$E\times B$ drift of a single particle in vacuum with unifiorm $E$ and $B$

The force on a charged particle in electric field $E$ and magnetic field $B$ is given by: $$m \dot{v} = q ( E + v \times B)$$ Then, the $E \times B$ drift velocity $v_E$ of a single particle of mass $m$ and charge $q$ in a uniform magnetic field $B$ and a uniform electric field $E$ is given by $$v_E= E \times B /B^2$$ and if $E$ and $B$ are perpendicular, $v_E=E/B$.

How do I understand the fact that $v_E \propto 1/B$ ?

if $B \sim 0$, $V_E \sim \infty$, more than the velocity of light. Furthermore,it is also plausible that $V_E >> v_{th}$, the thermal velocity.

• Are you still interested in an answer to this question? @Jagte – Gotaquestion Dec 1 '13 at 9:34

• Now average over time scales longer than the gyro-frequency, the $e^{i \omega_c t}$ component cancels out, leaving $E/B$ as the average velocity. That is what confuses me. – Jagte Sep 1 '13 at 20:37
• Comment Rewite: ExB velocity is the motion of the particle's guiding center agreed, but now average over time scales longer than the gyro-frequency, the $e^{i \omega_c t}$ component averages to zero thus leaving $E/B$ as the average velocity of a particle. That is what confuses me. Next assume $B$ is small, one has $V_E > v_{th}$ – Jagte Sep 1 '13 at 20:44
• I tried to understand this problem in the following way: Assume $E_x$ is the electric field $\perp$ to $B$. Then in a time $\delta t$, the gain in velocity in the limit $B \rightarrow 0$ would be $\delta V_x = (q E_x/m) \delta t$, which should be infinite as $\delta t \rightarrow \infty$. Substituting $\delta t=1/\omega_c$ (one gyration), $\delta V_x =E_z/B$. This means particle accelerates and de-accelerates in every gyration. Hence on average $V_E =E_z/B$ – Jagte Sep 2 '13 at 0:33