1) 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?
We define what is called the “loss cone angle”:
To do the problem, one must understand he's 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:
Θ (O) > (Θ)L
Thus, Θ (O) = (Θ)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 π òo Θ(L) 2 π sin(Θ) d Θ
= f(L) = òo Θ(L) sin(Θ) d Θ
From the problem, if N denotes the total electrons released, then you will have to find the fraction lost from:
And: f(L) = N - [1 - B(0)/ B(L)] 1/2
Bearing in mind, B(z=L) = B(z= -L) = 10 (B(0)
f(L) = N - [1 - 1/ 10] 1/2= N – [9/10] 1/2 =
Now let N = 1 (Since N > 1) be the total (normalized) released electrons, then the total fraction lost is:
f(L) = 1 – 0.948 = 0.052
Or: 5.2% of the released electrons lost.