General Solutions for a particle in a magnetic field

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The discussion centers on a particle's motion in a magnetic field, specifically addressing discrepancies between expected circular motion and the observed elliptical trajectory. The equations derived from the Lorentz force suggest circular motion, yet the results indicate an ellipse due to the parametric nature of the equations. The participant initially struggles with proving the motion is circular, but later realizes the importance of considering the explicit relationship between x and y as functions of time. Adjusting the scales of the axes did not resolve the issue, but further analysis clarified the misunderstanding. Ultimately, the participant acknowledges the role of parametric equations in the observed motion.
danielakkerma
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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:
<br /> \large<br /> \vec{F} = q(\vec{v}\times\vec{B})<br />
<br /> ma_x(t) = -qv_yB<br />
<br /> ma_y(t) = qv_xB<br />
Which in turn, with these initial conditions leads to:
With v0 = Sqrt(vx0^2+vy0^2);
Alpha derived from: Arctan[vy/vx] = alpha;
<br /> x(t) = x_0+ \frac{v_0(-\sin(\alpha)+\sin(\omega t + \alpha))}{\omega}<br />
<br /> y(t) = y_0+ \frac{v_0(\cos(\alpha)-\cos(\omega t + \alpha))}{\omega}<br />
<br /> \omega = \frac{qB}{m}<br />
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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You don't think it might not be caused by the fact that your axes have different scales?
 
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
 
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.
 
Hi,
Thanks again for your response;
I guess I did overreact, and that it should turn out alright; as for x^2+y^2..., I think the problem lied in my not taking the explicit form of y as a function of x. Thus, the problem resides, in this case, in the function being a parametric one.
Thanks again for all your help!
Couldn't have done it with you!
Beholden,
Bound,
Daniel
 
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