## Monday, October 17, 2022

### From Debye Length To Gyro-Period And Larmor Radius: Solving Basic Problems In Plasma Physics

The Debye length, the plasma parameter, loss cone angle, mirror machines, Larmor radius, gyrofrequency, guiding center and gyro-period are all elements of basic plasma physics.

Below I show how example problems involving these parameters are solved.  Most are taken from my book, Explorations in Mathematical Physics, Chapter 8,

1) A helium plasma is at a temperature of  10 6 K, and exhibits a number density of

s  10 10 /m3.   Find  the Debye length  l D,s  and the plasma parameter, L

Solution:

For helium plasma we have:   T sT He = 10

Debye length is:

l D =[kT  / 4p  n e  e2 ] 1/2

Where k is Boltzmann’s constant and e is the electronic charge.

For this ion-species based Debye length we may write:

l D,s =[kT s / 4p  Z s2 n s  e2 ] 1/2

Where the subscript ‘s’ refers to the ion species. Here, for helium:

Z s=  2,    e  =  1.6  x 10 -19  C

n s   =  10 10 /m3

l D,s   =

[(1.38 x 10-23  (10 K) (8.85 x 10 -12 F/m) / (4p)  22 (10 10/m3)  e2 ] ½

l D,s   =  0.30    m

The plasma parameter is defined:

L  =  s  l3 D

From the values given (T =10 K, and n s10 10 /m3)

L  =  s  l3 D   =  10 10 /m3) (0.30  m ) 3

L   =   2.9 x 10

(2) Consider a plasma mirror machine of length 2L with a mirror ratio of 10 so that B(L) = B(-L) = 10 B(0). E.g.

A group of N (N > 1) electrons with an isotropic velocity distribution is released at the center of the machine. Ignoring collisions and the effect of space charge, how many electrons escape?

Solution:

The mirror ratio is (Bmin / B max) = 10, meaning that the induction strength at those end points will be ten times the induction at the center point or apex of the magnetic loop or mirror machine.

We define what is called the “loss cone angle”:

sin (q L ) = ± Ö (Bmin / B max )

In the problem, Bmin = B(0) or the "zero level" for the magnetic induction, say at position L = 0. This doesn't mean the induction is zero at that point literally, however.

To do the problem, one must understand he's really being asked for the fraction of electrons lost. A special condition obtains which applies to the angle - for which the electrons will be TRAPPED only provided:

Θ (O) > (Θ)L

Thus, Θ (O) = (Θ)L

is said to be the "loss cone" of the system or machine.

If an isotropic particle distribution (in this case, electrons) is introduced at a position L = 0, the fraction of particles that will be lost to the mirror system is:

f(L) = 1/ 2 π
ò o  Θ(L)   2 π sin(Θ) d Θ

= f(L) =

ò o Θ(L)   sin(Θ) d Θ

= 1 - cos(Θ)L

From the problem, if N denotes the total of electrons released, then you will have to find the fraction lost from  :f(L) = N - [1 - B(0)/ B(L)]1/2

Bearing in mind, B(z=L) = B(z= -L)

Hence: f(L) = N - [1 - B(0)/ B(L)] 1/2

And,

B(z=L) = B(z= -L) = 10 (B(0))

This then yields:

f(L) = N - [1 - 1/ 10] 1/2

= N – [9/10] 1/2 = N – (0.9) 1/2

= N – 0.948

N > 1  the total released electrons, so the total  fraction lost is:

f(L) = 1 – 0.948 = 0.052  (e.g. 5.2 % of electrons lost)

(3) Make use of the revised force equation:

F z =   m  B z/ z

To show that:  dm/ dt » 0

Solution:

Consider component of motion along Bonly then the revised force equation is:

F z = m dvz/ dt =   - m  B z/ z

Multiply by velocity  vz  on the left and the equivalent dz/dt on right:

vz  (m dvz/ dt) =   - m  dz/ dt (B z/ z)

Where dz »  z

®  m vz dvz/ dt  =    - m  B z/ t

Where  B z/ t  is a variation of Bz as seen by the particle.

Since the energy of the particle must be conserved:

d/dt(½ m vz2    + ½ m v2 ) = d/dt (m vz2    + mB) = 0

Thence:

m vz dvz/ dt  =   d/dt (m vz2 )    =  - m  dB z/ dt

But: - m  dB / dt  +   d/dt (mB) = 0

Therefore: dm/ dt » 0

(4) If the perpendicular velocity component ( v) is 10 m/s for the electron, find its Larmor radius and its gyro-period. Thence or otherwise obtain the gyration energy

Solution:

Larmor radius:  r = m/ q [v / B] =  v/ (qB/ e) = vΩ e

r = (10 5  m/s) / 1.7 x 10 7  /s   =     0.0056 m or:  0.56 cm

Gyro-period: T = 2 p  / Ω e  =   p / 1.7 x 10 7  /s   =  3.5 x 10 -7  s

Gyration energy E = e  (v)2/ 2 =
(9.1 x 10 -31  kg) (10 5  m/s) 2  / 2

E =    4.5 x 10 -21  J

Or: in Electron volts:   E= (4.5 x 10 -21  J )/ (1.6 x 10 -19  J/eV) = 0.028 eV

(5) A proton moves in a uniform electric and magnetic field, with fields given by: E = 10 V/m (x^) and B = 0.0001 T (z^)where '^' denotes vector direction.

a) Find the gyrofrequency and the gyro-radius

b) Find the proton's E X B drift speed

c)Find the gyration speed v  and compare to the drift speed

d)Find the gyro-period, gyration energy and magnetic moment of the proton.

Solutions:

(a) The proton's gyro-frequency is just: Ω i  = qB/ m i

Ω i  = qB/ m i  =   [(1.6 x 10 -19  C) (0.0001 T) ]/ 1.7 x 10 -27  kg

Then:

Ω i   =  9.4 x 10 3 /s

The gyro radius r, is defined: r = v / (qB/m) =  v /   Ω

So we need to get v first.  This can be obtained from the energy of gyration:

E =  m(v)2/ 2 =     m m  B      where : m  = 8.5  x 10 -22  J/T

v = Ö (2 m m  Bi  )  =
Ö [2 (8.5  x 10 -22  J/T) (0.0001 T)/  1.7 x 10 -27  kg] = 10 m/s

Then, the radius of gyration is: r = v /   Ω   = 10 m/s/ 9.4 x 10 3 /s

And: r =  0.001 m =   0.1 cm   or 1 mm

b) The proton's (E X B) drift speed is just the magnitude:

[E/B]^y =  [ (10 V/m (x^)/ (0.0001 T (z^))  ]  =    10 5 m/s  ^y

c)The gyration velocity is just: v = Ö (2 m m  Bi  )  =   10 m/s

d) The gyro-period is: 2 p / Ω   = 2 p /(9.4 x 10 3 /s ) = 6.6 x 10 -4  s

The gyration energy E =    m m  B      where : m  = 8.5  x 10 -22  J/T

E = (8.5  x 10 -22  J/T) (0.0001 T) = 8.5  x 10 -26  J

The magnetic moment  (m ) of the proton for the problem was given but can also be worked out independently from:

E  = m m  B = m/2 (E/B) 2,

So: [E/B] = 10 5 m/s

And the magnetic moment of the proton is:

m m  = m/2 (E/B) 2 B  =  m/2 (10 5 m/s ) 2 0.0001 T

=  (1.7 x 10 -27  kg) (10 5 m/s ) 2 0.0001 T/ 2  = 8.5  x 10 -22  J/T

(6)  Find the guiding center positions for the electron referenced in Problem 4 if t = T/4.

Solution:

The electron (or ion)  moves along the B-field referenced to a guiding center, such that the following hold:

(a ) x – xo = - i v
exp (i Ω t)/ W = r sin ( t)

and

(b) y – yo = ± v
exp (i Ω t)/ W = r  cos ( t)

The preceding equation pair (a, b) describes a circular orbit around the guiding center (xo, yo), with the direction of gyration always such that the magnetic field generated by the charged particle is opposite to the externally imposed field.

The gyro-period is: T = 2 p  Ω

Bear in mind the gyration energy:

E  =
m m  B = m/2 (E/B2,

Guiding center positions:

x – xo =   r sin (Ω e t)   =    (0.0056 m)  sin (Ω e  T/4) =

=  (0.0056 m) sin [(1.7 x 10 7  /s) (3.5 x 10 -7  s/ 4)] = 0.0056 m

y – yo =   r cos (Ω e t)   =

(0.0056 m) sin [(1.7 x 10 7  /s) (3.5 x 10 -7  s/ 4)] = 0

To see how these values can be, note that: (Ω e T)   = 6.28 rad = 2p rad

So:  (Ω e T/ 4)   =1.57 rad =  p  / 2

But:  sin (p /2) = 1 and (cos p /2) = 0

So the value for x – xo is simply dictated by the value for  r