Irodov 3.392: The Cycloid of Crossed Fields – Amit & Deepali Sinha Foundation

Problem Irodov 3.392

In a right-handed $xyz$ frame a uniform electric field of strength $\vect E$ points along the positive $y$ axis and a uniform magnetic field of induction $\vect B$ along the positive $z$ axis. A particle of specific charge $q/m$ is at rest at the origin $O$ at the moment $t = 0$ . Treating the motion as non-relativistic, find:

the law of motion $x(t)$ , $y(t)$ of the particle, and the shape of its trajectory;
the length of the path between two successive points at which the particle’s velocity vanishes;
the mean value of the $x$ component of the particle’s velocity (the drift velocity).

The frame with z toward the viewer: E along +y and B along +z, both uniform; the particle starts at rest at the origin O. — The frame with $z$ toward the viewer: $\vect E$ along $+y$ and $\vect B$ along $+z$ , both uniform; the particle starts at rest at the origin $O$ .

Given

$\vect E \parallel +y$
$\vect B \parallel +z$
$q/m$
at rest at $O$ when $t = 0$

Find

(a)

x(t)

y(t)

and the trajectory’s shape · (b) the path between successive stops · (c) the drift velocity

\langle v_x\rangle

Solution

Step 1The equations of motion

The force is $q(\vect E + \vect v\times\vect B)$ . Neither field has a $z$ component of force on a particle moving in the $xy$ plane, and the particle starts at rest, so it stays in that plane. With $\vect B = B\,\vect k$ the magnetic part has the components

\vect v\times\vect B = \left(v_y B,\; -v_x B,\; 0\right)

Dividing by $m$ and writing $\omega$ for the cyclotron frequency:

\begin{aligned} \frac{dv_x}{dt} &= \omega v_y, \\ \frac{dv_y}{dt} &= \frac{qE}{m} - \omega v_x, \\ \omega &= \frac{qB}{m} \end{aligned} \tag{1}

Step 2The velocity components

Differentiating the first equation of (1) and substituting the second eliminates $v_y$ :

\frac{d^2v_x}{dt^2} = \omega\,\frac{dv_y}{dt} = \omega^2\left(\frac{E}{B} - v_x\right)

So $v_x - E/B$ oscillates at the frequency $\omega$ . The start from rest gives $v_x(0) = 0$ and, through the first equation of (1), $dv_x/dt = 0$ at $t = 0$ ; the same equation then returns $v_y$ :

\begin{aligned} v_x &= \frac{E}{B}\,(1 - \cos\omega t), \\ v_y &= \frac{1}{\omega}\,\frac{dv_x}{dt} = \frac{E}{B}\,\sin\omega t \end{aligned} \tag{2}

Step 3(a) The law of motion and the trajectory

Integrating (2) from $O$ at $t = 0$ :

\begin{aligned} x &= \int_0^t \frac{E}{B}\,(1 - \cos\omega t')\,dt' = \frac{E}{B\omega}\,(\omega t - \sin\omega t), \\ y &= \int_0^t \frac{E}{B}\,\sin\omega t'\,dt' = \frac{E}{B\omega}\,(1 - \cos\omega t) \end{aligned}

The common factor is a length; with $\omega$ from (1),

a = \frac{E}{B\omega} = \frac{mE}{qB^2}

which gives

\begin{gathered} x = a\,(\omega t - \sin\omega t), \quad y = a\,(1 - \cos\omega t), \\ a = \frac{mE}{qB^2}, \quad \omega = \frac{qB}{m} \end{gathered}

(3)

These are the equations of a cycloid: the particle moves as a point on the rim of a circle of radius $a$ that rolls without slipping along the $x$ axis, its centre advancing at the speed $a\omega = E/B$ .

The trajectory from rest at O: a cycloid of height 2a with cusps on the x axis. The dashed rolling circle of radius a turns clockwise as its centre moves along x. — The trajectory from rest at $O$ : a cycloid of height $2a$ with cusps on the $x$ axis. The dashed rolling circle of radius $a$ turns clockwise as its centre moves along $x$ .

Step 4(b) The path between two stops

The speed follows from (2), using $1 - \cos\omega t = 2\sin^2(\omega t/2)$ :

v^2 = v_x^2 + v_y^2 = \frac{E^2}{B^2}\left(2 - 2\cos\omega t\right) = \frac{4E^2}{B^2}\,\sin^2\frac{\omega t}{2}

So $v = (2E/B)\,\lvert\sin(\omega t/2)\rvert$ , which vanishes at the cusps, $\omega t = 0, 2\pi, 4\pi, \ldots$ Integrating the speed over one arch, from $t = 0$ to $t = 2\pi/\omega$ :

s = \int_0^{2\pi/\omega} \frac{2E}{B}\,\sin\frac{\omega t}{2}\,dt = \frac{2E}{B}\left[-\frac{2}{\omega}\cos\frac{\omega t}{2}\right]_0^{2\pi/\omega} = \frac{2E}{B}\cdot\frac{4}{\omega}

s = \frac{8mE}{qB^2} = 8a

(4)

Step 5(c) The drift velocity

Over one period $T = 2\pi/\omega$ the motion repeats itself shifted along $x$ by one arch, whose length along the axis is $x(T) = 2\pi a$ from the answer of (a). The mean of $v_x$ is that shift over the period:

\langle v_x\rangle = \frac{x(T)}{T} = \frac{2\pi a}{2\pi/\omega} = a\omega

\langle v_x\rangle = \frac{E}{B}

(5)

Check

Units: $mE/(qB^2)$ is $\text{kg}\cdot(\text{V}/\text{m})/(\text{C}\cdot\text{T}^2)$ ; with $\text{V}\cdot\text{C} = \text{J}$ and $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ this is $\text{J}\cdot\text{s}^2/(\text{kg}\cdot\text{m}) = \text{m}$ , a length, and $E/B$ is $(\text{V}/\text{m})/(\text{V}\cdot\text{s}/\text{m}^2) = \text{m}/\text{s}$ . The top speed, at $\omega t = \pi$ , is $2E/B$ , twice the drift, as for a point on the top of a rolling wheel. For a negative charge both $a$ and $\omega$ change sign: $y = a(1 - \cos\omega t)$ becomes negative, so the arches hang below the $x$ axis, while $x = a(\omega t - \sin\omega t)$ stays positive, so the drift is still along $+x$ , the direction of $\vect E\times\vect B$ .

Remark

In a frame moving along $x$ at $u = E/B$ the electric field disappears, since $\vect E + \vect u\times\vect B = E\,\vect j - E\,\vect j = 0$ . There the particle starts with velocity $-u$ in a pure magnetic field and simply circles with radius $mu/(qB) = a$ ; carried along at $u$ , that circle traces the cycloid. The drift $E/B$ does not depend on $q$ or $m$ , so in crossed fields electrons and ions of a plasma drift together and carry no net current along $\vect E\times\vect B$ .