# Electromagnetism - Lorentz Force

1. Feb 13, 2007

### Asrai

1. The problem statement, all variables and given/known data

A particle of charge q starts from rest at the origin of coordinates in a region where there is a uniform electric field of strenth E parallel to the x-axis, and a uniform magnetic field B parallel to the z-axis.

Find the equations of motion, and solve them to show that the coordinates of the particle at a time t later will be:

x = (E/B*omega)*(1 - cos(omega*t))

y = - (E/B*omega)*(omega*t - sin(omega*t))

z = 0

where omega = q*B/m. (The path of the circle is a cycloid.)

2. Relevant equations

The parametric equation of a cycloid:

x = constant*(1 - cos(omega*t))

y = constant*(omega*t - sin(omega*t))

The force acting on the particle:

F = q*E + q*vxB

3. The attempt at a solution

I've done some work on this problem and so far the equations of motion that I've got for the particle are as follows:

1) F(x) = q*E + q*v(y)*B -> x[double-dot] = q*E/m + omega*y[dot]

2) F(y) = -q*v(x)*B -> y[double-dot] = -q*B*x[dot]

I've tried integrating these equations once (eg. integrate 2)) and then substituting this into the other equation. This then gave me:

x[double-dot] + omega^2*x = E*B

And this is where I'm stuck. This has the form of a simple harmonic oscillator, except that the r.h.s. isn't zero, so I can't solve it. Also, I'm not even sure if everything that I've done so far is correct.

Any help on this would be very much appreciated!

2. Feb 13, 2007

### JK423

I found the same equations of motion but i kinda stuck there as well. I`ll check it again tomorrow :)

3. Feb 14, 2007

### Asrai

Thanks, that would be really helpful! I'm glad to know that I don't seem to be a complete moron though, or that I've overlooked some vital detail that will make the whole thing ridiculously easy.

4. Feb 14, 2007

### JK423

Solving the differential equation x[double-dot]+x*ω^2=qE/m (1)

X=X1+X2=(c1*cosωt+c2*sinωt)+qE/(m*ω^2) , c1,c2=constants

X1 is the solution of the : x[double-dot]+x*ω^2=0
X2 is a "special" solution that satisfies (1).

I think you can solve that now.
x(0)=0
x[dot](0)=0

Last edited: Feb 14, 2007