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?

__Solution__:

The mirror ratio is (B

We define what is called the “

_{min}/ B_{max}) = 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 (q

_{L}) = ± Ö (B_{min}/ B_{max})In the problem, B

To do the problem, one must understand he's really being asked for the

Θ (O) > (Θ)

Thus, Θ (O) = (Θ)

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 ò

= f(L) = ò

_{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 ò

_{o}^{Θ(L)}^{ }2 π sin(Θ) d Θ= f(L) = ò

_{o}^{Θ(L)}^{ }sin(Θ) d Θ= 1 - cos(Θ)

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

_{L}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}And: f(L) = N - [1 - B(0)/ B(L)]

^{1/2}

Bearing in mind, B(z=L) = B(z= -L) = 10 (B(0)

This yields:

f(L) = N - [1 - 1/ 10]

^{ 1/2}= N – [9/10]

^{ 1/2}=

N – (0.9)

Now let N = 1 (Since N

f(L) = 1 – 0.948 = 0.052

Or: 5.2% of the released electrons lost.

^{ 1/2}= N – 0.948Now 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.

2) Consider a similar mirror configuration for a solar coronal loop for which B

_{min}= 0.2 B_{max}. Find the loss cone angle for this loop and also determine whether particles will remain within it. Find the velocity ratio :v

_{^}/ v_{||}_{ }if B_{max}= 0.95 B_{z}_{}

__Solution__:

The loss cone angle is:

sin (q = ± Ö (0.2 B

_{L})**=**± Ö (B_{min}/ B_{max})_{max}/ B_{max})sin (q

_{L})**=**± Ö (0.2) = ± 0.447So that: (q

_{L}) = arcsin**(± 0.447)**(q

_{L}) = ± 26.55 degThe dual (±) sign means the angle is symmetric with respect to the magnetic axis of the system.

Whether particles remain in the loss cone angle depends on whether the condition:

sin (a ) > Ö (B is met.

_{min}/ B_{max})Hence, for all angles (a ) > 26.55 deg particles will not remain within the loss cone.

To get the velocity ratio : v

_{^}/ v_{||}_{ }:_{}Let: B

_{max}= 0.95 B_{z , }And: recall B_{z}= B_{min}_{}Then: B

_{max / }B_{min}= 0.95But: [v

_{^}^{2}/ v_{||}_{ }^{2}] = B_{z }/B_{max }=B

_{min}/ B_{max }= (1/0.95) = 1.05v

_{^}/ v_{||}_{ }= Ö (1.05) » 1.02_{}
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