Solutions To Space Plasma Physics (3) Problems

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,s  =

[(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  

  =  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.