Electron moving in constant magnetic field I'm revisiting an old problem which is to show that the trajectory of an electron in a constant magnetic field $\textbf{B} = (0,0,B)$ is a helix. I started with the Lorentz force equation
$$ \textbf{F}  = q(\textbf{v} \times \textbf{B}) $$
Then I decomposed $\textbf{F}$ into $\textbf{F}_x = m\ddot{\textbf{x}}$ and $\textbf{F}_y = m\ddot{\textbf{y}}$. Then equating them to the right-hand side of the equation I get two 2nd order differential equations and solving them gives me a circle in the xy plane. Adding the velocity in the z direction would make this a helix.
I would like to know if there is another way to solve this problem other than the approach I had taken (just out of curiosity)?
 A: There's the obvious answer of using Newton's Laws in cylindrical coordinates, but there are some amusing  techniques that use the Lagrangian formalism.  Both are drawn from "Thoughts on the magnetic vector potential" by Mark D. Semon & John R. Taylor (Am. J. Phys. 64 (11), 1996).  As a preliminary, recall that the Lagrangian for a charged particle in a magnetic field is given by
$$
\mathcal{L} = \frac{1}{2} m \vec{v}^2 + q \vec{v} \cdot \vec{A},
$$
where $\vec{A}$ is the vector potential.
Method 1
Consider the vector potential $\vec{A} = \frac{1}{2} \vec{B} \times \vec{r}$;  it's not hard to check that $\vec{\nabla} \times \vec{A} = \vec{B}$.  Let's use cylindrical coordinates $\{s, \phi, z\}$, and pick our rotation axis such that $\vec{B} = B \hat{z}$.  Given all of this, we have $\vec{A} = \frac{1}{2} B s \hat{\phi}$, and the Lagrangian is
$$
\mathcal{L} = \frac{1}{2} m \left( \dot{s}^2 + s^2 \dot{\phi}^2 + \dot{z}^2 \right) + \frac{1}{2} q B s^2 \dot{\phi}.
$$
The conjugate momenta for $\phi$, and $z$ are
$$p_\phi =  m s^2 \dot{\phi} + \frac{1}{2} q B s^2 \qquad p_z = m \dot{z}.
$$
Since the Lagrangian is independent of $\phi$ and $z$, these conjugate momenta are constants.  The Euler-Lagrange equation for $s$, meanwhile, is 
$$
m \ddot{s} = (m \dot{\phi} + q B) s \dot{\phi}
$$
If we want a solution to the above equations where $s = const.$, we must then have
either $s = 0$, $\dot{\phi} = 0$ (in which cases there is no motion in the plane perpendicular to $\vec{B}$) or $\dot{\phi} = - qB/m$.  Thus, we can recover the case of uniform circular motion with an angular speed at the cyclotron frequency.
Method 2
The vector potential given above is not the only $\vec{A}$ that corresponds to a uniform magnetic field $\vec{B}$.  Here are two others:
$$
\vec{A}_1 = - B y \hat{x} \qquad \mathcal{L}_1 = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) - q B y \dot{x}
$$
$$
\vec{A}_2 = B x \hat{y} \qquad \mathcal{L}_2 = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 + \dot{z}^2 \right) + q B x \dot{y}
$$
In the first Lagrangian, we have $p_x = m \dot{x} - qBy =$ const., while in the second, we have $p_y = m\dot{y} + qBx =$ const.  We can redefine our origin in $x$ and $y$ to set these constants to zero, with the resulting equations
$$
m \dot{x} = q B y \qquad m \dot{y} = -q B x
$$
Again, we have equations that can be to yield uniform circular motion.  Note, though, that they're lower order than the equations you would get from Newton's Laws, and so a bit easier to solve.
A: I don't know if it's another (different) way exactly, but you can use the old left-hand rule to get the direction of the force, and realise that there will be no force (from the magnetic field) in the $z$ direction due to the cross product. Analysis using elementary circular motion techniques might work too! Alternatively, use cylindrical or circular polar  coordinates!
