Download Motion in an Electromagnetic Field

3-D Motion in an Electromagnetic Field
Special case: E and B are static and uniform
Let’s let B = Bz, and E = Eyy + Ezz . This is more general than it may first appear. We simply
choose the z direction as that of the magnetic field. We then allow the electric field to have a
component parallel to and a component perpendicular to the magnetic field.
The electric and magnetic forces are then:
FE = qE = qEyy + qEzz ,
Fm = qv  B = -qx’B(y) + qy’B(x) .
Newton’s Second Law (F = ma) gives (in component form):
Fx = qBy’ = mx’’
Fy = qEy – qBx’ = my’’
Fz = qEz = mz’’ .
The z equation is unconnected and can be identified as that of constant force (and hence, constant
acceleration), and so can be solved for z(t):
z’(t) = vzo + (qEz/m)t , and
z(t) = zo + vzot + ½(qEz/m)t2 .
This implies that any electric field parallel to the magnetic field will accelerate the charged
particle as if the magnetic field were not there.
The x and y component equations, however, are linked (that is, there is an x’ in the y component
equation, and there is a y’ in the x component equation). How do we proceed in this kind of case?
Since there is a y’ term in the x component equation, we can differentiate the x component
equation with respect to time. This will turn the y’ term into a term involving y’’. We can then
use the y component equation to find an expression for y’’, and substitute this for the y’’ into the
x component equation. Doing the d/dt of the x component equation gives:
qBy’’ = mx’’’
Solving for y’’ in the y component equation gives:
y’’ = qEy/m – (qB/m)x’ .
Putting these two equations together gives:
mx’’’ = qB[qEy/m – (qB/m)x’] = q2EyB/m – (q2B2/m)x’ .
If we treat x’ as the basic quantity (rather than x), we see that the above equation is a second
order, linear, inhomogeneous differential equation:
m d2(x’)/dt2 + (q2B2/m)x’ = q2EyB/m .
m d2(x’)/dt2 + (q2B2/m)x’ = 0 ,
The homogeneous eq:
has the solution: x’H = Ax cos(ωt + θx) where ω = qB/m and Ax and θx are the two constants
from the second order differential equation;
The inhomogeneous eq:
has the solution:
m d2(x’)/dt2 + (q2B2/m)x’ = q2EyB/m
x’I = Ey/B = constant.
Therefore, the solution for x’(t) is:
x’(t) = Ax cos(ωt + θx) + Ey/B .
To find x(t), we use x’(t) = dx/dt, or dx = x’(t) dt, or
x(t) – xo =
t
t=0∫
xo∫
x(t)
dx =
t
t=0∫
x’(t) dt :
[Ax cos(ωt + θx) + Ey/B] dt , or
x(t) = xo + (Ey/B)t + (Ax/ω) sin(ωt + θx) – (Ax/ω) sin(θx) .
This equation has three constants, xo, Ax, and θx. We have two initial conditions: x(t=0) and
x’(t=0) which will allow us to find two of the three constants. Where did the third constant
enter in? It entered in when we differentiated the x component equation with respect to time. In
order to find the third relation to find all three constants, we need to solve for y(t) – which means
we need to do the same process we just did.
We first take d/dt of the y component equation: qEy – qBx’ = my’’ to get:
– qBx’’ = my’’’
and then substitute for x’’ from the x component equation:
qBy’ = mx’’ to get
-q2B2y’/m = my’’’ .
We recognize this as a homogeneous, second order, linear, differential equation in y’, with a
solution of:
y’(t) = Ay cos(ωt + θy) again with ω = qB/m .
And solving for y(t) just as we did above for x(t), we get:
y(t) = yo + (Ay/ω) sin(ωt + θy) – (Ay/ω) sin(θy) .
Again, we have three constants: yo, Ay, and θy; and we have two initial conditions. Overall, we
have six constants and four initial conditions. We need two more relations (equations) to
determine all of the constants. We find these in our original Newton’s Second Law component
equations.
The x component equation,
qBy’ = mx’’ = m dx’/dt , gives:
qB[Ay cos(ωt + θy)] = m d[Ax cos(ωt + θx) + Ey/B] /dt
= -mωAx sin((ωt + θx)
The y component equation,
qEy – qBx’ = my’’ = m dy’/dt , gives:
qEy – qB[Ax cos(ωt + θx) + Ey/B] = m d[Ay cos(ωt + θy)]/dt
= -mωAy sin(ωt + θy) .
We now recall that ω = qB/m so that the first (x component) equation gives
Ay cos(ωt + θy) = -Ax sin((ωt + θx) .
The only way this can be true for all times is for Ax = Ay = A; and for θy = θx + 90o .
Note: recall that sin(θ+90o) = cos(θ); and sin(θ+180o) = cos(θ+90o) = -sin(θ) .
This gives us:
x’(t) = A cos(ωt + θ) + Ey/B
y’(t) = -A sin(ωt + θ) .
We can use the two initial conditions, one on x’(t=0) = vxo and one on y’(t=0) = vyo, to solve for
both A and θ:
x’(t=0) = vxo = A cos(θ) + Ey/B,
or
cos(θ) = [vxo – Ey/B]/A
y’(t=0) = vyo = -A sin(θ), or sin(θ) = -vyo/A
so we get:
tan(θ) = sin(θ)/cos(θ) = {-vyo/A} / {[vxo – Ey/B]/A} = -vyo / [vxo – Ey/B] .
We can then use the easier of the two equations, sin(θ) = -vyo/A, to solve for A:
A = -vyo / sin(θ) .
Analysis of the result:
The sine and cosine terms of vx and vy indicate a circular motion of frequency, ω = qB/m, and
the extra Ey/B in the vx expression indicates a drift velocity in the x-direction which is
perpendicular to both E and B .