1. ## Lorentz force law

According to the Lorentz force law, an electron moving in a constant magnetic field follows a path $\displaystyle c(t)$ which satisfies $\displaystyle \ddot c = \dot c \times B$, where $\displaystyle B$ is a constant vector on $\displaystyle \mathbb{R}^3$. Prove that this path is a helix.

2. Originally Posted by mathman88
According to the Lorentz force law, an electron moving in a constant magnetic field follows a path $\displaystyle c(t)$ which satisfies $\displaystyle \ddot c = \dot c \times B$, where $\displaystyle B$ is a constant vector on $\displaystyle \mathbb{R}^3$. Prove that this path is a helix.
We can choose to align the z axis with the magnetic field B. Then:

$\displaystyle B=B_z$, $\displaystyle B_x=0$, $\displaystyle B_y=0$

So:

$\displaystyle \ddot c_x = \dot c_y B_z$

$\displaystyle \ddot c_y = -\dot c_x B_z$

$\displaystyle \ddot c_z = 0$

Solving the third equation by integrating twice with respect to t: $\displaystyle c_z = kt$ where k is a constant.

Now if you differentiate the second equation and then substitute for $\displaystyle \ddot c_x$ from the first then you will get:
$\displaystyle \ddot \dot c_y=-\dot c_yB_z^2$

which has the general solution:
$\displaystyle \dot c_y=-a sin(B_zt)+b cos(B_zt) = csin(B_zt+\phi)$

So
$\displaystyle c_y=\frac{-c cos(B_zt + \phi)}{B_z}$

Now using equation 1
$\displaystyle \ddot c_x=\dot c_y B_z= cB_zsin(B_zt+\phi)$

So
$\displaystyle c_x= \frac {-csin(B_zt+\phi)}{B_z}$

Now using $\displaystyle c_x$, $\displaystyle c_y$, $\displaystyle c_z$ we can write:

$\displaystyle \bar c = \frac {-csin(B_zt+\phi)}{B_z} \hat e_x-\frac {ccos(B_zt+\phi)}{B_z} \hat e_y+kt\hat e_z$

This is the equation of a helix.