Solving Basic Plasma Physics Problems (2)

Before going on to solve another basic type of plasma physics problem, let's look at the previous one to do with plasma orbital theory and gyro-motion. Again, the question was:

A proton moves in a uniform electric and magnetic field, with fields given by:E = 10 V/m (x^) and B = 0.0001 T (z^)where '^' denotes vector direction.

a) Find the gyrofrequency and the gyro-radius

b) Find the proton's E X B drift speed

c)Find the gyration speed v(o) and compare to the drift speed

d)Find the gyro-period, gyration energy and magnetic moment of the proton.

Let's first work on (a). We need basic data also to solve this set, including the mass of the proton: m(p) = 1.7 x 10^-27 kg, and the proton charge: q = 1.6 x 10^-19 C

The proton's gyro-frequency is just: W(g) = qB/m(p)

W(g) = (1.6 x 10^-19C) (0.0001T)/ {1.7 x 10^-27 kg} = 9.4 x 10^3 s^-1

The gyro radius r, is defined: r = v(perp)/ W(g)

so we need to get v(perp) first.

For the proton, given the information provided in the blog, this can be obtained from the energy of gyration: E(g) = u(m) B, so v(perp) = [2u(m)B/ m(p)]^1/2  = [100 m/s]^1/2 = 10 m/s

Then, the radius of gyration is: r = (10 m/s)/ (9.4 x 10^3 s^-1) = 0.0011m or 0.1 cm

If u(m) is a constant of the motion, as we expect it to be, then the local perpendicular gyration velocity is just v(perp).

(b) The proton's (E X B) drift speed is just the magnitude [E/B]^y = 10^5 m/s (^y)

(c) and so is 3 orders of magnitude greater than the perpendicular gyration velocity.

(d) The gyro-period is T(g) = 2 pi/ W(g) =

2 pi/ (9.4 x 10^3 s^-1) = 3.5 x 10^-7 s

The gyration energy E(g) = u(m) B = 8.5 x 10^-26 J

Therefore, the magnetic moment of the proton is:

u(m) = E(g)/ B = (8.5 x 10^-26 J)/ (0.0001T) = 8.5 x 10^-22 J/T


--
Now, in the new problem we examine a plasma magnetic mirror (see the model shown)

Consider a plasma mirror machine of length 2L with a mirror ratio of 10 so that B(z=L) = B(z=-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?

The mathematical basis for a typical plasma mirror machine is shown, and note the end points (at z = ± L) are “pinched” and thus of narrower bore than at the midpoint. 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 (Θ)_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, 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 INT (0 to Θ(L)) 2 π sin(Θ) d Θ

= f(L) = INT (0 to Θ(L)) sin(Θ) d Θ

= 1 - cos (Θ)_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(z=L) = B(z= -L)

We shall finish solving this in the next instalment, but readers are invited to try their hand in the meantime!