- #1
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!
I wondered whether anyone could point me in the right direction (vector pun not intended)? Thanks!
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