Charged particle motion in
magnetic field
Particle motion in curved magnetic fieldlines
We divide the equation of motion into a velocity component along the magnetic
field and perpendicular to the magnetic field. Suppose that the fieldlines are
circular.
K
K
ˆ
dv⊥ dv& ˆ db
dv
K G
=m
+
m
b + v& = q v⊥ × B
dt
dt
dt
dt
bˆ = ϕˆ = − sin ϕ xˆ + cos ϕ yˆ
dv
bˆ : component m & = 0
dt
→ v& = constant
dbˆ
dϕ
dϕ
= −(cos ϕ xˆ + sin ϕ yˆ )
=−
ρˆ
dt
dt
dt
v
dϕ v&
dbˆ
= →
= − & ρˆ
ρ
dt ρ
dt
We divide the velocity into a periodic part and a drift part
G
~G
G
v⊥ (t ) = v⊥ (t ) + Vd
The perpendicular component of the equ.of motion
G
ˆ
G G
dv⊥
db
G G
+m
= q v⊥ × B + qVd × B
m v&
dt
dt
G
2
G
G
G
m
v
dv⊥
G
= q v⊥ × B + qVd × B + & ρˆ
m
ρ
dt
Taking the time − average of this equation gives
G G m v&2
ρˆ = 0
qVd × B +
ρ
Solving for the drift velocity we get the so called curvature drift velocity
G
G m v ρˆ × B
Vd =
2
qρ B
2
&
Generalising this expression to a curved field linewith curvature R we
have
G
G m v Rˆ × B
VR =
2
R qB
2
&
For a curved magnetic field line the magnetic field is non-uniform we must
then add the grad B-drift to get the total drift velocity. For a vacuum field we
have locally
∇B
Rˆ
=− 3
B
R
The local total drift of a particle in a curved irrotational field
then becomes finally
G
G
G m Rˆ × B 2 1 2
( v& + v⊥ )
V∇B + VR =
3
2
2
q R B
Example: Particle motion in dipole magnetic field (Problem 2.8 in Chen)
−5
Suppose the earth’s magnetic field is 3 ⋅10 T
at the equator and falls off
as 1/r3 , as for a perfect dipole. Let there be an isotropic population of 1eV
protons and 30-keV electrons, each with density n=107m-3 at r=5 earth radii in
the equatorial plane.
a) Compute the ion and electron grad B drift velocities.
b) Does an electron drift eastward or westward?
c) How long does it take for an electron to encircle the earth?
d) Compute the ring current in A/m2
Problem 2.8
The grad B drift is given by
G
G
1 mv⊥2 G
V∇B =
B×∇ B
3
2 qB
RE 3
) RE = earth radius 6.4 ⋅106 m
r
B0 = 3.0 ⋅10−5 T
B(r ) = B0 (
3
B( r )
R
∇B = −3 B0 E4 rˆ = −3
rˆ
r
r
G
B( r )
B( r ) 2 ˆ
φ
B × ∇B = B(r ) zˆ × −3
rˆ = −3
r
r
1 2 1 2 3
Isotropic protons kineticenergy = m v⊥ + m v = κT
2
2
2
1
Convention(Chen)1eV = κ T
κT per degrees of freedom
2
1
m v⊥2 = 1eV = 1.6 ⋅10−19 J
2
Problem 2.8
G
G 1 mv⊥2
1 mv⊥2 G
B(r ) 2 ˆ
V∇B =
(−3
B×∇ B =
)φ =
3
3
2 qB
2 q
B(r ) r
1 mv⊥2 − 3 ˆ 1 mv⊥2 − 3 r 2 ˆ
(
)φ =
(
)φ
=
3
2 q B(r )r
2 q B0 RE
evaluated at r = 5 RE we get
G
1 mv⊥2 − 3 ⋅ (5 RE ) 2 ˆ
V∇B (5 RE ) =
φ
3
B0 RE
2 q
3 ⋅ 25 ˆ
75
=−
φ =−
φˆ = −0.39 φˆ m / s
−5
6
B0 RE
3 ⋅10 6.4 ⋅10
The protons drift in the westward direction
Problem 2.8
evaluated at r = 5 RE we get
G
1 mv⊥2 − 3 ⋅ (5 RE ) 2 ˆ
V∇B (5 RE ) =
φ
3
2 q
B0 RE
3 ⋅ 25 ˆ
75
ˆ = −0.39 φˆ m / s
=−
φ =−
φ
3 ⋅10 −56.4 ⋅10 6
B0 RE
For the electrons at r = 5 RE we get
G
1 mv⊥2 − 3 ⋅ (5 RE ) 2 ˆ
V∇B (5 RE ) = −
φ
3
2 e
B0 RE
3 ⋅ 25 ˆ
3 ⋅10 4 75
ˆ = 11719 φˆ m / s
= 30000
φ=
φ
B0 RE
3 ⋅10 −56.4 ⋅10 6
Problem 2.8
c) How long does it take for an electron to encircle the earth?
2 π (5 RE ) 10 ⋅ π ⋅ 6.4 ⋅106
T=
=
= 17157 sek
G
11719
V∇B
=
17157
hr = 4.8 hr
3600
d) Compute the ring current in A/m2
G
G
G
j = n (eV∇B , p − eV∇B ,e ) = 107 ⋅1.6 ⋅10−19 (−.39 − 11719)φˆ =
= −0.18 ⋅10−7 φˆ A / m 2
Particle motion in magnetic mirrorrs
G
1 ∂
∂
(r Br ) + Bz = 0
From ∇⋅ B = 0 we have
∂z
r ∂r
∂Bz
is given at r = 0 and does not vary much
∂z
with r then we have approximately
If
1 ⎛ ∂B ⎞
Br ≈ − r ⎜ z ⎟
2 ⎝ ∂z ⎠r =0
Magnetic mirrors cont.
The z-component of the equation of motion can be written
dv z
q
q r ∂Bz
q rL ∂Bz
= − vθ Br = vθ (
) r = 0 = − v⊥ (
) r =0
2 ∂z
dt
m
m 2 ∂z
m
From which we get
dv z
q v⊥ mv⊥ ∂Bz
mvz
(
) r =0 v z
= −m
dt
m 2 qBz ∂z
θ
z
+
B
Ion with guiding centre
at the z-axis
dv z
mv⊥2 ∂Bz
=−
mvz
(
) r =0 v z
dt
2 Bz ∂z
Magnetic mirrors cont.
d 1 2
mv ∂Bz
dz
W⊥ dBz
( mvz ) = −
(
) r =0
=−
dt 2
2 Bz ∂z
dt
Bz dt
2
⊥
mv⊥2 ∂Bz
dz
W⊥ dBz
d 1 2
=−
(
) r =0
( mv z ) = −
dt
Bz dt
dt 2
2 Bz ∂z
The kinetic energy is conserved for a particle in a magnetic field
so
d 1 2
d 1 2
( mv z ) + ( mv⊥ ) = 0
dt 2
dt 2
therefore
dW⊥
W⊥ dBz
d 1 2
d 1 2
=−
( mv z ) = − ( mv⊥ ) = −
dt
Bz dt
dt 2
dt 2
Magnetic mirrors cont.
d 1 2
d 1 2
dW⊥
W⊥ dBz
( mvz ) = − ( mv⊥ ) = −
=−
dt 2
dt 2
dt
Bz dt
rearranging
dW⊥ dBz
=
W⊥
Bz
W⊥
µ=
= constant The magnetic moment is an
Bz
approximate constant of motion
The magnetic moment of a particle
+
B
The magnetic dipole moment m is defined as
G
m = I S zˆ
For a charged particle we can define the magnetic moment as
m v⊥ 2 m v⊥2 W⊥
q
q2 B
Ω
S=
=
µ=IS = S =q
π(
) =
T
m 2π
qB
B
2π
2B
Trapped particles in a mirror field
z
Suppose we have a particle at z=0 and r=0 with initial velocity
v⊥ = v⊥ ,0
µ=
m v⊥2 ,0
2 Bz (0)
v& = v&,0
the magnetic moment of the particle is then
which is a constant of motion, therefore
where the primed velocity is the velocity at
an arbitrary position. B’ the corresponding
magnetic field at this point.
We also have energy conservation so
1
1
1
1
B′
2
2
2
2
2
′
+
=
+
m
v
m
v
m
v
m v&′2
v & ,0 ≤ (
− 1 ) v ⊥ ,0
⊥ ,0
& ,0
⊥
2
2
2
2
B0
m v⊥′2
m v⊥′2
=
=
µ=
2 Bz (0) 2 Bz ( z ) 2 B′
m v⊥2 ,0
Trapped particles
The conservation of magnetic moment and
kinetic energy gives
m v⊥′2
m v⊥′2
=
=
µ=
2 Bz (0) 2 Bz ( z ) 2 B′
m v⊥2 ,0
1
1
1
1
m v⊥2 ,0 + m v&2,0 = m v⊥′2 + m v&′2
2
2
2
2
v &′ 2 = v &2,0 + (1 −
B′ 2
) v ⊥ ,0
B0
We note that there is a possibility to have the parallell velocity zero at
maximum B if the following condition of the initial parallel velocity fulfills the
condition
B′
v ≤ ( − 1)v⊥2 ,0 These particles are therefore trapped in the magnetic mirror
B0
2
& ,0
Trapped particles
We define a pitch-angle Θ as follows
Loss cone
tan Θ =
v&2,0
v⊥ ,0
v&,0
B′
v ≤ ( − 1)v⊥2 ,0
B0
2
& ,0
The maximum parallel initial velocity for the
particle to be trapped is
v&,0,max ≤ (
Bmax
− 1)1/ 2 v⊥ ,0 = ( Rm − 1)1/ 2 v⊥ ,0
B0
where Rm =
Bmax
is the so called mirror ratio
B0
Using the definition of the pitch-angle this condition can be written
tan Θ ≥
1
Rm − 1
Particles that dont fulfill this condition are lost
Particles trapped in the earth’s
magnetic field
Adiabatic invariants
The magnetic moment µ is an example of an adiabatic invariant.
The definition of an adiabatic invariant is
rL
~
µ (t ) = µ + O(exp(−1 / ε ) f (t ) where ε = << 1
L
The error is the second termwhich when Taylor-expanded for small ε gives
exp(−1 / ε ) = 0 ⋅ ε + 0 ⋅ ε 2 + 0 ⋅ ε 3 + 0 ⋅ ε 4 + ...
This means that the error i smaller than any power of the small parameter ε.
Adiabatic invariants
ω (t ) =
g
l (ε t )
If the length of the pendelum is modified slowly then
W (t )
= const
ω (t )
Adiabatic
invariant
∆ω
ω
<< 1 over a period
ε << 1
Problem 2.11
•
A plasma with an isotropic velocity
distribution is placed in a magnetic mirror
trap with mirror ratio Rm=4. There are no
collisions, so the particles in the loss cone
simply escape and the rest remains trapped.
What fraction is trapped?
z
vz
Isotropic velocity distribution
f ( v) = F ( v )
F is constant on a sphere with radius
v
vy
Kinetic energy is a constant of motion
vx
1
1
1
1
m v⊥2 ,0 + m v&2,0 = m v⊥′2 + m v&′2
2
2
2
2
v
Problem 2.11
1
1
1
1
m v⊥2 ,0 + m v&2,0 = m v⊥′2 + m v&′2
2
2
2
2
(1)
The magnetic moment is a constant of motion ->
m v⊥′2
m v⊥′2
µ=
=
=
2 Bz (0) 2 Bz ( z ) 2 B′
m v⊥2 ,0
(2)
Combining (1) and (2) we get
B′ 2
v&′ = v + (1 − )v⊥ ,0
B0
2
2
& ,0
Particles are reflected at a point where the parallel
velocity becomes zero
B′ 2
0 = v + (1 − )v⊥ ,0
B0
2
& ,0
Problem 2.11
B′ 2
-> 0 = v + (1 −
)v⊥ ,0
B0
2
& ,0
The maximum parallel velocity at the origin for the particle to be trapped is then
Bm
v = ( − 1) v⊥2 ,0
B0
2
& ,0
v⊥2 ,0
v&2,0
v⊥ ,0
v&,0
=
Bm
Rm =
B0
1
Bm
−1
B0
= tan θ m =
1
Rm − 1
Particles that are lost from the mirror are therefore the particles located
in the two loss cones.
Problem 2.11
vz
v
vy
vx
The number of particles per volume in the two loss cones is given by
∞ θ m 2π
∫
loss cones
∞
F ( v ) dvx dv y dvz = 2 ∫ ∫
0 0 0
= 4π ∫ F (v) v 2 dv (1 − cos(θ m ))
0
∫
∞
F (v) v sin(θ ) dv dθ dϕ = 4π ∫ F (v) v 2 dv
2
0
[ − cos(θ )]0
θm
Problem 2.11
The number of particles per volume in the two loss cones is given by
∞ θ m 2π
∫
loss cones
F ( v ) dvx dv y dvz = 2 ∫ ∫
∫
0 0 0
∞
F (v) v sin(θ ) dv dθ dϕ = 4π ∫ F (v) v 2 dv
2
[ − cos(θ )]0
θm
0
∞
= 4π ∫ F (v) v 2 dv (1 − cos(θ m ))
0
The total number of particles per volume is
∫
All velocity space
∞ π 2π
F ( v ) dvx dv y dvz = ∫ ∫
∫ F (v ) v
0 0 0
∞
2
sin(θ ) dv dθ dϕ = 4π ∫ F (v) v 2 dv
0
The fraction of particles lost and trapped particles in the mirror is therefore
Losses = 1 − cos(θ m )
Trapped = cos(θ m )
Problem 2.11
In the present problem the mirror ratio is Rm=4
v⊥ ,0
v&,0
= tan θ m =
⇒ θm =
1
1
=
Rm − 1
3
π
6
π
3
Losses = 1 − cos( ) = 1 −
6
2
3
Trapped fraction =
2