Expert: Philip A. Stahl - 4/4/2007

Question
Can you help with ideas to solve this problem?

Consider a plasma mirror machine of length 2L with a mirror ratio of 10 so that B(L) = B(-L) = 10 B(0). A group of N (N > 1) electrons with an isotropic velocity distribution is released at the center of the machine. Ignoring collisions and the effect of space charge, how many electrons escape?

Answer
Hello, Jani

----------------!--------------
     
----------------0-------------- B = (L)
B = (-L)

<--------- L -------------- -->

The mathematical basis for a typical plasma mirror machine is shown above, but make the adjustment that the end points (at B = ± L) are “pinched” and thus of narrower bore than at the midpoint. (This is hard to show in a sketch so I leave it to you to do it mentally!)

Then this yields higher magnetic induction, B, at those points which will be “B_max”.

The mirror ratio is  (B_max/ B_min) = 10, meaning that the induction strength at those end points will be ten times the induction at the center point or apex of the magnetic loop or mirror machine.

We define what is called the “loss cone angle”:

sin (THETA)_L  =   ± [B_min/  B_max]^1/2

In the problem, B_min = B(0) or the "zero level" for the magnetic induction, say at position L = 0. This doesn't mean the induction is zero at that point literally, however.

To do the problem, you have to understand you are really being asked for the *fraction* of electrons lost.

A special condition obtains which applies to the angle - for which the electrons will be TRAPPED only provided:

THETA (O) >  (THETA)_L

Thus, THETA (O) =  (THETA)_L

is said to be the "loss cone" of the system or machine.

If an isotropic particle distribution (in this case, electrons) is introduced at a position L = 0, the fraction of particles that will be lost to the mirror system is:

f(L) =  1/ 2 pi INT (0 to THETA(L)) 2 pi sin(THETA) d THETA


= f(L) =  INT (0 to THETA(L)) sin(THETA) d THETA

=  1 - cos (THETA)_L


Now, from the problem, if N denotes the total electrons released, then you will have to find the fraction lost from:

f(L) =  N - [1 -  B(0)/ B(L)]^1/2

bearing in mind, B(L) = B(-L) and that you have the mirror ratio already!

Hope this helps!

Note the integral 'INT' has limits inside the brackets to the immediate right.