Monday, December 19, 2022

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 10 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/ 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

= 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 =   p  /2 Ω  =  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   p  Ω e  ) = (0.0056 m) sin p

=  (0.0056 m) 0 = 0

y – yo =    r  cos (Ω e   p  Ω e  )   =

(0.0056 m) cos p   =   - 0.0056 m