## Sunday, November 11, 2018

### 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:

i  = qB/ m =   [(1.6 x 10 -19  C) (0.0001 T) ]/ 1.7 x 10 -27  kg
i   =  9.4 x 10 3 /s

And the electron gyrofrequency is:

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

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/ e

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

Gyro-period: T = 2 p  / 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 – xo =   r sin (e t)   =    (0.0056 m)  sin (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 (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: (e T)   = 6.28 rad = 2p rad

So:  (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