Lorentz equation?

[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?

[Tom Mattson]

Originally posted by mindcircus
But the cross product is really confusing me, and I don't really know how to simplify from there.


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).


Am I going in the right direction?


Yes.

[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.

[turin]

Originally posted by mindcircus
I think I should make a=dv/dt, and solve for v, then take the derivative to get the trajectory.Did you mean "antiderivative?"

[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.

[drag]

Greetings !
Originally posted by mindcircus
If the velocity and force are perpendicular, it's centripetal motion. But I don't really know what to do with that.
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.

[discoverer02]

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