Expert: Philip A. Stahl - 1/18/2008

Question
Help! I am fit to be distracted with this problem. Can you help?  A proton moves in a uniform electric and magnetic field, with fields given by:

E = 10 V/m (x^)     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.

Thanks for any assistance or advice!

Answer
Hi, Jani

I am not going to solve this for you - but rather provide you with what I regard as ample background (and a few hints) to do it yourself!

Let's look first at the proton particle orbit aspects:

Here we enter plasma “orbit theory” and consider a charged particle (say of charge q) in a uniform and constant magnetic field (B).

The governing equation of motion is:

m (dv/dt) = q(v X B)

The motion here is such that v is always perpendicular to the force acting on the particle so  v ^ F, implying circular motion.

Thus:

dv/dt = q/ m [v X B] is a centripetal acceleration.

e.g. a = (-r^) (v_perp^)^2/ r

Note that the velocity has two components:

v =   v_par    +   v_perp

where the first term denotes the velocity *along B* which stays constant so that

d(v_par)/ dt = 0

Meanwhile:   v_perp^2/ r  =   q/ m [v_perp  * B]

(setting the centripetal force = to the magnetic force producing it)    

the quantity r is none other than the *gyro-radius*. Solving for it one finds:

r =   m/ q [v_perp / B]  =  v_perp/  (qB/m)

where the denominator denotes the GYRO-FREQUENCY

W(g) =   qB/m

While on this topic it is useful to consider this as well:

The "gyration velocity" is simply the magnitude:

V(o) =  E/B

Thus, the proton gyrates at a velocity equal to the drift speed.

The gyro-period is:  T(p)  =   2 pi/ W(g)

Now, given all the preceding, you should be able to work out the rest, with just a few pointers.

Bear in mind the gyration energy E(g) = u(m)B  =  m/2 (E/B)^2, where u(m) is the magnetic moment.

Also, the ans. for E(g) will be in joules (J) and recall 1 eV = 1.6 x 10^-19 J


Let's now examine the 'E cross B' drift:

Consider the field graphic below, superposed on x-y axes.


       
^ y
!     ----------> electrons (-q)
!
!    ----------> Ions (+q)          B  (out)
!
!
!--------------------------->x

E is in +y direction.

Both electrons and ions (e.g. protons) undergo a corkscrew motion around the centers of their paths (guiding centers) and have drift velocities in the same direction in this field orientation.

This drift motion occurs in crossed (E X B) fields and hence this is called “E cross B” drift.

Bear in mind when you go to calculate E X B you need to take the vector directions into account (using the right hand rule) based on the information supplied in your problem.

From all this, you ought to be able to work out the assorted parts of the problem without too much difficulty. If not, please get back to me for more pointers!