Irodov 3.395: A Resonant Push Unwinds a Spiral – Amit & Deepali Sinha Foundation

Problem Irodov 3.395

In a right-handed $xyz$ frame a constant uniform magnetic field of induction $B$ points along the positive $z$ axis, and a uniform electric field along the $y$ axis varies with time as $E = E_m\cos\omega t$ , where $\omega = qB/m$ . A particle of specific charge $q/m$ starts from 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. What is the approximate shape of its trajectory?

The frame with z toward the viewer: the oscillating E along y (drawn for ω t > 0) and the constant B along +z; the particle starts at rest at the origin O. — The frame with $z$ toward the viewer: the oscillating $\vect E$ along $y$ (drawn for $\cos\omega t > 0$ ) and the constant $\vect B$ along $+z$ ; the particle starts at rest at the origin $O$ .

Given

$\vect B \parallel +z$
$E_y = E_m\cos\omega t$
$\omega = qB/m$
$q/m$
at rest at $O$ , $t = 0$

Find

x(t)

y(t)

and the approximate shape of the trajectory

Solution

Step 1The equations of motion

The force $q(\vect E + \vect v\times\vect B)$ has no $z$ component for motion in the $xy$ plane, so the particle stays in that plane. With $\vect B = B\,\vect k$ , $\vect v\times\vect B = (v_y B, -v_x B, 0)$ , and writing $a$ for the amplitude of the electric acceleration:

\begin{aligned} \frac{dv_x}{dt} &= \omega v_y, \\ \frac{dv_y}{dt} &= a\cos\omega t - \omega v_x, \\ a &= \frac{qE_m}{m} \end{aligned} \tag{1}

Step 2One complex equation

Combine the components into $w = v_x + i v_y$ . Adding the first equation of (1) to $i$ times the second:

\begin{align*} \frac{dw}{dt} &= \omega v_y + i\,a\cos\omega t - i\omega v_x = -i\omega\left(v_x + i v_y\right) + i\,a\cos\omega t, \\ \frac{dw}{dt} + i\omega w &= i\,a\cos\omega t \tag{2}\end{align*}

Step 3The velocity

Multiplying (2) by $e^{i\omega t}$ makes the left side a derivative, and the cosine splits into exponentials:

\frac{d}{dt}\left(w\,e^{i\omega t}\right) = i\,a\cos\omega t\;e^{i\omega t} = \frac{i a}{2}\left(1 + e^{2i\omega t}\right)

Integrating from rest, $w = 0$ at $t = 0$ , and multiplying back by $e^{-i\omega t}$ :

\begin{aligned} w\,e^{i\omega t} &= \frac{i a}{2}\,t + \frac{a}{4\omega}\left(e^{2i\omega t} - 1\right), \\ w &= \frac{i a t}{2}\left(\cos\omega t - i\sin\omega t\right) + \frac{i a}{2\omega}\,\sin\omega t \end{aligned}

The real and imaginary parts are the two velocity components:

\begin{aligned} v_x &= \frac{a t}{2}\,\sin\omega t, \\ v_y &= \frac{a t}{2}\,\cos\omega t + \frac{a}{2\omega}\,\sin\omega t \end{aligned} \tag{3}

Step 4The law of motion

Integrating (3) from $O$ , with the integral $\int t\sin\omega t\,dt = -t\cos\omega t/\omega + \sin\omega t/\omega^2$ and its cosine companion, both found by parts:

\begin{aligned} x &= \frac{a}{2}\left[-\frac{t\cos\omega t}{\omega} + \frac{\sin\omega t}{\omega^2}\right], \\ y &= \frac{a}{2}\left[\frac{t\sin\omega t}{\omega} + \frac{\cos\omega t - 1}{\omega^2}\right] + \frac{a}{2\omega^2}\left(1 - \cos\omega t\right) \end{aligned}

In $y$ the two non-growing terms cancel, leaving

\begin{gathered} x = \frac{a}{2\omega^2}\left(\sin\omega t - \omega t\cos\omega t\right), \\ y = \frac{a}{2\omega}\,t\sin\omega t, \\ a = \frac{qE_m}{m} \end{gathered}

(4)

Step 5The shape of the trajectory

Once $\omega t \gg 1$ the terms growing with $t$ dominate, and the motion is a circling whose radius $r$ grows in proportion to the time:

x \approx -\frac{a t}{2\omega}\cos\omega t, \qquad y \approx \frac{a t}{2\omega}\sin\omega t, \qquad r \approx \frac{a t}{2\omega}

Each turn adds $\pi a/\omega^2$ to the radius: the trajectory is an unwinding spiral, close to an Archimedean one, turning clockwise (for $q > 0$ ) about the origin.

The law of motion plotted over two and a half turns: from rest at O the particle sets off along y and spirals outward clockwise, the gap between turns staying the same. — The law of motion plotted over two and a half turns: from rest at $O$ the particle sets off along $y$ and spirals outward clockwise, the gap between turns staying the same.

Check

Early times: for $\omega t \ll 1$ , $\sin\omega t - \omega t\cos\omega t \approx (\omega t)^3/3$ , so $x \approx a\omega t^3/6$ and $y \approx a t^2/2$ ; the particle starts off along $y$ as if the electric force $qE_m$ acted alone, which it does before any speed builds up for $B$ to act on. Units: $a$ is an acceleration, so $a t/\omega$ and $a/\omega^2$ are lengths. Speed: from (3), $v_x^2 + v_y^2 \approx (a t/2)^2$ at late times, and $r\omega \approx a t/2$ is the same speed, consistent with circling at the frequency $\omega$ .

Remark

The field $E_m\cos\omega t$ along a line is the sum of two fields of strength $E_m/2$ rotating in opposite senses at the frequency $\omega$ . The one that rotates with the particle stays in step with its velocity and pushes it forward on every turn, so the speed grows steadily at the rate $qE_m/(2m)$ ; the other rotates against it and its push averages out. This is cyclotron resonance, the principle of the cyclotron and of the resonant heating of plasmas; off resonance the speed would only beat, never grow.