Particle motion in a magnetic field

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Homework Help Overview

The discussion revolves around the motion of a charged particle in a magnetic field, particularly focusing on the integration of the equation of motion and the effects of gravitational forces. Participants are exploring the implications of these forces on the trajectory of the particle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate the equation of motion while considering the effects of a magnetic field and gravity. They express uncertainty about the role of certain velocity components and seek guidance on deriving the equation of a helix. Other participants engage by suggesting a frame transformation to simplify the analysis and discuss the resulting motion.

Discussion Status

The discussion is active, with participants exploring different interpretations of the motion involved. Some guidance has been offered regarding the transformation of frames and the conceptual understanding of helical motion, but no consensus has been reached on deriving specific equations.

Contextual Notes

Participants are navigating the complexities of integrating motion equations under the influence of both magnetic and gravitational forces, with some expressing confusion about the implications of certain terms in their equations.

etotheipi
Homework Statement
A particle of mass ##m##, charge ##q## and position ##\mathbf{r}(t)## moves in a magnetic field ##\mathbf{B}## pointing horizontally, and a uniform gravitational field ##\mathbf{g}## pointing vertically downward.

Derive that the particle undergoes helical motion with constant drift ##\bot## to ##\mathbf{B}##.
Relevant Equations
N/A
The equation of motion can be integrated w.r.t. ##t## since ##\frac{d}{dt} (\mathbf{r} \times \mathbf{B}) = \dot{\mathbf{r}} \times \mathbf{B} + \mathbf{0}## $$\int (q\dot{\mathbf{r}} \times \mathbf{B} + m\mathbf{g}) dt = \int m\ddot{\mathbf{r}}(t) dt$$ $$\frac{q}{m} \mathbf{r} \times \mathbf{B} + t\mathbf{g} + \mathbf{c} = \dot{\mathbf{r}}$$ I thought it might help to put a basis to it, so I let ##\mathbf{B} = B\mathbf{\hat{x}}## and ##\mathbf{g} = -g\mathbf{\hat{y}}##, so that $$\dot{\mathbf{r}} = \frac{qB}{m} \mathbf{r} \times \mathbf{\hat{x}} -gt \mathbf{\hat{y}} + \mathbf{c}$$The ##\frac{qB}{m} \mathbf{r} \times \mathbf{\hat{x}}## is a component of the velocity perpendicular to the ##\mathbf{x}## axis which I assume contributes to the 'circular' part of the motion. It's not obvious to me what the ##\mathbf{c} - gt\mathbf{\hat{y}}## component of velocity will cause. I don't think I can integrate this equation again either.

I wondered whether anyone could point me in the right direction (vector pun not intended)? Thanks!
 
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Thanks for the reply! It's easier to visualise without the constant term, so yes let's say we boost into another frame with ##\dot{\mathbf{r}}' = \dot{\mathbf{r}} - \mathbf{c}## so that we are left with ##\dot{\mathbf{r}'} = \frac{qB}{m} \mathbf{r}' \times \mathbf{\hat{x}} -gt \mathbf{\hat{y}}##. In the absence of the weight, ##\dot{\mathbf{r}}' \cdot \ddot{\mathbf{r}}' = 0## so we would end up with helical motion.

With the weight added back in, conceptually I can see why it will be helical motion with constant vertical translation, though I wonder if there is any way of deriving the equation of the helix to show this?
 
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sorry about previous post, i completely misread your question
 
Replusz said:
sorry about previous post, i completely misread your question

Ah no worries, I thought it was useful :smile:. Didn't even notice!
 
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