Irodov 3.388: One Parabola for Every Speed – Amit & Deepali Sinha Foundation

Problem Irodov 3.388

A uniform electric field of strength $E$ and a uniform magnetic field of induction $B$ both point along the positive $y$ axis of a right-handed $xyz$ frame. A narrow beam of identical ions of specific charge $q/m$ , with a spread of speeds, enters the fields at the origin $O$ moving along the positive $x$ axis. A flat screen perpendicular to the $x$ axis stands at the distance $l$ from $O$ . Find the equation of the trace the ions leave on the screen, and show that for $z \ll l$ it is a parabola.

The frame with z toward the viewer. Both uniform fields, E and B, point along +y; the ions enter at the origin O with velocity v_0 along +x. — The frame with $z$ toward the viewer. Both uniform fields, $\vect E$ and $\vect B$ , point along $+y$ ; the ions enter at the origin $O$ with velocity $\vect v_0$ along $+x$ .

Given

$\vect E \parallel \vect B \parallel +y$
$q/m$
$\vect v_0$ along $+x$ at $O$ , speeds differing
screen $x = l$

Find

the trace

z(y)

on the screen · its form for

z \ll l

Solution

Step 1The motion along and across the fields

The magnetic force has no $y$ component, so along $y$ the ion feels only $qE$ and starts with $v_y = 0$ :

y = \frac{qE}{2m}\,t^2 \tag{1}

Across the fields the speed stays $v_0$ and the projection on the $xz$ plane is a uniform circle of radius $R$ with angular velocity $\omega$ , on which $q v_0 B$ is the centripetal force:

R = \frac{m v_0}{qB}, \qquad \omega = \frac{v_0}{R} = \frac{qB}{m} \tag{2}

The initial force $q v_0 B\,(\vect i\times\vect j) = q v_0 B\,\vect k$ puts the centre at $z = R$ on the $z$ axis, so after turning through $\theta = \omega t$ the projection is at

x = R\sin\theta, \qquad z = R\,(1 - \cos\theta) \tag{3}

Step 2Where an ion meets the screen

An ion reaches the screen when $x = l$ . Ions of different speed move on circles of different radius, but all of them start at $O$ tangent to the $x$ axis.

The projection on the xz plane, seen with B into the page: the circle of radius R about C from O to the hit point P on the screen, the angle θ turned at C, and the chord OP at θ/2 to the x axis. — The projection on the $xz$ plane, seen with $\vect B$ into the page: the circle of radius $R$ about $C$ from $O$ to the hit point $P$ on the screen, the angle $\theta$ turned at $C$ , and the chord $OP$ at $\theta/2$ to the $x$ axis.

Setting $x = l$ in (3) and dividing $z$ by $x$ eliminates the radius, so the speed drops out of the hit point’s direction:

\frac{z}{l} = \frac{1 - \cos\theta}{\sin\theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}

\frac{z}{l} = \tan\frac{\theta}{2}

(4)

This is the tangent-chord angle: the chord $OP$ makes the angle $\theta/2$ with the tangent at $O$ , the $x$ axis. Only ions with $R > l$ reach the screen, and they arrive before a quarter turn, so $0 < z < l$ .

Step 3The height of the hit point

The time of flight to the screen is $t = \theta/\omega$ , the same for the $y$ motion. Putting it into (1) with $\omega$ from (2):

y = \frac{qE}{2m}\cdot\frac{\theta^2 m^2}{q^2B^2} = \frac{mE}{2qB^2}\,\theta^2

Each ion’s speed now enters only through $\theta$ . Eliminating $\theta$ with (4), $\theta = 2\arctan(z/l)$ :

y = \frac{2mE}{qB^2}\arctan^2\frac{z}{l}

Solving for $z$ gives the trace on the screen:

z = l\tan\sqrt{\frac{qB^2\,y}{2mE}}

(5)

Fast ions land near $O$ , slow ones (with $R$ just over $l$ ) near $z = l$ ; all of them lie on this one curve.

Step 4The small-

z

limit

For $z \ll l$ the angle $\theta$ is small and the tangent may be replaced by its argument:

z \approx l\sqrt{\frac{qB^2\,y}{2mE}} \quad\Longrightarrow\quad z^2 = \frac{qB^2l^2}{2mE}\,y

y = \frac{2mE}{qB^2l^2}\,z^2

(6)

a parabola with its vertex at the point where the undeflected beam would hit.

The trace on the screen, the height y against z for ions of all speeds, and the parabola it approaches for z l: the two agree near the origin, and the parabola lies above the trace farther out. — The trace on the screen, the height $y$ against $z$ for ions of all speeds, and the parabola it approaches for $z \ll l$ : the two agree near the origin, and the parabola lies above the trace farther out.

Check

Units: $qB^2y/(mE)$ is $(\text{C}\cdot\text{T}^2\cdot\text{m})/(\text{kg}\cdot\text{V/m})$ ; with $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ and $\text{V}\cdot\text{C} = \text{kg}\cdot\text{m}^2/\text{s}^2$ this is a pure number, as the argument of a tangent must be. Endpoints: the slowest ions to arrive have $R \to l$ , $\theta \to \pi/2$ , so $z \to l$ and $y \to \pi^2 mE/(8qB^2)$ , which the trace gives, since $\tan\sqrt{\pi^2/16} = \tan(\pi/4) = 1$ . As $E \to 0$ every ion lands at $y = 0$ , spread along $z$ by speed, as for $B$ alone.

Remark

This is J. J. Thomson’s parabola method. Speed is spread along the curve, but the curve itself depends on $q/m$ only: ions of another specific charge trace another parabola, $y \propto (m/q)\,z^2$ . A photographic plate in place of the screen therefore sorts a beam by $q/m$ whatever the spread of speeds, which is how the two isotopes of neon were first separated.