# General Solutions for a particle in a magnetic field

1. Dec 5, 2011

### danielakkerma

Hello everyone!
I encountered a curious problem while trying to solve the case of a particle with v(v_x0, v_y0, 0), and B(0, 0, Bz); The elucidation of the differential equations obtained through the Lorentz force in this case, should coincide with those obtained through a simplification granted by using the Centripetal force, but here, instead of circular motion I get an unattractive ellipse :(.
Assuming the particle moves only on the x-y plane, and starts at (x0, y0, 0), the Lorentz force yields the following:
$\large \vec{F} = q(\vec{v}\times\vec{B})$
$ma_x(t) = -qv_yB$
$ma_y(t) = qv_xB$
Which in turn, with these initial conditions leads to:
With v0 = Sqrt(vx0^2+vy0^2);
Alpha derived from: Arctan[vy/vx] = alpha;
$x(t) = x_0+ \frac{v_0(-\sin(\alpha)+\sin(\omega t + \alpha))}{\omega}$
$y(t) = y_0+ \frac{v_0(\cos(\alpha)-\cos(\omega t + \alpha))}{\omega}$
$\omega = \frac{qB}{m}$
Plugging in some random values, leads to the attached image, while we all know that motion in a magnetic field should be accompanied by uniform circular motion;
Where have I gone wrong?
Thanks,
Daniel
P.S
This is not related in anyway, to homework.

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2. Dec 5, 2011

### Born2bwire

You don't think it might not be caused by the fact that your axes have different scales?

3. Dec 5, 2011

### danielakkerma

Firstly, let me thank you for a very prompt reply!
Yes, I have thought of that, as a matter of fact, but the problem is, that attempting to prove that curve is a circle, i.e, by placing it in (x-a)^2+(y-b)^2=R^2, doesn't produce the desired result. In other words, after expanding the left-hand-side, the time dependent component does not vanish, further suggesting that this is somehow, not a circle.
Anyhow, adjusting the scales does little good :(.
What else could there be?
Thankful as always,
Daniel

4. Dec 5, 2011

### Born2bwire

When I make a plot it looks like a circle to me. And from what I can tell x^2+y^2 = R^2, where R is a constant. You forget that x and y have constant offsets dependent upon x_0, y_0 and \alpha. If you retain the time dependent parts you can easily see that they create a circle.

5. Dec 5, 2011

Hi,