Solutions To Space Plasma Physics Problems (1)

1)      Find the ion and electron gyrofrequencies for an ion and electron in solar plasma with a magnetic induction (field strength) of B = 0.0001 T.

Solution:   The ion gyrofrequency will be:

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

Ω _i   =  9.4 x 10 ³ /s

 And the electron gyrofrequency is:

 Ω _e  = qB/ m _e  =   [(1.6 x 10 ^-19  C) (0.0001 T) ]/ 9.1 x 10 ^-31  kg

Ω _e  =  1.7 x 10 ⁷  /s

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

 

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

3)     Thence or otherwise obtain the gyration energy in eV.

(1.6 x 10 ^-19  J = 1 eV)

 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  

Or: in Electron volts:  

 

 (4.5 x 10 ^-21  J )/ (1.6 x 10 ^-19  J/eV) = 0.028 eV

4)     Find the guiding center positions for the electron referenced above (previous problems) if t = T/4.

 Guiding center positions:

 x – x_o =   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 – y_o =   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 – x_o is simply dictated by the value for  r