Irodov 3.383: Two Focusing Fields Weigh the Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.383

Charged particles, accelerated from rest through a potential difference $V$ and moving at non-relativistic speeds, leave a point $A$ on the axis of a long straight solenoid as a slightly divergent beam travelling along the axis. As the solenoid’s magnetic induction is raised, the beam comes to a focus at the distance $l$ from $A$ for two successive values of the induction, $B_1$ and $B_2$ . Find the specific charge $q/m$ of the particles.

The solenoid's uniform field B along its axis, and the slightly divergent beam leaving the point A on the axis with speed v (divergence exaggerated). The beam is focused on the axis at the distance l from A. — The solenoid’s uniform field $\vect B$ along its axis, and the slightly divergent beam leaving the point $A$ on the axis with speed $v$ (divergence exaggerated). The beam is focused on the axis at the distance $l$ from $A$ .

Given

$V$
focus at distance $l$ from $A$
successive inductions $B_1$ , $B_2$

Find

the specific charge

q/m

Solution

Step 1The speed of the particles

The accelerating field does work $qV$ on each particle, which starts from rest, so all of them reach the same speed $v$ :

\frac{m v^2}{2} = qV \quad\Longrightarrow\quad v = \sqrt{\frac{2qV}{m}} \tag{1}

Step 2Every particle returns to the axis after one period

A particle leaving $A$ at a small angle $\alpha$ to the axis has $v_\perp = v\sin\alpha$ across $\vect B$ and $v_\parallel = v\cos\alpha$ along it. The force $q\,\vect v\times\vect B$ turns $v_\perp$ round a circle of radius $R = m v_\perp/(qB)$ that passes through the axis, and leaves $v_\parallel$ alone. One turn takes

T = \frac{2\pi R}{v_\perp} = \frac{2\pi m}{qB}

After each whole turn the projection of the particle is back at its start, so the particle is back on the axis. Over one period it has advanced along the axis by $v_\parallel T$ , and for a slightly divergent beam $\cos\alpha \approx 1$ :

h = v T = \frac{2\pi m v}{qB} \tag{2}

In between, the particle’s distance from the axis, the chord of its circle from the start, rises and falls back to zero once per step. The step is the same for every particle of the beam: the whole beam converges on the axis at the distances $h$ , $2h$ , $3h$ , and so on from $A$ .

Step 3Two successive focusing fields

A focus falls at the distance $l$ when $l$ holds a whole number of steps $h$ . As $B$ rises, $h$ shrinks as $1/B$ , so the successive fields $B_1 < B_2$ fit $n$ and $n + 1$ steps into $l$ . The figure shows the distance of a particle from the axis along its path for $n = 1$ .

The distance r_⊥ of a particle from the axis against the distance z along it, for the two fields, drawn for n = 1: at B_1 one turn spans l, at B_2 two turns do, each half as wide. — The distance $r_\perp$ of a particle from the axis against the distance $z$ along it, for the two fields, drawn for $n = 1$ : at $B_1$ one turn spans $l$ , at $B_2$ two turns do, each half as wide.

Writing both conditions with (2):

\begin{aligned} \frac{l}{h_1} &= \frac{lqB_1}{2\pi m v} = n, \\ \frac{l}{h_2} &= \frac{lqB_2}{2\pi m v} = n + 1 \end{aligned}

Subtracting the first line from the second removes the unknown $n$ :

\frac{lq\,(B_2 - B_1)}{2\pi m v} = 1

(3)

Step 4The specific charge

Putting $v$ from (1) into (3) and squaring both sides:

\begin{aligned} lq\,(B_2 - B_1) &= 2\pi m\sqrt{\frac{2qV}{m}}, \\ l^2q^2\,(B_2 - B_1)^2 &= 4\pi^2m^2\cdot\frac{2qV}{m} = 8\pi^2 m q V \end{aligned}

Dividing by $m q\, l^2 (B_2 - B_1)^2$ :

\frac{q}{m} = \frac{8\pi^2 V}{l^2\,(B_2 - B_1)^2}

(4)

Check

Units: $V/(l^2B^2)$ is $\text{V}/(\text{m}^2\,\text{T}^2)$ , and with $\text{T} = \text{V}\cdot\text{s}/\text{m}^2$ this is $\text{m}^2/(\text{V}\cdot\text{s}^2)$ , which is $\text{C}/\text{kg}$ since $\text{V}\cdot\text{C} = \text{kg}\cdot\text{m}^2/\text{s}^2$ . Special case $n = 1$ : then $B_2 = 2B_1$ and one step spans $l$ , so $B_2 - B_1 = B_1$ and the result becomes $l^2 q B_1^2 = 8\pi^2 m V$ , which is $l = 2\pi m v/(qB_1)$ with $v = \sqrt{2qV/m}$ : exactly one turn over $l$ , as it must be.

Remark

One focusing field alone would leave the whole number $n$ unknown, which is why the method uses two successive fields: their difference is the field that adds exactly one turn over $l$ . This is the longitudinal-field method for $e/m$ ; the same speed-blind period lets a solenoid act as a magnetic lens that images $A$ onto the axis, the principle of the first electron microscopes.