## Wednesday, January 27, 2016

### Solution To Space Plasma Physics Problems (3)

1) The plasma parameter is defined:

L  =  n o  l3 D

From the values given (T =10 6 K, and n = 10 10 /m3)

L  =  n o  l3 D    =   ( 10 10 /m3) (0.30  m ) 3

L  =   2.9 x 10 8

The number of particles is defined:

N  =   4p n o  l3 D  / 3

N =  (4p)(9.2 x 10 6 ) / 3  =  1.2 x 10 9

2)For the solar atmosphere, first find the Debye length:

l D,s =[kT s / 4p  Z s2 n s  e2 ] 1/2

For which Z= 1 (hydrogen bulk plasma)   Then:

l D,s =[kT s / 4p n s  e2 ] 1/2

l D,s   =

[(1.38 x 10-23  (10 4 K) (8.85 x 10 -12 F/m)   / (4p)  (10 20/m3)  e2 ] ½

l D,s   =  1.9 x 10 - 7   m

Then the plasma parameter is:

L  =  n o  l3 D    =   ( 10 20 /m3) (1.9 x 10 - 7   m) 3

L  =  0.74

The number of particles is:

N  =   4p n o  l3 D  / 3

N =  (4p)(0.74) / 3 = 3.1   »   3.0

Note that L, N are drastically less than for the helium plasma.

3)  For the laser fusion plasma we have for the Debye length:

l D,  =

[(1.38 x 10-23  )(10 7 K) (8.85 x 10 -12 F/m)   / (4p)  (10 29/m3)  e2 ] ½

l D,s   =  1.9 x 10 - 10   m

For the tail of Earth’s magnetosphere we have for the Debye length:

l D,  =

[(1.38 x 10-23  (10 7 K) (8.85 x 10 -12 F/m)   / (4p)  (10 6/m3)  e2 ] ½

l D,s   =  61.6  m

For the laser fusion plasma we have for the plasma parameter:

L  =  n o  l3 D   =

( 10 29 /m3) (1.9 x 10 - 10   m) 3

L  =  0.74

N = 4p n o  l3 D  / 3  =  3

For the tail of Earth’s magnetosphere we have for the plasma parameter:

L  =   ( 10 6 /m3) (61.6  m) 3

L  =  2.3 x 10 11

The number of particles in the Debye sphere:

N = 4p n o  l3 D  / 3  =  9.7 x 10 11

The values L, N are vastly larger for the tail of Earth’s magnetosphere

4)Can a plasma with n o = 10 6 /m3    be maintained at an electron temperature of 100K?  (Hint: Calculate the density limit using the plasma parameter).

For these numbers the plasma parameter is:

L  =  n o  l3 D

And the Debye length is:

l D =10 0.84   (ÖT/ Ön)  =

10 0.84   (Ö100K / Ö(10 6 /m3)  =     0.069 m

So that:  L  =  n o  l3 D  =     ( 10 6  /m3) (0.069 m)

=  3.0 x 10 9

The number of particles is:

N  =   4p (3.0 x 10 9 )   / 3  = 1.2  x 10 10

The value N is much larger than unity and hence  is consistent with the principle of shielding so the plasma can be maintained at the given electron temperature.

5) In the limit 1/N << 1 and 1/N ® 0  we say a plasma is “collisionless”. Do any of the plasmas cited in the previous problems qualify as collisionless?

The values 1/N for the given plasmas are as follows:

helium plasma: 1/N = 1/ 3.8 x 10 7   =   2.6  x 10 - 8

solar atmosphere: 1/N = 1/3.1 = 0.32

laser fusion plasma:  1/N = 1/3.1 = 0.32

tail of Earth’s magnetosphere: 1/N = 10 – 12

The values 1/N show  that only the helium and magnetospheric plasmas can be considered as collisionless.