= Solution
The <Lorentz force> determines the motion of a <charged particle>:
$$
\boxed{\frac{d\mathbf p}{dt}=q(\mathbf E+\mathbf v\times\mathbf B).}
$$
At nonrelativistic <speeds>, $\mathbf p=m\mathbf v$. The <electric field> does work at rate $q\mathbf E\cdot\mathbf v$, while the magnetic part of the <Lorentz force> is perpendicular to <velocity> and does no work. A uniform <electric field> gives constant <acceleration> $q\mathbf E/m$; with a transverse initial <velocity>, the trajectory is a <parabola>.
In a uniform <magnetic field>, the parallel <velocity> is constant, while the perpendicular component undergoes <circular motion>. The angular <cyclotron frequency>, orbit radius and helical pitch are
$$
\boxed{\Omega_c=\frac{|q|B}{m},\qquad r_L=\frac{mv_\perp}{|q|B},\qquad p_{\rm helix}=\frac{2\pi v_\parallel}{\Omega_c}.}
$$
The sign of $q$ determines the sense of rotation. For perpendicular uniform <electric fields> and <magnetic fields>, the guiding centre has the <electric drift> velocity
$$
\boxed{\mathbf v_D=\frac{\mathbf E\times\mathbf B}{B^2}.}
$$
Indeed, $\mathbf E+\mathbf v_D\times\mathbf B=0$, so the remaining motion is cyclotron rotation about that drifting centre, with possible motion along the <magnetic field>. The drift is independent of <electric charge> sign and <mass>. At relativistic <energies>, retain $\mathbf p=\Gamma m\mathbf v$ and the corresponding relativistic orbit radius rather than assuming constant nonrelativistic <acceleration>.
Back to article page
