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PLASMA PHYSICS
II. MOTION OF IONS AND ELECTRONS IN E AND B FIELDS
We consider the paths of ions and electrons in E and B fields for some simple cases.
The Lorentz force on a point charge is F = q( E + v × B) .
E is measured in V m−1. B in T (often in gauss. 10000 gauss = 1 T)
1. E = constant, uniform
Suppose E = Ex$ . The Lorentz force equation becomes
dv x
q
= E
dt
m
dv y
=0
dt
dv z
= 0.
dt
This describes a constant acceleration along x.
2. B = constant, uniform
Suppose B = B$z .
Here is how we might produce a uniform magnetic field.
dv x
q
= vy B
(1)
dt
m
dv y
q
= − vx B
(2)
dt
m
dv z
(3)
= 0.
dt
1
d
of (1), substitute using (2)
dt
d 2 vx
q dvy
q q
=
B =  vx B B .
2

dt
m dt
mm
Take
Write
qB
m
the (angular) cyclotron frequency or gyrofrequency. (Note the symbol Ω is often used.)
ωc =
d 2 vx
+ ω c 2 vx = 0 .
2
dt
Similarly,
d 2 vy
dt
The solutions can be written as
2
+ ω c 2 vy = 0 .
vx = − v⊥ sinω c t
vy = m v⊥ cosω c t
(The signs and phase angles have been chosen to match the sketches below. The upper sign is
for a positive charge, the lower for a negative.)
Integrate again
v⊥
cosω c t = rL cosω c t
ωc
v
y = m ⊥ sinω c t = m rL sinω c t
ωc
x=
rL =
v⊥
ωc
is called the Larmor radius, radius of gyration, or gyroradius.
So a charge in a constant, uniform B moves in a circle with constant speed. Note that the
cyclotron frequency does not depend on how fast the charge is moving.
You do. Check that the directions of motion are correct and that the equations above match the
sketches.
You do. Calculate the cyclotron frequency (in Hz) for (a) hydrogen ions and (b) electrons in a
magnetic field of 1 T.
2
If the charge has a vz, this z-component of the motion is unchanged. The charge moves in a
helical path.
3. E constant, uniform. B constant, uniform.
Suppose B = B$z .
dv x
q
=
Ex + vy B
dt
m
dv y
q
=
Ey − vx B
dt
m
dv z
q
= Ez .
(3)
dt
m
(
)
(1)
(
)
(2)
(3) gives constant acceleration along z.
You do. Suppose E has a z-component only. Describe the motion and sketch the path for this
case.
(1) and (2) are manipulated as before, they give
Ey
d 2 vx
2
2
+
ω
v
=
ω
and
c
x
c
dt 2
B
d 2 vy
E
+ ω c 2 vy = −ω c 2 x ,
2
dt
B
with solutions
Ey
v x = − v ⊥ sinω c t +
B
E
v y = m v ⊥ cosω c t − x .
B
The path of an electron is a combination of uniform circular motion plus a drift, called an
E × B drift.
1
E×B
v E× B =
.
E y x$ − E x y$ =
B
B2
Note that the drift term is independent of the charge and its sign, so all the charges will drift
together. The paths are cycloids.
(
)
3
If E⊥B then v E × B =
E
.
B
Here is an example where the E × B drift can cause a plasma to rotate.
4. B = constant, non-uniform.
We will consider two distinct cases.
Case (a):
4
Bz = B0 (1 + αz )
α
x
2
α
By = − B0 y
2
Bx = − B0
where α is small.
You do. Show Maxwell’s equations ∇. B = 0 and ∇ × B = 0 are satisfied as long as we include
these small Bx and By terms.
The Lorentz force equation becomes
dv x
q
=
v y Bz − v z By =
dt
m
dv y
q
= (v z Bx − v x Bz ) =
dt
m
dv z
q
=
v x By − v y Bx =
dt
m
(
(
)
α
q
 v y B0 + v y B0αz + v z B0 y
2 
m
)
α
q
 − v z B0 x − v x B0 − v x B0αz

2
m
α
α
q
 − v x B0 y + v y B0 x 

2
2 
m
We will write v = v 0 + v 1 where 0 indicates the uniform constant or zero-order part and 1 a
small first-order correction, of the same order as α, and substitute in the equations. This is a
standard approach and we will use it frequently.
The zero-order equations. If we write down the zero-order terms, i.e., the terms in v 0 and
those that do not contain α, the equations that remain describe motion in a constant, uniform
B. This was discussed earlier.
You do. Show this.
The first-order equations. We first solve the zero-order equations to obtain
v x0 , v y0 , v z0 , x 0 , y 0 , z 0 ; then we write down the first order terms, i.e., the terms in v 1 and those
containing α; then substitute for v x0 , v y0 , v z0 , x 0 , y 0 , z 0 .
This gives
dv1x
q
α
=  v1y B0 + v⊥0 cosω c t B0αvz0t + vz0 B0 rL sinω c t 

dt
m
2
dv1y
dt
=
q 0 α

 − vz B0 rL cosω c t − v1x B0 − v⊥0 sinω c t B0αvz0t 

m
2
dv1z
q
α
α
=  v⊥0 sinω c t B0 rL sinω c t + v⊥0 cosω c t B0 rL cosω c t 

dt
m
2
2
which can be written as
5
dv 1x
α
= ±ω c v 1y − αv ⊥0 v z0ω c t cosω c t − v z0 vz0 sinω c t
2
dt
1
dv y
α
= m v ⊥0 v z0 cosω c t m ω c v 1x ± αv ⊥0 v z0ω c t sinω c t
2
dt
1
dv z
α 2
= − v ⊥0
2
dt
upper sign ions, lower sign electrons, with solutions
α 0 0
α
v ⊥ v z ω c t 2 cosω c t − v ⊥0 v z0 t sinω c t
2
2
α
α
v 1y = m v ⊥0 v z0 t cosω c t ± v ⊥0 vz0ω c t 2 sinω c t
2
2
α 2
v 1z = − v ⊥0 t.
2
v 1x = −
Adiabatic invariant
1 2
mv ⊥
2
is a constant of the motion or adiabatic invariant. Adiabatic carries the idea of slowlyB
changing.
(
You do. Show this is true to first order in α. Start with v ⊥2 = ( v x0 + v 1x ) + v y0 + v 1y
2
)
2
and
substitute using the solutions above. (Note. Chen p 31 gives an alternative derivation.)
You do. Use the definition of rL and this result to show that the magnetic flux encircled by
orbit Φ M = BA is constant.
It follows that the magnetic moment of the gyrating charge is constant. The magnetic moment
q
is defined as µ = iA where i = , where q is the charge and t is the time for one gyration and
t
A is the area encircled by the orbit.
1 2
mv⊥
e
µ=
πrL 2 = 2
.
2π
B
ωc
So µ is constant.
Magnetic mirror
As an electron spirals into a higher B region, v ⊥ increases and rL decreases. Since the total
1
energy mv 2 is a constant, vz must decrease. Eventually vz = 0 and the electron reverses
2
direction. It has been reflected by a magnetic mirror.
6
e.g., magnetic mirror used to trap plasma in an experimental device.
1 2
mv ⊥
2
To do a magnetic mirror calculation use
= constant and conservation of energy
B
1 2 1 2
mv ⊥ + mv z = constant.
2
2
Case (b):
Bz = B0 (1 + αx )
Bx = B0αz
You do. Show Maxwell’s equations are satisfied.
B0αx describes the gradient.
B0αz describes the curvature. In the derivations below, the curvature terms are underlined.
dv x q
= (v y B0 + v y B0αx )
dt
m
dv y q
=
v z B0αz − v x B0 − v x B0αx
dt
m
dv z
q
= (− v y B0αz )
dt
m
(
)
7
Proceed as before.
dv 1x
2
= ±ω c v 1y − αv ⊥0 cos 2ω c t
dt
dv 1y
2
2
= ± αω c v z0 t m ω c v 1x ± αv ⊥0 sinω c t cosω c t
dt
dv 1z
= αω c v ⊥0 v z0 t cosω c t
dt
upper sign ions, lower sign electrons, with solutions
2
αv 0
2
v = − ⊥ sin 2ω c t + αv z0 t
2ω c
1
x
2
2
αv 0
αv 0 αv 0
v = m ⊥ cos2ω c t ± ⊥ ± z
ωc
2ω c
2ω c
2
1
y
αv ⊥0 v z0
(cosω c t − 1 − ω c tsinω c t )
v =
ωc
1
z
Consider v 1y . There are constant drift terms.
2
αv 0
± ⊥ due to the gradient
2ω c
2
αv 0
± z due to the curvature.
ωc
They combine to give
vd =
 v ⊥0 2
2
±α 
+ v z0 
 2

ωc
.
This is the expression we will use.
It is perhaps unfortunate that these drifts are in the same direction. We cannot devise a B such
that they cancel.
We can express α in terms of the gradient of the magnetic field or the radius of curvature Rc
of the field lines.
(i) From the equations for B above, α =
(ii) From the sketch above, α =
1
Rc
You do. Show this.
8
∇B z
.
B0
gradient drift
It is easy to see why a gradient gives rise to a drift. Consider the path of a charge where the B
field is large above the line and small below it. Above, the Larmor radius is small and below,
it is large.
We can sketch the drift.
The drift, for positive ions, is in the direction of − ∇B × B or R c × B .
e.g. in a toroidal magnetic field
B=
µ 0 Ni
2πr
1
1
and ∇B ≈ − 2 . i.e., ∇B increases as you go radially in towards the axis.
r
r
∇B
1
=− .
In this case α =
B
r
so B ≈
9
e.g., radiation belts in the earth’s magnetic field. This illustrates the magnetic mirror as well.
5. Magnetic field with time variation
Drift and mirroring equations do not allow the long range prediction of trajectories,
particularly if there is no symmetry. It is nice to have constants of the motion or invariants.
Again it can be shown that even when the magnetic field varies in time, the magnetic moment
µ is constant. This is the first adiabatic invariant.
e.g. Adiabatic compression as a method of heating a plasma
Suppose a plasma is trapped by a magnetic field. If the magnetic field is increased then v⊥
increases. Collisions will distribute this extra energy. The plasma is heated.
There are two other invariants. They are illustrated by the following example.
(1) µ = constant
(2) Longitudinal (or second) adiabatic invariant
J=
∫
over a path
back and forward
between mirrors
v ⋅ dl = constant
So if the location of the mirrors changes slowly with time, due to the solar wind, this remains
constant.
(3) Third adiabatic invariant
The guiding centre may precess going from one field line to another. But the field lines all lie
on a flux surface - a barrel-shaped surface such that the enclosed flux is constant.
10
Exercises for Chapter 2
B = constant, uniform
1.
Calculate cyclotron frequencies and Larmor radii for
(i) 18 keV deuteron in a fusion reactor. B = 5.7 T
(ii) 5 eV electron in a plasma CVD source. B = 200 gauss.
(iii) 10 keV electron in the earth’s magnetic field. B = 0.5 gauss.
E constant, uniform. B constant, uniform.
2.
In a low temperature plasma device called a magnetron, B is typically 300 gauss, the
potential difference V is 500 V over 2 mm in the region of interest and the E is perpendicular
to the B.
Estimate the drift velocity of the electrons.
B = constant, non-uniform. Case (a):
3.
(a) In a magnetic mirror where the magnetic field is B0, the trajectory makes an angle
θ0 with the magnetic field line.
B0
.
Show that reflection occurs where the magnetic field is B =
sin 2θ 0
(b) Suppose now Bmax is the maximum value of the magnetic field.
B0
then there is no reflection. In a magnetic mirror device,
Bmax
this would mean the particle was not trapped - it would be lost. This angle defines a cone in
velocity space - the loss cone
Show that if θ 0 < sin −1
4.
In our gyrotron millimetre-wave source most of the electrons travel from the electron
gun through the resonant cavity. But some electrons are reflected.
At the electron gun (z = −0.30 m) the magnetic field is 0.52 T, at the resonant cavity (z
= 0 m) it is 12.0 T.
Consider a particular electron that is reflected at the cavity. It leaves the gun with an
energy of 10 keV. Its guiding centre is 3.00 mm from the axis.
Calculate
(i) v⊥ at the cavity,
(ii) v⊥ and hence vz at the gun,
(iii) the distance to the guiding centre from the axis at the cavity,
(iv) the Larmor radius at the cavity,
(v) the Larmor radius at the gun.
B = constant, non-uniform. Case (b):
1
5.
In a small experimental plasma device a toroidal B is produced by uniformly winding
120 turns around a toroidal vacuum vessel, and passing a current of 250 A through it. The
major radius is 0.6 m.
A plasma is produced in hydrogen by a radiofrequency field. The electron temperature
is 80 eV and the ion temperature is 10 eV. The two temperature distributions are Maxwellian.
The plasma density at the centre of the vessel is 1016 m−3.
(i) Calculate the B field at the centre of the vessel
(ii) Calculate the total drift for both ions and electrons at the centre of the vessel. On a
sketch, show the directions of these drifts.
3
R 
6.
The earth’s magnetic field, in the equatorial plane, is B = 3 × 10  e  T .
 r 
At 5 Re in one of the Van Allen radiation belts, the electrons have an energy of 30 keV
and the protons an energy of 1 eV.
−5
(i) Calculate the total drift for both protons and electrons. On a sketch show the
directions of these drifts.
(ii) If the plasma density is 105 m−3, calculate the ring current density.
Re = 6.37 × 106 m.
7.
For the Van Allen belt in Exercise 6, estimate the following times.
(i)
cyclotron period,
(ii)
time between mirror reflections,
(iii) time to drift once around the earth.
2