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.
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
(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 – 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
No comments:
Post a Comment