[플라즈마물리]Single Particle Motion - E cross B drift

Problem Consider a particle of charge $q$ and mass $m$, initially at rest at (0,0,0), in the presence of a static magntic field $\vec{B}=B\hat{z}$ and $\vec{E}=E\hat{y}$.

(a) Taking $E, B > 0$, sketch the orbit of the particle when $q>0$.

(b) Derive an exact expression for the orbit $[x(t),y(t),z(t)]$ or the particle. Express your answer in terms of $E$, $B$, and $\omega_c$

(c) Find the drift velocity after averaging the motion in time. If there were many particles of various charges and masses present, would there be any net current?

(d) Suppose the electric field were replaced by a force $F$ in the $y$-direction. What would be the drift velocity?

Answer (a)

Answer (b)

$\vec{B}=B\hat{z}$ and $\vec{E}=E\hat{y}$. From Lorentz equation,

$$m\frac{dv}{dt}=q(\vec{E} + \vec{v}\times \vec{B})$$

$$\begin{aligned} \dot{v_x}&= \quad \quad \quad \quad \frac{q}{m}v_y B \nonumber \\ \dot{v_y}&= \frac{q}{m}\vec{E_y} - \frac{q}{m}v_xB \nonumber \\ \dot{v_z}&= 0\end{aligned}$$

In z-direction $$z(t) = constant = 0$$

$$\begin{aligned} \frac{dv_x}{dt} &= \frac{qB}{m}v_y = \omega_c v_y \nonumber \\ \frac{dv_y}{dt} &= \frac{qE}{m}-\frac{qB}{m}v_x = \frac{qE_y}{m}-\omega_c v_x \nonumber\end{aligned}$$

$$\begin{aligned} \frac{d^2 v_y}{dt^2} &= -\omega_c^2 v_y \nonumber \\ \frac{d^2 v_x}{dt^2} &= -\omega_c^2 \left(v_x-\frac{E}{B} \right) \nonumber\end{aligned}$$

since $\frac{E}{B}$ is constant, $$\frac{d^2}{dt^2} \left[ v_x-\frac{E}{B} \right] = \frac{d^2 v_x}{dt^2} = -\omega_c^2 \left(v_x-\frac{E}{B} \right)$$

$$\begin{aligned} v_x &= iv_{\perp}e^{i\omega_c t} + \frac{E}{B}\nonumber \\ v_y &= v_{\perp}e^{i\omega_c t} \nonumber \end{aligned}$$

$$\begin{aligned} x(t) &= x(0) + r_L \cos{\omega_c t} +\frac{E}{B}t \nonumber \\ y(t) &= y(0) + r_L \sin{\omega_c t} \nonumber \end{aligned}$$

Answer (c)

$$<v_d> = \frac{\int_{-\infty}^{\infty}v_d f(v) dv}{\int_{-\infty}^{\infty}f(v) dv} = \frac{E}{B}$$

There will be no current because $\vec{E} \times \vec{B}$ drifts the ions and electrons in the same direction.

Answer (d)

$$\begin{aligned} \vec{v_f}&= \frac{1}{q} \frac{\vec{F} \times \vec{B}}{B^2} \nonumber \\ &=\frac{FE}{qB}\hat{y}\end{aligned}$$

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