Solutions:
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 -22 J/T) (0.0001 T)/ 1.7 x 10 -27 kg] = 10 m/s
Ö [2 (8.5 x 10 -22 J/T) (0.0001 T)/ 1.7 x 10 -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 (x^)/ (0.0001 T (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 -22 J/T) (0.0001 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) 2 B = m/2
(10 5 m/s ) 2 0.0001 T
= (1.7 x 10 -27 kg) (10 5
m/s ) 2 0.0001 T/ 2
= 8.5 x 10 -22 J/T
2) The mirror ratio is (Bmin / 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”:
We define what is called the “loss cone angle”:
sin (q L ) = ± Ö (Bmin
/ B max )
In the problem, Bmin = 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 òo Θ(L) 2 π sin(Θ) d Θ
= f(L) = òo Θ(L) sin(Θ) d Θ
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
= 1 - cos(Θ)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
bearing in mind, B(z=L) = B(z= -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
bearing in mind, B(z=L) = B(z= -L)
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.
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 L ) = ± Ö (Bmin
/ 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 (Bmin
/ B max )
Hence, for all angles (a ) > 26.55
deg or
< (-26.55 deg) particles will not remain within
the cone.
Find the velocity ratio v ^ / v || if Bmax = 0.95 B z
Let: Bmax = 0.95 B z , And: recall Bz = Bmin
Then: Bmax
/ Bmin = 0.95
But: [v ^2 / v || 2 ]
= Bz /Bmax
v
^ / v || = Ö (1.05) » 1.02
4) Rewrite: 1 - w i 2 / w 2 -- w e 2 /( w - w e ) 2
As: 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 real and imaginary parts, i.e.g
( 1/2) 5/3 (m e / m i) 1/3 w e » 0.1 w e
(This is the frequency seen in the ion rest frame)
The roots for the full expression can be written as real and imaginary parts, i.e.g
w
= w r + +i g
For which: w r = ( 1/2) 5/3 (m e / m i) 1/3 w e » 0.1 w e
(This is the frequency seen in the ion rest frame)
And: g = (
1/2) 5/3 Ö 3 (m e / m i)
1/3 w e
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