*Solutions to Problems for Part (2):*

1a)The proton's gyro-frequency is just:

*Ω*_{i}**= qB/ m**_{i}*Ω*

_{i}

**= qB/ m**

_{i }= [(1.6 x 10

^{-19}C) (0.0001 T) ]/ 1.7 x 10

^{-27}kg

*Ω*

_{i}= 9.4 x 10

^{3}/s

The gyro radius r, is defined: r = v

_{⊥}/ (qB/m) =**v**_{⊥}**/**Ω
So we need to get v

_{⊥}first. This can be obtained from the energy of gyration:
E = m(

**v**⊥)^{2}/ 2 = m_{m}**B**_{ }where :*m*_{m }*= 8.5 x 10*^{-22}J/T**v**⊥ = Ö (2 m

_{m}

**B**/ m

_{i }) =

Ö [2 (

*8.5 x 10*0.0001 T)/ 1.7 x 10

^{-22}J/T) (^{-27}kg] = 10 m/s

Then, the radius of gyration is: r =

**v**_{⊥}**/***Ω*_{i }= 10 m/s/ 9.4 x 10^{3}/s
r = 0.001 m = 0.1 cm or 1 mm

b) The proton's (

**E X B**) drift speed is just the magnitude:
[

**E**/**B**]^**y**= [ (*10 V/m (*(0.0001 T**x**^)/*(*) ]**z**^)*=*10^{5}m/s**^y**
c)The gyration velocity is just:

**v**⊥ = Ö (2 m_{m}**B**/ m_{i }) = 10 m/s
(See part (a) – worked out to get r)

So the drift speed is 3 orders of magnitude greater.

d) The gyro-period is: 2 p /

*Ω*_{i }= 2 p /(9.4 x 10^{3}/s ) = 6.6 x 10^{-4}s
The

*gyration energy E*= m_{m}**B**_{ }where :*m*_{m }*= 8.5 x 10*^{-22}J/T
E = (

*8.5 x 10*0.0001 T) =^{-22}J/T) (*8.5 x 10*^{-26}J
The

*magnetic moment (*m_{m }*) of the proton for the problem was given but can also be worked out independently from:*
E = m

_{m}B = m/2 (*E/B*)**,**^{2}
So: [E/B] = 10

^{5}m/s and:
m

_{m}= m/2 (*E/B*)**B = m/2 (10**^{2}^{5}m/s )**0.0001 T**^{ 2}
= (1.7 x 10

^{-27}kg) (10^{5}m/s )**0.0001 T/ 2 =**^{ 2}*8.5 x 10*^{-22}J/T
2) 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:f(L) = N - [1 - B(0)/ B(L)]

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

_{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 then yields:

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

^{ 1/2}= N – [9/10]

^{ 1/2}= N – (0.9)

^{ 1/2}= N – 0.948

Now let N = 1 (Since N

__>__1) be the total (normalized) released electrons, then the total lost is:

f(L) = 1 – 0.948 = 0.052

e.g. 5.2% of the released electrons are lost.

3) 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) = ± 0.447
So that: (q

_{L}) = arcsin**(± 0.447)**
(q

_{L}) = ± 26.55 deg
The 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.

Find the velocity ratio : v

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

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

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

_{^}^{2}/ v_{||}_{ }^{2}] = B_{z }/B_{max }= B_{min}/ B_{max }= (1/0.95) = 1.05
v

_{^}/ v_{||}_{ }= Ö (1.05) » 1.02_{}
4)

1 - w

*Solution:*We make use of:1 - w

_{ i}^{ 2 }/ w_{ }^{2 }-^{ }w_{ e }^{2 }/( w - w_{ e })^{ 2 }
Which we rewrite : 1 - w

_{ i}^{ 2 }/ w_{ }^{2 }-^{ }1^{ }/( 1 - w / w_{ e })^{ 2}
» 1 - w

_{ i}^{ 2 }/ w_{ }^{2 }-^{ }( 1 + 2w / w_{ e })
Therefore: 0 = 1 - w

_{ i}^{ 2 }/ w_{ }^{2 }-^{ }( 1 + 2w / w_{ e })
And: 0 = - w

_{ i}^{ 2 }/ w_{ }^{2 }-^{ }2w / w_{ e}
Or: w

_{ i}^{ 2 }/ w_{ }^{2 }= -^{ }2w / w_{ e}
But: w

_{ }^{3 }= - ½_{ }*(*w_{ i}*)*^{ 2 }(w_{ e})_{}
Rewrite as: w /w

_{ e}= ( -1/2)^{ 1/3 }(m_{ e}/ m_{ i})^{ 1/3 }
Since: Im(w /w

_{ e}) = (m_{ e}/ m_{ i})^{ 1/3 }
And recall: w

_{ i}^{ 2 }/ w_{ }^{2 }= -^{ }2w / w_{ e }=_{ }- 2(m_{ e}/m_{ i})
This leads to roots of (-1)

^{ 1/3 }which can be written formulaically:
(-1)

^{ 1/3 }= exp [i(2n + 1) p/ 3
So that we obtain for the roots: ½ + iÖ 3/ 2 (n = 0)

-1 for n = 1

And : ½ - iÖ 3/ 2 (n = 2)

(See also: http://brane-space.blogspot.com/2013/03/taking-complex-roots-1_12.html)

The roots for the full expression can be written as

For which the real part of the frequency seen in the rest frame is therefore:

The above represents an instability for what we call the “

The roots for the full expression can be written as

*real*and imaginary parts, i.e.g
w

_{ }^{ }= w_{ r }+ +i_{ }**g**For which the real part of the frequency seen in the rest frame is therefore:

**w**_{ r}**=**( 1/2)^{ 5/3 }(m_{ e}/ m_{ i})^{ 1/3 }w_{ e }» 0.1 w_{ e}_{}
And:

**g****=**( 1/2)^{ 5/3 }Ö 3^{ }(m_{ e}/ m_{ i})^{ 1/3 }w_{ e }*two stream*” problem, applicable for when a plasma does not consist of Maxwellian electrons or ions, i.e. following the Maxwell distribution._{}

^{}

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