# Lorentz equation?

1. Feb 12, 2004

### mindcircus

The motion of a charged particle in an electromagnetic field can be obtained from the Lorentz equation for the force on a particle in such a field. If the electric field vector is E and the magnetic field vector is B, the force on a particle of mass m that carries a charge q and has a velocity v is given by
F=qE+qv X B
(The X is the cross product.)
If there is no electric field and if the particle enters the magnetic field in a direction perpendicular to the lines of magnetic flux, show that the trajectory is a circle with radius
r=(mv)/(qB)=v/(omega)
where omega=qB/m, which is the cyclotron frequency.

Okay, if there's no electric field, then I drop the qE term, meaning F=qv X B. I set this equal to F=ma.
qv X B = ma
I think I should make a=dv/dt, and solve for v, then take the derivative to get the trajectory. But the cross product is really confusing me, and I don't really know how to simplify from there. Am I going in the right direction?

2. Feb 12, 2004

### Tom Mattson

Staff Emeritus
After you take the cross product, note that the force is perpendicular to the velocity. What kind of motion does that give rise to?

(Hint: consider a ball on a string, being swung around parallel to the ground--you have the same type of situation).

Yes.

3. Feb 12, 2004

### mindcircus

If the velocity and force are perpendicular, it's centripetal motion. But I don't really know what to do with that. I know velocity is the cross product of the radius and omega (d.theta/dt). Because I'm trying to find the radius, I can't use this, I think?

Otherwise, when I try to do the math, I get a natural log, which I know I shouldn't be getting.

qvxB=qvB*sin 90=qvB
F=ma=qvB
m(dv/dt)=qvB
(qB/m)*dt=dv/v
Integrate.
(qB/m)t=ln v

I'm not using the fact that it's centripetal motion, but I don't really know how to tie this in.

4. Feb 12, 2004

### turin

Did you mean "antiderivative?"

5. Feb 12, 2004

### mindcircus

Yes, sorry, that's what I meant. And that's what I'm trying to do, I wasn't typing carefully. I have to integrate v to get r.

6. Feb 13, 2004

### drag

Greetings !
If the qE component is zero, you still have
circuilar motion due to the magnetic field.
I can't tell you how to solve for polar coordinates
but for cartesian coordinates all you need to do
is write the equations for two axes, and you get:
( qB/m ) Vy = ax
-( qB/m ) Vx = ay (the second minor is negative)
Then you use the derivative of either equation and
put the result in the other.
The solutions, depending on the enitial conditions
at t=0 (giving you the enitial angle = q and x0 and y0)are:
x = x0 + Rx sin(wt+q)
y = y0 + Ry cos(wt+q)

Live long and prosper.

7. Feb 13, 2004

### discoverer02

Doesn't the fact centripetal acceleration = v^2/r make this exercise a whole lot easier?