Irodov 3.384: A Chord Measures the Landing Spot – Amit & Deepali Sinha Foundation

Problem Irodov 3.384

An electron starts from a point $A$ on the axis of a long straight solenoid, whose uniform field has induction $B$ , with a non-relativistic velocity $v$ directed at the angle $\alpha$ to the axis. A screen perpendicular to the axis stands at the distance $l$ from $A$ . Find the distance $r$ from the axis to the point where the electron strikes the screen.

The solenoid's uniform field B along its axis, the electron leaving the point A on the axis with velocity v at the angle α to it, and the screen normal to the axis at the distance l from A. — The solenoid’s uniform field $\vect B$ along its axis, the electron leaving the point $A$ on the axis with velocity $\vect v$ at the angle $\alpha$ to it, and the screen normal to the axis at the distance $l$ from $A$ .

Given

$v$ at angle $\alpha$ to the axis, from $A$ on the axis
$B$
screen at distance $l$

Find

the distance

r

from the axis to the point of impact

Solution

Step 1Split the motion along and across the axis

The magnetic force $-e\,\vect v\times\vect B$ is perpendicular to $\vect B$ , so the velocity component along the axis stays constant; across the axis the velocity keeps its magnitude and turns round a circle. Resolving $\vect v$ :

v_\parallel = v\cos\alpha, \qquad v_\perp = v\sin\alpha \tag{1}

The force $e v_\perp B$ is the centripetal force of that circle, which fixes its radius $\rho$ and the angular velocity $\omega = v_\perp/\rho$ :

\begin{align*} \omega &= \frac{v_\perp}{\rho} = \frac{eB}{m}, \\ \frac{m v_\perp^2}{\rho} &= e v_\perp B \quad\Longrightarrow\quad \rho = \frac{m v\sin\alpha}{eB} \tag{2}\end{align*}

The circle starts on the axis, so it passes through the axis.

Step 2The angle turned before the screen

The electron reaches the screen when its constant axial motion has covered $l$ , and in that time its projection turns through the angle $\varphi$ round the circle. With $v_\parallel$ from (1):

t = \frac{l}{v\cos\alpha}

\varphi = \omega t = \frac{eBl}{m v\cos\alpha}

(3)

Step 3The distance from the axis

Seen along the axis, the projection of the electron starts at the axis point $O$ and moves round the circle of radius $\rho$ about its centre $C$ . When it strikes the screen, at $P$ , the radius $CP$ has turned by $\varphi$ from $CO$ .

The motion seen along the axis, with B toward the viewer. The electron's projection runs from the axis point O round a circle of radius ρ about C; at impact it is at P, the radius having turned by φ. The chord OP is r. — The motion seen along the axis, with $\vect B$ toward the viewer. The electron’s projection runs from the axis point $O$ round a circle of radius $\rho$ about $C$ ; at impact it is at $P$ , the radius having turned by $\varphi$ . The chord $OP$ is $r$ .

The triangle $OCP$ is isosceles with two sides $\rho$ and the apex angle $\varphi$ , so its base is

r = 2\rho\left|\sin\frac{\varphi}{2}\right|

Putting in $\rho$ from (2) and $\varphi$ from (3):

r = \frac{2 m v\sin\alpha}{eB}\left|\sin\frac{eBl}{2 m v\cos\alpha}\right|

(4)

The modulus keeps $r$ positive once the projection has gone more than halfway round, $\varphi > \pi$ .

Check

Units: $mv/(eB)$ is a length (it is $\rho/\sin\alpha$ ), and $eBl/(mv)$ is a pure number, as an angle must be. Weak field: as $B \to 0$ the sine’s argument is small and the sine may be replaced by its argument,

r \to \frac{2mv\sin\alpha}{eB}\cdot\frac{eBl}{2mv\cos\alpha} = l\tan\alpha,

the straight line at the angle $\alpha$ , as it must be with no field. And $r = 0$ whenever $\varphi$ is a whole number of turns, $2\pi n$ : the electron is back on the axis.

Remark

However long the solenoid, the electron never strays farther than $2\rho = 2mv\sin\alpha/(eB)$ from the axis, the diameter of its projected circle: a strong axial field confines a beam to a thin tube round the axis. As $l$ grows, the landing point sweeps round the circle through $O$ and back to the axis once per turn; the zeros, at $eBl/(mv\cos\alpha) = 2\pi n$ , are the focusing condition of a magnetic lens, which also serves to measure $e/m$ .