
B
1
⋅
(
3
(
1
+
3
)
2
)
{\displaystyle {\begin{aligned}B_{y}&=B_{1_{y}}+B_{2_{y}}\\B_{y}&=0+{\frac {\sqrt {3}}{2}}\cdot B_{1}\cdot (1+{\sqrt {3}})\\B_{y}&=B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}})\end{aligned}}}
Donc
B
=
(
B
1
⋅
(
3
(
1
+
3
)
2
)
)
2
+
(
B
1
⋅
(
3
(
1
+
3
)
2
)
)
2
B
=
(
2
⋅
B
1
⋅
(
3
(
1
+
3
)
2
)
)
2
B
=
2
⋅
B
1
⋅
(
3
(
1
+
3
)
2
)
B
=
B
1
⋅
(
3
(
1
+
3
)
2
)
{\displaystyle {\begin{aligned}B&={\sqrt {(B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}}))^{2}+(B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}}))^{2}}}\\B&={\sqrt {(2\cdot B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}}))^{2}}}\\B&={\sqrt {2}}\cdot B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}})\\B&=B_{1}\cdot ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{\sqrt {2}}})\end{aligned}}}
On détermine d'abord l’angle
(
v
→
,
B
→
)
{\displaystyle ({\vec {v}},{\vec {B}})}
ou
(
B
→
1
,
B
→
)
{\displaystyle ({\vec {B}}_{1},{\vec {B}})}
car
v
→
=
k
B
1
→
{\displaystyle {\vec {v}}=k{\vec {B_{1}}}}
avec
k
>
0
{\displaystyle k>0}
.
On sait que
(
B
→
1
,
B
→
2
)
=
α
=
π
3
{\displaystyle ({\vec {B}}_{1},{\vec {B}}_{2})=\alpha ={\frac {\pi }{3}}}
et
B
2
=
B
1
⋅
(
1
+
3
)
{\displaystyle B_{2}=B_{1}\cdot (1+{\sqrt {3}})}
.
B
→
1
⋅
B
→
2
=
B
1
⋅
B
2
⋅
cos
(
π
3
)
=
1
2
⋅
B
1
2
⋅
(
1
+
3
)
{\displaystyle {\vec {B}}_{1}\cdot {\vec {B}}_{2}=B_{1}\cdot B_{2}\cdot \cos({\frac {\pi }{3}})={\frac {1}{2}}\cdot {B_{1}}^{2}\cdot (1+{\sqrt {3}})}
B
→
1
⋅
(
B
→
1
+
B
→
2
)
=
B
→
1
⋅
B
→
=
B
1
⋅
B
⋅
cos
(
B
→
1
,
B
→
)
{\displaystyle {\vec {B}}_{1}\cdot ({\vec {B}}_{1}+{\vec {B}}_{2})={\vec {B}}_{1}\cdot {\vec {B}}=B_{1}\cdot B\cdot \cos({\vec {B}}_{1},{\vec {B}})}
B
→
1
⋅
B
→
1
+
B
→
1
⋅
B
→
2
=
B
1
⋅
B
1
2
+
2
B
→
1
⋅
B
→
+
B
2
2
⋅
cos
(
B
→
1
,
B
→
)
{\displaystyle {\vec {B}}_{1}\cdot {\vec {B}}_{1}+{\vec {B}}_{1}\cdot {\vec {B}}_{2}=B_{1}\cdot {\sqrt {{B_{1}}^{2}+2{\vec {B}}_{1}\cdot {\vec {B}}+{B_{2}}^{2}}}\cdot \cos({\vec {B}}_{1},{\vec {B}})}
B
1
2
+
(
1
+
3
)
⋅
B
1
2
⋅
1
2
=
B
1
⋅
B
1
2
+
(
1
+
3
)
B
1
2
+
(
1
+
3
)
2
B
1
2
⋅
cos
(
B
→
1
,
B
→
)
{\displaystyle {B_{1}}^{2}+(1+{\sqrt {3}})\cdot {B_{1}}^{2}\cdot {\frac {1}{2}}=B_{1}\cdot {\sqrt {{B_{1}}^{2}+(1+{\sqrt {3}}){B_{1}}^{2}+(1+{\sqrt {3}})^{2}{B_{1}}^{2}}}\cdot \cos({\vec {B}}_{1},{\vec {B}})}
1
+
1
2
(
1
+
3
)
=
1
+
(
1
+
3
)
+
(
1
+
3
)
2
⋅
cos
(
B
→
1
,
B
→
)
{\displaystyle 1+{\frac {1}{2}}(1+{\sqrt {3}})={\sqrt {1+(1+{\sqrt {3}})+(1+{\sqrt {3}})^{2}}}\cdot \cos({\vec {B}}_{1},{\vec {B}})}
3
+
3
2
=
6
+
3
3
⋅
cos
(
B
→
1
,
B
→
)
{\displaystyle {\frac {3+{\sqrt {3}}}{2}}={\sqrt {6+3{\sqrt {3}}}}\cdot \cos({\vec {B}}_{1},{\vec {B}})}
(
3
+
3
2
)
2
=
(
6
+
3
3
)
⋅
cos
2
(
B
→
1
,
B
→
)
{\displaystyle ({\frac {3+{\sqrt {3}}}{2}})^{2}=(6+3{\sqrt {3}})\cdot \cos ^{2}({\vec {B}}_{1},{\vec {B}})}
6
+
3
3
2
=
(
6
+
3
3
)
⋅
cos
2
(
B
→
1
,
B
→
)
{\displaystyle {\frac {6+3{\sqrt {3}}}{2}}=(6+3{\sqrt {3}})\cdot \cos ^{2}({\vec {B}}_{1},{\vec {B}})}
cos
2
(
B
→
1
,
B
→
)
=
1
2
{\displaystyle \cos ^{2}({\vec {B}}_{1},{\vec {B}})={\frac {1}{2}}}
cos
(
B
→
1
,
B
→
)
=
1
2
{\displaystyle \cos({\vec {B}}_{1},{\vec {B}})={\frac {1}{\sqrt {2}}}}
(
B
→
1
,
B
→
)
=
π
4
{\displaystyle ({\vec {B}}_{1},{\vec {B}})={\frac {\pi }{4}}}
On calcule la force
F
{\displaystyle F}
subie par une particule de charge
q
{\displaystyle q}
animée d'une vitesse
v
→
{\displaystyle {\vec {v}}}
qui se trouve en
M
{\displaystyle M}
à l'instant
t
{\displaystyle t}
.
F
1
(
t
)
=
q
⋅
v
⋅
B
⋅
sin
(
v
→
,
B
→
)
{\displaystyle F_{1}(t)=q\cdot v\cdot B\cdot \sin({\vec {v}},{\vec {B}})}
Pour
v
1
{\displaystyle v_{1}}
:
F
1
(
t
)
=
q
⋅
v
1
⋅
B
⋅
sin
(
v
→
,
B
→
)
=
1
,
6
⋅
10
−
19
×
0
,
01
×
t
×
0,001
×
(
3
(
1
+
3
)
2
)
=
3
,
8
⋅
10
−
25
N
{\displaystyle F_{1}(t)=q\cdot v_{1}\cdot B\cdot \sin({\vec {v}},{\vec {B}})=1{,}6\cdot 10^{-19}\times 0{,}01\times t\times 0{,}001\times ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}})=3{,}8\cdot 10^{-25}\ {\text{N}}}
De même :
F
1
(
t
)
=
q
⋅
v
2
⋅
B
⋅
sin
(
v
→
,
B
→
)
=
1
,
6
⋅
10
−
19
×
10
×
t
×
0,001
×
(
3
(
1
+
3
)
2
)
=
3
,
8
⋅
10
−
21
N
{\displaystyle F_{1}(t)=q\cdot v_{2}\cdot B\cdot \sin({\vec {v}},{\vec {B}})=1{,}6\cdot 10^{-19}\times 10\times t\times 0{,}001\times ({\frac {{\sqrt {3}}(1+{\sqrt {3}})}{2}})=3{,}8\cdot 10^{-21}\ {\text{N}}}