Homework #2 (Ch. 3)
APPH E6101x
Plasma Physics 1
toroidal current, which is induced by a big transformer with the plasma
torus as secondary winding. The rotational transform of the field lines
counteracts the losses arising from plasma drifts.
• In a stellarator, the rotational transform is effected by external helical currents. Modern stellarators use modular coils which produce both the confining magnetic field and the rotational transform.
field is approximately 30 µT. Calculate the dipole moment at the Earth center that
would generate such a magnetic field.
3.5 (a) Calculate the gradient of the Earth magnetic field at the magnetic equator
at an altitude of 500 km and the radius of curvature of a magnetic field line, Rc =
|Bθ /(dBθ /dr )|.
(b) What is the speed of the gradient drift and curvature drift for electrons, which
have 3 eV kinetic energy in parallel and perpendicular motion?
3.6 Determine the trajectory [x(t), y(t)] of an electron in crossed fields B =
(0, 0, Bz ) and E = (E x , 0, 0), when the electron is initially at rest, v(t = 0) = 0.
Problems
3.1 Consider a cylindrical straight wire of radius a with a homogeneous distribution
of current density inside. Use Ampere’s law to derive the azimuthal magnetic field
70Hϕ (r ) for r < a and r ≥ a.
3 Single Particle Motion in Electric and Magnetic Fields
3.2 Consider now a cylindrical discharge tube, in which the plasma density profile
and the associated current distribution is parabolic:
!
"
r2
j (r ) = j0 1 − 2 .
a
3.7 The vector of the magnetic field is tangent to the field line. Therefore, the differential equation for a magnetic field line is
ds
= eB .
dt
Here, s = (x, y, z) is a point on the field line and t a parameter, which makes
tick-marks along the trajectory. Write the defining equation for the field line in
components, eliminate t, and show that the equation for a magnetic field line in
the x–z plane reads
What is the magnetic field distribution Hϕ (r ) for r < a in this case?
3.3 (a) What is the electron cyclotron frequency resulting from the Earth magnetic
field at the author’s location? (c.f. Sect. 3.1.3)
(b) What is the gyroradius of an electron with 10 eV kinetic energy in this field?
3.4 (a) The magnetic field created by a dipole of magnetic moment M = Mez reads
in cartesian coordinates:
B(r) =
µ0 3r(r · M) − r 2 M
.
4π
r5
Find the corresponding components Br and Bθ in spherical coordinates (r, θ).
(b) In the equatorial ionosphere the horizontal component of the Earth magnetic
field is approximately 30 µT. Calculate the dipole moment at the Earth center that
would generate such a magnetic field.
3.5 (a) Calculate the gradient of the Earth magnetic field at the magnetic equator
at an altitude of 500 km and the radius of curvature of a magnetic field line, Rc =
|Bθ /(dBθ /dr )|.
(b) What is the speed of the gradient drift and curvature drift for electrons, which
have 3 eV kinetic energy in parallel and perpendicular motion?
3.6 Determine the trajectory [x(t), y(t)] of an electron in crossed fields B =
(0, 0, Bz ) and E = (E x , 0, 0), when the electron is initially at rest, v(t = 0) = 0.
3.7 The vector of the magnetic field is tangent to the field line. Therefore, the differential equation for a magnetic field line is
ds
= eB .
dt
Here, s = (x, y, z) is a point on the field line and t a parameter, which makes
tick-marks along the trajectory. Write the defining equation for the field line in
components, eliminate t, and show that the equation for a magnetic field line in
the x–z plane reads
Problems
dz
Bz
.
=
dx
Bx
71
Solve this differential equation for the dipole field given in Problem 3.4 by separating
the variables and show that the field line is given as
z(x) =
!
2/3
x0 x 4/3 − x 2 ,
where x0 marks the intersection of the field line with the x-axis.
(4π/3)1/3 e2 n i
1
=
4π
ε0 kB Ti
3
Γi =
−2/3
4π
n i λ3Di
3
1 −2/3
N
3 Di
=
3.5 At the magnetic equator, θ = 90◦ and the magnetic field has only a θcomponent Bθ = (µ0 M/4π)r −3 . (a) Hence, dBθ /dr = −3(µ0 M/4π)r −4 and
Rc = r/3 = (RE + 500 km)/3 = 2, 290 km. (b) v R = 2v∇ B = 6W/(qr B). At
H = 500 km altitude, the equatorial magnetic field of 30 µT has decreased by a
factor [RE /(RE + H ]3 = 0.80 to 24 µT. Then v R = 0.11 m s−1 .
%
%
2.6 Starting from the definitions S = −kB i n i ln n i and U = i n i Wi we use
the thermodynamic definition of temperature 1/T = ∂ S/∂U .
1
∂ S ∂λ
=
=
T
∂λ ∂U
=
−kB λ
−kB
% ∂ni
i
∂λ
% & ∂ni
i
%
i
(λWi − ln Z ) +
% ∂ni
i
∂λ
% ∂ni
∂λ
∂n i
∂λ
Wi
Wi + kB (ln Z − 1)
i
Using
∂λ
Wi
% ∂ni
i
'
3.6 Starting from the equations of motion
v̇x = −
Solutions
= kB λ .
3.2 The current I (r ) flowing through a circle of radius r < a determines the magnetic field
n i = 1, we obtain the result λ = (kB T )−1 .
I (r ) = 2π
Problems of Chapter 3
2πr Hϕ = I
r2
a2
⇒
Hϕ =
Ir
,
2πa 2
which increases linearly to a maximum Hϕ = I /(2πa) at r = a. For r > a the
encircled current is always I and Hϕ = I /(2πr ) becomes a decreasing function of
Solutions
367
radius.
3.2 The current I (r ) flowing through a circle of radius r < a determines the magnetic field
I (r ) = 2π
0
′
′
′
r j (r )d r = 2π j0
"
r2
r4
− 2
2
4a
#
⇒
Hϕ = j0
"
r
r3
−
2 4a 2
#
.
3.3 B = 49.4 µT. (a) ωce = 8.7 × 106 s−1 . (b) rce = 0.22 m.
3.4 (a) Use M · r = Mr cos θ to define the angle θ. The unit vector er = r/r .
3r 2 (r · M) − r 2 (r · M)
µ0
4π
r6
µ0 M sin θ
Bθ = |er × B| =
.
4π r 3
Br = B · er =
=
µ0 M 2 cos θ
4π
r3
90◦ ,
(b) Solve the last line for M, insert θ =
Bθ = 30 µT and use the Earth radius
plus about (100–200) km altitude for the ionosphere to obtain M ≈ 8 × 1022 A m−2 .
90◦
3.5 At the magnetic equator, θ =
and the magnetic field has only a θcomponent Bθ = (µ0 M/4π)r −3 . (a) Hence, dBθ /dr = −3(µ0 M/4π)r −4 and
Rc = r/3 = (RE + 500 km)/3 = 2, 290 km. (b) v R = 2v∇ B = 6W/(qr B). At
H = 500 km altitude, the equatorial magnetic field of 30 µT has decreased by a
factor [RE /(RE + H ]3 = 0.80 to 24 µT. Then v R = 0.11 m s−1 .
3.6 Starting from the equations of motion
v̇x = −
e
E − ωce v y
me
and v̇ y = ωce vx
we see that the electron velocity performs harmonic oscillations at the frequency
ωce and v$x = v⊥ sin(ωce t). There is no cosine term because vx (0) = 0. Then,
v y = ωce vx dt + C = v⊥ [1 − cos(ωce t)] to fulfill v y (0) = 0. From v̇x (0) = 0 we
obtain v⊥ = −E/B. Hence,
vx = −
E
!r
0
r ′ j (r ′ )d r ′ = 2π j0
"
r2
r4
− 2
2
4a
#
⇒
Hϕ = j0
"
r
r3
−
2 4a 2
#
E
sin(ωce t) ,
B
vy =
E
[1 − cos(ωce t)]
B
E
E
3.4 (a) Use M · r = Mr cos θ to define the angle θ. The unit vector er = r/r .
µ0 3r 2 (r · M) − r 2 (r · M)
µ0 M 2 cos θ
=
4π
4π
r6
r3
µ0 M sin θ
Bθ = |er × B| =
.
4π r 3
vx = −
E
sin(ωce t) ,
B
3.5 At the magnetic equator, θ = 90◦ and the magnetic field has only a θcomponent Bθ = (µ0 M/4π)r −3 . (a) Hence, dBθ /dr = −3(µ0 M/4π)r −4 and
Rc = r/3 = (RE + 500 km)/3 = 2, 290 km. (b) v R = 2v∇ B = 6W/(qr B). At
H = 500 km altitude, the equatorial magnetic field of 30 µT has decreased by a
factor [RE /(RE + H ]3 = 0.80 to 24 µT. Then v R = 0.11 m s−1 .
3.7 The differential equation for a field line with the dipole source at the origin
reads
dz
2z
1x
=
−
.
dx
3x
3z
Set z = w x to obtain dz/dx = w + x(dw/dx). Inserting into the differential
368
equation
x
Solutions
1
11
dw
= − w−
dx
3
3w
⇒
wdw
1 dx
=−
3 x
w2 + 1
z=
and v̇ y = ωce vx
we see that the electron velocity performs harmonic oscillations at the frequency
ωce and v$x = v⊥ sin(ωce t). There is no cosine term because vx (0) = 0. Then,
v y = ωce vx dt + C = v⊥ [1 − cos(ωce t)] to fulfill v y (0) = 0. From v̇x (0) = 0 we
obtain v⊥ = −E/B. Hence,
E
vx = − sin(ωce t) ,
B
E
[cos(ωce t) − 1] ,
x=
Bωce
E
v y = [1 − cos(ωce t)]
B
E
E
y=− t+
sin(ωce t)
B
Bωce
3.7 The differential equation for a field line with the dipole source at the origin
reads
dz
2z
1x
=
−
.
dx
3x
3z
⇒
1
1
ln(w2 +1) = − ln(x)+C .
2
3
This results in w = (−1 + c1 x −2/3 )1/2 and z = x(−1 + c1 x −2/3 )1/2 . The maximum
2/3
distance xmax of the field line is determined by z = 0, which defines c1 = xmax ,
hence the shape of the field line becomes
3.6 Starting from the equations of motion
e
v̇x = − E − ωce v y
me
E
[1 − cos(ωce t)]
B
E
E
y=− t+
sin(ωce t)
B
Bωce
vy =
E
[cos(ωce t) − 1] ,
Bωce
x=
Br = B · er =
(b) Solve the last line for M, insert θ = 90◦ , Bθ = 30 µT and use the Earth radius
plus about (100–200) km altitude for the ionosphere to obtain M ≈ 8 × 1022 A m−2 .
and v̇ y = ωce vx
we see that the electron velocity performs harmonic oscillations at the frequency
ωce and v$x = v⊥ sin(ωce t). There is no cosine term because vx (0) = 0. Then,
v y = ωce vx dt + C = v⊥ [1 − cos(ωce t)] to fulfill v y (0) = 0. From v̇x (0) = 0 we
obtain v⊥ = −E/B. Hence,
.
3.3 B = 49.4 µT. (a) ωce = 8.7 × 106 s−1 . (b) rce = 0.22 m.
3.1 Ampere’s law states that the integral of the magnetic field strength H along
( a
closed path equals the current flowing though the area bounded by this path, H ·
d s = I . Choose a circle of radius r centered at the axis of the wire for the path.
Then, for any r < a the current flow through this circle is the fraction I r 2 /a 2 and
we obtain
!r
367
∂λ
e
E − ωce v y
me
!
2/3
xmax x 4/3 − x 2 .
Problems of Chapter 4
4.1 Let A be the normalizing factor of the velocity distribution f M (|v|). Then
0= A
#
"
#
"
#
"
d 2
mv 2
mv
mv 2
v exp −
= A 2v − v 2
exp −
,
dv
2kB T
kB T
2kB T
which requires that the expression in parantheses vanishes. This gives the desired
result vT = (2kB T /m)1/2 .
4.2 The mean thermal velocity is the first moment of the distribution of speeds
vth = 4π
Set z = w x to obtain dz/dx = w + x(dw/dx). Inserting into the differential
equation
"
m
2πkB T
#3/2 $∞
0
#
"
# $∞
"
8kB T 1/2
mv 2
dv =
v 3 exp −
ye−y dy
2kB T
πm
0
% &' (
=1
4.3 Reduce the degree of the velocity moment by partial integration
$∞
0
2
v 4 e−av dv = −
1
2a
$∞
2
v 3 (−2ave−av )dv
0
$∞
1 ) 3 −av 2 *∞ 3
2
v e
+
v 2 e−av dv
=−
2a %
&' 0( 2a 0
3 )
=0
2
*∞
3
$∞
2
3
+
π