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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.

Qmechanic
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Jagte
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1 Answers1

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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.

Maxim Umansky
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