Solutions To Single Particle Plasma Dynamics Problems

 1)Given the situation shown below for a particle in plasma:

And: v _o = v _xo x^ + v _yo y^ (for initial velocity)

Solve the equation of motion: F = q(E + v X B)

Solution:

m dv/dt = q(E + v X B)

v _x   =   qB (v _y)/m   +  q E _x /m

v _y  =  - qB (v _x)/m  

Or: 

v _y  =  - qB/m [qB (v _y)/m   +  q E _x /m]

=  - W _e ² (v _y)   -  W _e ² (E _x) / B 

v _x   =   - W  _e ² (v _x) -  W  _e ² (E _x ) / B 

 

2) (a)If the perpendicular velocity component ( v⊥) is 10⁵  m/s for an electron in a plasma, find its Larmor radius, gyration energy and its gyro-period.

Solution.

Larmor radius:  r  = m/ q [v⊥ / B] =  v⊥/ (qB/ m _e) = v⊥/ Ω _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  =   2 p / 1.7 x 10 ⁷  /s   =  3.5 x 10 ^-7  s

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

 E =    4.5 x 10 ^-21  J  

 = 0.028 eV

(b) Find the guiding center positions for the electron referenced in (a) if t = T/2.

Solution:

  The gyro-period is: T = 2 p  / Ω  

Then: T/2 =   2 p  /2 Ω  =  p / Ω  

Bear in mind the gyration energy:

E  = m _m  B = m/2 (E/B) ²,

Guiding center positions:

 x – x_o =   r sin (Ω _e t)   =  (0.0056 m)  sin (Ω _e   p  / Ω _e  ) = (0.0056 m) sin p

=  (0.0056 m) 0 = 0

y – y_o =    r  cos (Ω _e   p  / Ω _e  )   =    

(0.0056 m) cos p   =   - 0.0056 m