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The force on a charged particle in electric field $E$ and magnetic field $B$ is given by: \begin{equation} m \dot{v} = q ( E + v \times B) \end{equation} 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 \begin{equation} v_E= E \times B /B^2 \end{equation} 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.

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  • $\begingroup$ Are you still interested in an answer to this question? @Jagte $\endgroup$ – Gotaquestion Dec 1 '13 at 9:34
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This ExB velocity is the motion of the particle's guiding center; the motion of the physical particle itself is the sum of the guiding center drift and gyration motion around the guiding center. Once you sum up both components the velocity will not be unphysical.

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  • $\begingroup$ 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. $\endgroup$ – Jagte Sep 1 '13 at 20:37
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    $\begingroup$ 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 $V_E > v_{th}$ $\endgroup$ – Jagte Sep 1 '13 at 20:44
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    $\begingroup$ If B goes to zero then the gyro-frequency w=qB/mc goes to zero, so the period is infinite and you cannot average over it. $\endgroup$ – Maxim Umansky Sep 1 '13 at 23:18
  • $\begingroup$ 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 $V_E =E_z/B$ $\endgroup$ – Jagte Sep 2 '13 at 0:33
  • $\begingroup$ My point is that "averaging" can be done if the particle completes at least one gyro-circle during the time range relevant to your problem. You probably agree that if gyro-period is 1e-6 s but you are looking at processes on the scale 1e-9 s then gyro-averaging can not be done - correct? For B->0 the gyro-period becomes infinite, longer than any time of initerest. $\endgroup$ – Maxim Umansky Sep 2 '13 at 3:59

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