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Chapter 1 and 2: Introduction and single particle motion
MEMORANDUM-1
1. Fundamental constants
Electronic charge
Electron mass
Proton mass
Boltzmann’s constant
Permittivity of vacuum
Permeability of vacuum
e = 1.6 · 10−19 C
me = 9.1 · 10−31 kg
mp = 1.67 · 10−27 kg
K = 1.38 · 10−23 J/K
µ0 = 4π10−7 H/m
²0 = 8.85 · 10−12 F/m
2. Constants
1eV = 1.6 · 10−19 J
1eV = K · 11600 K
Ion mass mi = Amp
Ion charge q = Ze
KT = 1.6 · 10−19 TeV
(TeV = temperature in eV)
(A = mass in a.m.u.)
(Z = charge number)
3. Formulas
¡ε
0 kTe
ne2
¢1/2
¡ T ¢1/2
Debye length
λD =
Number of particles in a Debye sphere
ND = 34 πλ3D n = 1.72 · 1012 neV
1/2
Electron cyclotron frequency
ωc =
|q|B
m
= 1.76 · 1011 B
Ion cyclotron frequency
ωc =
|q|B
m
= 0.96 · 108 ZB
A
Larmor radius (electron)
rL =
mv⊥
|q|B
= 3.37 · 10−6
Larmor radius (ion)
rL =
mv⊥
|q|B
= 1.44 × 10−4
Bolzmann’s distribution for electrons
ne = n0 eeφ/KTe
= 7430
eV
n
T
1
√
3/2
E⊥eV
B
√
AE⊥eV
ZB
(m)
Chapter 1 and 2: Introduction and single particle motion
4. Drift Velocities
E-field drift of the guiding center
vE =
E×B
B2
General force drift
vf =
1 F ×B
q B2
Grad-B drift
v ∇B = ± v⊥2rL B×∇B
B2
Curvature drift
vR =
2
mv||
RC ×B
2
2
qB
RC
5. Magnetic mirror (∇B ||B)
2
mv⊥
2B
Magnetic moment
µ=
Loss cone
sin2 θm =
= const
B0
Bm
=
6. Adiabatic invariants
1st
µ=
2nd
J=
2
mv⊥
2B
Rb
= const
v|| ds = const
a
Units
Electric field
[E]= V/m
Magnetic induction = magnetic flux density
[B]= T (Wb/m2 0 N/ Am)
2
vE =
1
Rm
E
B
Chapter 1 and 2: Introduction and single particle motion
Problem 1-7
Compute λD and ND for the following cases
(a) A glow discharge, with n = 1016 m−3 , KT e = 2 eV.
(b) The earth’s ionosphere, with n = 1012 m−3 , KTe = 0.1 eV.
(c) A θ -pinch, with n = 1023 m−3 , KTe = 800 eV.
—————————————————————————————————————————————
(a)
µ
λD = 7430
TeV
n
¶1/2
µ
= 7430
2
1010
¶1/2
≈ 10−4 m = 100
µm
3/2
ND = 1.72 × 1012
23/2
TeV
12
=
1.72
×
10
≈ 48650 >> 1
n1/2
(1016 )1/2
(b)
µ
λD = 7430
TeV
n
¶1/2
µ
= 7430
0.1
1012
¶1/2
≈ 2.35
mm
3/2
ND = 1.72 × 1012
3/2
TeV
12 (0.1)
=
1.72
×
10
≈ 54390 >> 1
n1/2
(1012 )1/2
(c)
µ
λD = 7430
TeV
n
¶1/2
µ
= 7430
3/2
ND = 1.72 × 1012
800
1023
¶1/2
≈ 6.65 · 10−7
m
3/2
TeV
12 (800)
=
1.72
×
10
≈ 1.23 · 105 >> 1
n1/2
(1023 )1/2
3
Chapter 1 and 2: Introduction and single particle motion
Problem 1-10
A spherical conductor of radius a is immersed in a plasma and charged to a potential φ0 . The electrons remain
Maxwellian and move to form a Debye shield, but the ions are stationary during the time frame of the
experiment. Assuming φ0 << KTe /e , derive an expression for the potential as a function of r in terms of a, φ0
and λD .
—————————————————————————————————————————————
Poisson’s equation:
ε0 ∇2 φ = −ρ
(1)
Density of charged particles:
ρ = e(ni − ne )
(2)
ni = n0 = const
(3)
Boltzmann’s distribution:
ne = n0 eeφ/KTe
(4)
Using Equation (3) and (4) into Equation (2) we get,
¡
eφ/KTe
ρ = en0 1 − e
¢
µ
≈ en0
eφ
1−1−
KTe
as far as eeφ/KTe ≈ 1 + eφ/KTe
Thus Poisson’s equation yields:
∇2 φ =
Laplacian (in spherical coordinates):
4
φ
λ2D
¶
= −ε0
φ
λ2D
(5)
(eφ << KTe ).
(6)
Chapter 1 and 2: Introduction and single particle motion
1 ∂
∇ φ= 2
r ∂r
2
Substitution: φ =
ψ
r
µ
µ
¶
1
∂
∂φ
1
∂ 2φ
r
+ 2
sin θ
+ 2 2
∂r
r sin θ ∂θ
∂θ
r sin θ ∂ϕ2
µ
¶
1 d
φ
2 dφ
⇒ 2
r
= 2
r dr
dr
λD
2 ∂φ
¶
(7)
yields,
ψ
d2 ψ
= 2
2
dr
λD
⇒
(8)
General solution:
ψ = Ae−r/λD + Ber/λD → Ae−r/λD ⇒ φ =
A −r/λD
e
r
Boundary conditions:
φ|r=a = φ0
⇒
A −a/λD
= φ0
e
a
⇒
a
φ = φ0 e(a−r)/λD
r
5
A = aφ0 ea/λD
⇒
(9)
Chapter 1 and 2: Introduction and single particle motion
Problem 2-2
In the TFTR (Tokamak Fusion Test Reactor) at Princeton, the plasma will be heated by injection of 200-keV
neutral deuterium atoms, which, after entering the magnetic field, are converted to 200-keV D ions (A=2) by
charge exchange. These ions are confined only if rL << a , where a = 0.6 m is the minor radius of the toroidal
plasma. Compute the maximum Larmor radius in a 5-T field to see if this is satisfied.
—————————————————————————————————————————————
√
−4
rL = 1.44 × 10
√
5
AE⊥eV
−4 2 · 2 · 10
= 1.44 × 10
≈ 0.02 m = 2 cm
ZB
5
Thus, the ions are well confined.
Figure 1: Illustration of a tokamak.
6
Chapter 1 and 2: Introduction and single particle motion
Problem 2-7
An unneutralized electron beam has density ne = 1014 m−3 and radius a = 1 cm and flows along a 2-T magnetic
field. If B is in the +z direction and E is the electrostatic field due to the beam’s charge, calculate the magnitude
and direction of the E × B drift at r = a.
—————————————————————————————————————————————
Drift velocity:
vE =
E×B
B2
(1)
Calculation of E-field:
Maxwell’s equation:
∇·E =
ρ
ε0
(2)
Integration over a volume:
Z
Z
∇ · E dV =
ρ
dV
ε0
LHS :
RHS :
R
R
E · dS = Er · 2πrl
ρ
ε
(dS = n · dS)
(3)
dV =
e
− en
πr2 l
ε0
(3) ⇒ Er = −
(ρ = −ene )
ene
r
2ε0
Calculation of drift velocity:
vE =
E×B
B2
At r = a:
E|r=a = −r̂
ene
a,
2ε0
7
B = ẑB
(4)
Chapter 1 and 2: Introduction and single particle motion
⇒ vE =
E×B
ene a
ene a
θ̂
=−
r̂ × ẑ =
2
B
2ε0 B
2ε0 B
And inserting numbers:
vE = |v E | =
ene a
1.6 · 10−19 · 1014 · 0.01
=
≈ 4.5 · 103 m/s = 4.5 km/s
2ε0 B
2 · 8.85 · 10−12 · 2
8
Chapter 1 and 2: Introduction and single particle motion
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 l oss cone simply escape, and the rest remain trapped. What fraction
is trapped?
—————————————————————————————————————————————
Loss fraction:
ηloss =
2∆Ω
nloss
∆Ω
=
=
ntotal
4π
2π
(1)
∆Ω
2π
(2)
Trapped fraction:
ηtrap = 1 − ηloss = 1 −
∆Ω=solid angle; calculation of ∆Ω:
∆Ω =
∆S
r2
(3)
dS = 2πρ · rdθ = 2πr2 sin θ · dθ
⇒
Zθm
2πr2 sin θ · dθ = 2πr2 (1 − cos θm )
∆S =
⇒
0
∆Ω =
2πr2 (1 − cos θm )
= 2π(1 − cos θm )
r2
⇒
(4)
Inserting (4) into (2) gives:
ηtrap = 1 −
∆Ω
2π(1 − cos θm )
=1−
= cos θm
2π
2π
9
⇒
(5)
Chapter 1 and 2: Introduction and single particle motion
Loss cone:
sin2 θm =
sin θm =
ηtrap
1
1
B0
=
=
Bm
Rm
4
1
2
⇒
θm =
⇒
π
= 30◦
6
√
π
3
= cos =
≈ 0.87
6
2
Figure 2: Example of trapped population of particles in the Earth’s magnetic field, called the radiation belts. These
particles do not only gyrate and bounce, they alo undergo a drift motion as an effect of the gradient and curvature of
the dipole magnetic field. Electrons drift eastward and ions drift westward and consitutes a current, called the ring
current.
10
Chapter 1 and 2: Introduction and single particle motion
Problem 2-17
A 1-keV proton with v|| = 0 in a uniform magnetic field B = 0.1 T is accelerated as B is slowly increased to
1 T. It then makes an elastic collision with a heavy particle and changes direction so that v⊥ = v|| .The B-field is
then slowly decreased back to 0.1 T. What is the proton’s energy now?
—————————————————————————————————————————————
Magnetic moment:
µ=
2
mv⊥
= const
2B
region (1):
2
mv⊥0
mv0 2⊥
=
2B0
2B
⇒
2
2
v0 ⊥ = v⊥0
B
B0
(1)
region (2):
2
2
+ v||2
Energy conservation ⇒ v0 ⊥ = v⊥
(∆E ∝
mp
E
mph
<< 1) (elastic collision; heavy particle)
v⊥ = v|| ⇒
 2
1 2
1 2 B
 v⊥ = 2 v0 ⊥ = 2 v⊥0
B0

2 B
v||2 = 21 v⊥0
B0
11
(2)
Chapter 1 and 2: Introduction and single particle motion
region (3):
2
mv⊥
2B
=


2
mv⊥f
⇒
2B0

2
2
2 B0
= [using (2)] = 21 v⊥0
v⊥f
= v⊥
B
2
v||f
=
(3)
v||2
Initial energy is,
1 2
E0 = mv⊥0
,
2
and the final energy:
¢ 1
1 ¡ 2
2
Ef = m v||f
+ v⊥f
= m
2
2
µ
1 2 B
1 2
v⊥0
+ v⊥0
2
B0 2
¶
µ
µ
¶
¶
1 2
B
1
B
= mv⊥0 1 +
= E0
1+
4
B0
2
B0
Inserting numbers:
1
Ef = E0
2
µ
B
1+
B0
¶
12
=
11
E0 = 5.5 keV
2
(4)
Chapter 1 and 2: Introduction and single particle motion
Problem 2-20
Suppose the magnetic field along the axis of a magnetic mirror is given by Bz = B0 (1 + α2 z 2 ) .
2
, at what value of z is the electron reflected?
(a) If an electron at z = 0 has a velocity given by v2 = 3v||2 = 1.5v⊥
(b) Write the equation of motion of the guiding center for the direction parallel to the field.
(c) Show that the motion is sinusoidal, and calculate its frequency.
(d) Calculate the longitudinal invariant J corresponding to this motion.
—————————————————————————————————————————————
(a) z reflected:
µ=
2
mv⊥
= const
2B
(1)
2
2
v⊥0
= v2
3
2
2
v⊥ref
l = v
(2)
(v|| = 0 at reflection point.)
(3)
Equation (1), (2), and (3) gives,
2
2
mv⊥ref
mv⊥0
l
=
2B0
2Bref l
3
Bref l = B0
2
⇒
From problem definition of Bz and (4) ⇒
2
α2 zref
l =
(b) Equation of motion:
mz̈ = Fz
Lets find this force (see p. 30-32 in Chen):
∇·B =0
13
1
2
(4)
⇒
1
zref l = √
2α
(5)
Chapter 1 and 2: Introduction and single particle motion
·
¸
1 ∂Bz
∂Bz
⇒ Br = − r
( assuming
given at r = 0 and does not vary much with r)
2
∂z
∂z
The component of the Lorentz force of interest is:
1
Fz = −vθ Br = qvθ r
2
µ
∂Bz
∂z
¶
Averaged over one gyrations we obtain:
Fz = −µ
∂Bz
= −2µB0 α2 z
∂z
Thus the equation of motion becomes:
mz̈ = −2µB0 α2 z
(6)
(c) Sinusoidal motion:
New notation:
r
Ω=
s
2µB0 α2
=
m
2
√
2B0 α2 mv⊥0
= αv⊥0 = α 2v||0
m 2B0
Results in the equation of motion according to Equation (6):
z̈ + Ω2 z = 0
⇒
z = A sin Ωt + B cos Ωt
Initial condition:
z(t = 0) = 0
⇒
B=0
(d) Longitudinal invariant:
14
⇒
z = A sin Ωt
(7)
Chapter 1 and 2: Introduction and single particle motion
zZref l
J=
v|| dz
(8)
−zref l
v|| = ż = AΩ cos Ωt,
v|| (t = 0) = v||0
z=
v|| = v||0 cos Ωt = v||0
⇒
A=
v||0
Ω
⇒
v||0
sin Ωt
Ω
(9)
s
p
1 − sin2 Ωt,
⇒
v|| = v||0
1−
z2
2
zref
l
(10)
Thus, (10) into (8) yields:
zZref l
J = v||0
−zref l
s
1−
πv||0
z2
π
√
dz
=
v
z
=
ref
l
||0
2
zref
2
2 2α
l

Z1 √
Table’s integral (β):
−1
15

1 − x2 dx =
π
2
(11)