## Tuesday, October 19, 2010

### 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!

Unknown said...

Could you show me how to get a v(prep)=100m/s ?

In this content you refer about gyro energy which in plasma physic problem1 : E(g)=u(m)B=m/2(E/B)^2 ,i think E/B is a drift speed that you calculate as 10^5 in ans.(b).

I think you may calculate v(prep)from E(g)=u(m)B that means we should know u(m) to get E(g) but this isn't make sense cause you just calculate u(m) in ans.(d) from E(g)=8.5x10^-26J (I wonder how did you get a value E(g)=8.5x10^-26J ...)

That why I confused about how did you get 100m/s for v(prep).

with many thanks in advance

Copernicus said...

The energy of gyration is defined:

E(g)= u(m) B =

(8.5 x 10^-22 J/T) (0.0001T) =

(8.5 x 10^-26 J)/

Then v(perp)

= 2u(m)B/ m(p) =

2(8.5 x 10^-26 J)/ 1.7 x 10^-27 kg

= 17 x 10^-26 J/1.7 x 10^-27 kg

= 10 x 10^1 m/s = 100 m/s

Copernicus said...

REM: v(perp) refers to the component of GYRATION velocity.

Copernicus said...

Sorry! The soln. should read:

v(perp) = [2u(m)B/ m(p)]^1/2 =

[100 m^2/s^2]^1/2

= 10 m/s

Hope this helps!

Unknown said...

thank you very much for your reply, anyway I still confuse about how to get E(g)...

From your reply ,to calculate E(g), you know u(m)=8.5x10^-22J/T before you calculate the gyration energy then you use E(g)=u(m)B, right?
then how do you find u(m)=8.5x10^-22J/T (suppose we don't know energy yet we are going to find energy from u(m) if we use this formula)

...Anyway, in your answer (d) you show that u(m) can find from E(g) (that mean you know E(g) before you find u(m))then I think you may find E(g)=8.5 x 10^-26 J from other formula , then substitute E(g) and B in E(g)=u(m)B and finally get u(m).

I wonder how do I get E(g)=8.5 x 10^-26 J from the parameters, electric and magnetic field, you gave in this question.

with many thanks in advance

Copernicus said...

You wrote:

"From your reply ,to calculate E(g), you know u(m)=8.5x10^-22J/T before you calculate the gyration energy then you use E(g)=u(m)B, right?"

- Ans. Correct

"then how do you find u(m)=8.5x10^-22J/T (suppose we don't know energy yet we are going to find energy from u(m) if we use this formula)"

The magnetic moment (e.g. see any statistical physics text, such as Kittel's 'Thermal Physics') is always found as the product of the energy by the magnetic intensity B.

We know the energy is: E(g) =8.5x10^-26J

And the magnetic intensity is B = 0.0001T

So that:

u(m) B =

(8.5 x 10^-22 J/T) (0.0001T) =

8.5 x 10^-26 J

It cannot be found from the E, B parameters alone.

At least, I am unaware of any such method.

Copernicus said...

Adendum: It ought to have been clarified in the problem that:
u(m)=8.5x10^-22J/T

is based on a (presumed) work done on the dipole in a rotation from (e.g.) pi/2 to and angle theta with the field yielding

u(m) B or often u(m)H

Thus, u(m) is given in this problem.

Details on how it is derived can be found in Kittel, 'Thermal Physics', p. 25.

Unknown said...

Thank you very much for your reply

Copernicus said...

You're welcome. You may also find this discussion (especially the diagrams) useful:

http://brane-space.blogspot.com/2012/02/uses-of-hermitian-matrices-in-physics.html

Unknown said...

MHD power generators may possibly be a more efficient way of converting heat into
electricity. Think of one as consisting of a simple rectangular channel of (x-) width a,
(y-) height b, in which the plasma flows under pressure in the z-direction. Take the
plasma density and velocity to be uniform. A uniform magnetic field, B, is applied in
the y-direction and the walls at x = 0, a are electrodes where the electric current density
(density j, assumed uniform) is picked off at a voltage difference φ. Use the MHD equations
to answer the following questions.
(a) If the resistivity, η, of the plasma is negligible, what is the plasma velocity?
(b) If the pressure is po at z = 0, what is its value as a function of z?
(c) How much electric power is generated per unit length of the channel?
1
(d) What is the rate of doing work per unit channel length by the plasma pressure force?
(e) If η is not negligible but can be considered fixed, and the flow velocity and B-field are
also fixed but the current density can be varied, what is the maximum electric power
per unit length that can be generated?
Can you give me a bit idea about this question part 4 and 5

Copernicus said...

Thanks for your interesting insights!