Expert: Philip A. Stahl - 4/5/2007

Question
I have been trying to work out this past exam problem but with no success. Can you provide any hints or clues as to solutions?

An electron with speed v_o moves along the axis of a magnetic tube of length 2L. The axial magnetic field has the form: B(z) =  B(o) exp k [z]


a) Find an equation for da/ dt where a is the pitch angle of the trajectory

b) Find the minimum value of a(o) where a(o) is the value of a at z = 0, that is needed to ensure the electron is trapped.

c) Find the bounce period as a function of a(o).

Thanks!

Answer
Hello,

This problem actually isn't too difficult to work, and is largely a variant on the previous problem we addressed.

Recall from that answer, that we found:

the "loss cone angle":

sin (THETA)_L  =   ± [B_min/  B_max]^1/2

and the "mirror ratio" is [B_max/ B_min)

In this current problem you have from the information:

B(z) =  B(o) exp k [z]

where I take it that the z is in an absolute value sign.

Since the total length of tube is 2L the value of B-max will be at z = L, so:

B_max = B(z=L) = B(L)

The minimum is, of course, at B(z = 0)

Then:

B(z) =  B(o) exp k [z] = B(z) =  B(o) exp k [0]= B(o)

since: exp k [0]= 1

The mirror ratio will be:

R =   B(L)/ B(o)

Now let's look at the parts of the problem and feasible approaches.

a) the key here is to re-cast the pitch angle (or de facto loss-cone angle) in terms of velocities.

Since the adiabatic invariant for particle motion is a constant of the motion:

u =  1/2  [mv ^2 /B]  

we have:

[v_ perp ^2 /B_max]  =     [v ||^ 2 /B(0)] =    const.  or

 [v_perp ^2 / v ||^ 2  ]  =  B(0) /B_max

Using the above, it ought to be possible for you to show:

a =  arctan [v_perp / v ||]

and since the only variation is along the axis (z) and

v || = dz/ dt (while you may assume v_perp = const.)


you should be able to obtain da/ dt

where v_perp  denotes the perpendicular component of the velocity and v || is the parallel component (e.g. parallel to the z-axis in the problem)


b) To solve this, you will need to set up a ratio of the pitch angles (applied to a and a_o) and relate to the mirror ratio for this problem.

it should be easy to show:

sin^(a)/ sin^2 (a_o) = =   B(L)/ B(o)


Bear in mind here that the maximum value (B_max) for the magnetic bottle determines the minimum value of a_o needed for confinement.

Further working, and using the results obtained for the previous problem, it will be seen that:

sin (a_o) >  1/ R^1/2  =  (B(o)/ B_max)^1/2 = [B(o)/B(L)]^1/2

from which the value needed to ensure the electron is trapped can easily be obtained.


c) The key to this solution is to realize that a particle (like the electron confined in a magnetic bottle, say a line-tied solar coronal loop) "bounces" back and forth with a bounce period T_B given by the integral over one "bounce cycle":

Thus,

T_B =  (INT)  dz/ v || (z)

where (INT) denotes a closed path integral.

This is equal to:

4 INT (z= 0 to z = L)  dz/ v [1 - B(z)/B(0) sin^2 (a_o)]^1/2

where 'v' is the total partricle speed

(Note the integral 'INT' has limits inside the brackets to the immediate right)

From the preceding, you should be able to obtain the bounce period as a function of a(o).

Hope this eases your problem!