- #1
Barbara.Ba
- 2
- 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];
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];
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