Solutions to Chen's Plasma Physics

Since B ∼ r −3 , we obtain
So
∇B = ∂B
∂r ˆr = − 3 r 4 = −3B r
v ∇B = 3E
eBr =
⇒
3Er3
eB 0 Rer 3 = 3Er2
eB 0 Re
3
=
| ∇B
B | = 3 r
3E(eV )r2
B 0 R 3 e
where I have used the fact that B = B 0 R 3 e/r 3 . Now we can plug in the numbers:
(47)
(48)
v ∇B,e = 3 × 30 × 103 × (5 × 6 × 10 6 ) 2
3 × 10 −5 × (6 × 10 6 ) 3 = 1.3 × 10 4 m/s ✷ (49)
v ∇B,i = 3 × 1 × (5 × 6 × 106 ) 2
3 × 10 −5 × (6 × 10 6 = 0.42 m/s ✷ (50)
) 3
b) The magnetic field is azimuthal, from north to south, i.e. the −ˆθ direction. The gradient of the
magnetic field is clearly in the radial direction, so we have B × ∇B ∼ ˆr × ˆθ = ˆφ, which is eastward.
This is for the electrons. The ions, which come without the minus sign to cancel the minus sign in
equation 47, will go in - ˆφ, which is westward. ✷
c) Well, it has to travel a distance L = 2π(5R e ), with the velocity v ∇B,e . So,
T =
L 2π × 5 × 6.4 × 106
=
v ∇B,e 1.3 × 10 4 ≈ 4.5 hours ✷ (51)
d) The current density is given by J = nev, so, using the grad-B velocity in this expression, we get
J = nev ∇B,e = 10 7 × 1.6 × 10 −19 × 1.3 × 10 4 = 2 × 10 −8 A/m 2 ✷ (52)
2-10. A 20 − keV deuteron in a large mirror fusion device has a pitch angle θ = 45 o at
the mid plane, where B = 0.7 T . Compute it’s Larmor radius.
The Larmor radius is given by
r L = mv ⊥
(53)
qB
In natural units, a deuteron has m = 2 and q = +1. Furthermore, v ⊥ = vsin(θ). To find v, we
convert the energy to velocity:
√
√
r L = m 2E
eB m sin(θ) = 2 × 1.6 × 10−27 2 × 20 × 10 3 × 1.6 × 10 −19
1.6 × 10 −19 × 0.7 2 × 1.6 × 10 −27 sin(45 o ) = 0.3 m ✷ (54)
2-12. A cosmic ray proton is trapped between two moving magnetic mirrors with
R m = 5 and initially has W = 1 keV and v ⊥ = v || at the mid plane. Each mirror moves
toward the mid plane with a velocity v m = 10 km/sec (Fig. 2-10).
a) Using the loss cone formula and the invariance of µ, find the energy to which the
proton will be accelerated before it escapes.
b) How long will take to reach that energy?
1. Treat the mirrors as flat pistons and show that the velocity gained at each bounce
is 2v m .
2. Compute the number of bounces necessary.
3. Compute the time T it takes to traverse L = 10 10 km that many times. Factor of
two accuracy will suffice.
a) The loss cone formula is sin 2 (θ m ) = 1/R m , where θ m is the angle of the magnetic mirror and
R m is the mirror ratio B max /B 0 . We can also write down a formula for sin(θ m ) in terms of the
parallel and perpendicular velocities:
sin(θ m ) = 1 √
Rm
=
v ⊥,f
√
v
2
⊥,f
+ v 2 ||,f
=
1
√
1 + (
v ||,f
v ⊥,f
) 2 (55)