A particle moving in an electromagnetic field

AI Thread Summary
A charged particle with mass m and charge q moves in an electric field E and a magnetic field B, with initial conditions provided. The equations of motion derived indicate a correct solution for E_0 = 0, but a problematic case arises for B_0 = 0 due to its presence in the denominator of the solutions. The discussion highlights the importance of crossed electric and magnetic fields in applications like particle beam filtering and magnetron microwave tubes. Despite initial concerns about divergence in the solutions, the use of Mathematica confirmed the limits were correctly handled. The conversation emphasizes the need for clarity in the mathematical treatment of such physical scenarios.
Barbara.Ba
Messages
2
Reaction score
0
Dear Friends

A particle having charge q and mass m moves in a region where
an electric field
E = ( 0 , E_0 , 0 )
and a magnetic one
B = ( 0 , 0 , B_0 )
exist.

For t = 0 its position is
R = ( 0 , 0 , 0 )
and its velocity is
V = ( V_0 , 0 , 0 ).

I have worked out the equations of motion

(
you can see them and their solution in a fine graphical form at
http://i786.photobucket.com/albums/yy150/simona1989_photo/img.jpg

In text only mode they are:

m x''[t]==B0 q y'[t]
m y''[t]==q (E0-B0 x'[t])
m z''[t]==0

x[t] = ( B0 E0 q t-m (E0-B0 V) Sin[(B0 q t)/m] ) / ( B_02 q)
y[t] = ( 2 m (E0-B0 V) Sin[(B0 q t)/(2 m)]2 ) / ( B_02 q )
z[t] = 0
)

but I get a strange and obviously wrong phenomenon in the solution:
they correctly give the solution of the case E_0 = 0 but they have no meaning for B_0 = 0 (because B_0 appears into denominator of solutions).

Please note that the equations of motions of the general case ( E_0 and B_0 both not zero ) reduces normally at the right equations of motions for E_0 = 0 and B_0 = 0.

Please can you suggest an hint for the reason of that, or the wrong step in my reasoning ?

If you feel confortable with Mathematica, may be you want see, at bottom of this message, the Mathematica code I have used to confirm my hand written solutions.

Thanks for your attention!

Warmest regards
Barbara.Ba

------------------------------------------------------------------------------
(* position at t *)
pos = {x[t], y[t], z[t]};

(* velociy' and acceleration t *)
vel = (D[#, {t, 1}]) & /@ pos;
ac = (D[#, {t, 2}]) & /@ pos;

(* the given fields *)
cE = {0, E0, 0};
cB = {0, 0, B0};

(* the Lorentz force for the given fields *)
F = q (cE + Cross[vel, cB]);

(* the equations of motions *)
eq = Table[m ac[[k]] == F[[k]], {k, 1, Length[ac]}];

(* position and velocity for t = 0 *)
cond = {x[0] == 0, y[0] == 0, z[0] == 0, x'[0] == V, y'[0] == 0, z'[0] == 0};

(* this solves the equations of motions *)
sol = DSolve[{eq, cond}, pos, t] // Simplify;

(* output *)
Print["\n\n--- equations of motions ---\n\n", eq // TableForm];
Print["\n\n--- solutions ---\n\n", Transpose[sol] // TableForm];
 
Last edited by a moderator:
Physics news on Phys.org
This is a good question. A crossed field (crossed dc E and B fields) is used to filter out charged particles with different velocities in a particle beam with defined momentum. This is called a Wein filter. See
http://titan.triumf.ca/equipment/beamline/wienfilter.shtml
(correct spelling is Wein). Also search web.
Also, crossed fields are used in magnetron microwave tubes, like the ones used in microwave ovens. The electrons are accelerated away from the filament by the E field, and rotate around the cavity, deflected into a circular orbit by the B field.

Bob S
 
Sorry for mistake.

Despite B_0 into the denominator the whole functions doesn't diverge to infinity.
I got the right limit by means of Mathematica.

Warmest Regard
Barbara
(Rome)
 
Thread 'Motional EMF in Faraday disc, co-rotating magnet axial mean flux'
So here is the motional EMF formula. Now I understand the standard Faraday paradox that an axis symmetric field source (like a speaker motor ring magnet) has a magnetic field that is frame invariant under rotation around axis of symmetry. The field is static whether you rotate the magnet or not. So far so good. What puzzles me is this , there is a term average magnetic flux or "azimuthal mean" , this term describes the average magnetic field through the area swept by the rotating Faraday...
Back
Top