Irodov 3.391: A Straight Beam Fixes the Speed – Amit & Deepali Sinha Foundation

Problem Irodov 3.391

A beam of non-relativistic charged particles crosses a region $A$ of length $a$ along the beam. In this region a uniform electric field of strength $E$ and a uniform magnetic field of induction $B$ act, both perpendicular to the beam and to each other, and the beam passes through without deviation. A screen $S$ stands across the beam at the distance $b$ beyond the exit edge of the region. When the magnetic field is switched off, the spot of the beam on the screen shifts by $\Delta x$ . Find the specific charge $q/m$ of the particles.

The region A of length a with crossed E (plate signs) and B, and the screen S a distance b beyond it. With both fields on the beam runs straight; with B off it lands Δ x higher. — The region $A$ of length $a$ with crossed $\vect E$ (plate signs) and $\vect B$ , and the screen $S$ a distance $b$ beyond it. With both fields on the beam runs straight; with $\vect B$ off it lands $\Delta x$ higher.

Given

$E$
$B$
$a$
$b$
$\Delta x$

Find

the specific charge

q/m

Solution

Step 1The straight passage fixes the speed

With both fields on, the beam is not deflected, so the electric force $qE$ and the magnetic force $qvB$ on each particle are equal and opposite:

qE = qvB \quad\Longrightarrow\quad v = \frac{E}{B} \tag{1}

Step 2Crossing the field region with

B

off

Without the magnetic field the only force is $qE$ , constant and across the beam, so the speed $v$ along the beam is unchanged and the particle spends the time $\tau$ in the region. Across the beam it starts from rest and accelerates uniformly at $qE/m$ , which gives the sideways displacement $y_1$ and the sideways velocity $v_\perp$ at the exit:

\begin{aligned} \tau &= \frac{a}{v}, \\ y_1 &= \frac{qE}{2m}\,\tau^2 = \frac{qEa^2}{2mv^2}, \\ v_\perp &= \frac{qE}{m}\,\tau = \frac{qEa}{mv} \end{aligned} \tag{2}

Step 3The straight drift to the screen

Beyond the region the particle moves in a straight line for the time $b/v$ , so its sideways velocity adds $v_\perp b/v$ to the displacement $y_1$ . Adding the two parts from (2):

\Delta x = y_1 + v_\perp\,\frac{b}{v} = \frac{qEa^2}{2mv^2} + \frac{qEab}{mv^2}

\Delta x = \frac{qEa\,(a + 2b)}{2mv^2}

(3)

With B off: a parabola between the plates over the length a, then a straight line over b to the screen, leaving the region at the angle θ to the axis. — With $\vect B$ off: a parabola between the plates over the length $a$ , then a straight line over $b$ to the screen, leaving the region at the angle $\theta$ to the axis.

Step 4The specific charge

Putting the speed (1) into (3), the electric field partly cancels:

\Delta x = \frac{qEa\,(a+2b)\,B^2}{2mE^2} = \frac{q}{m}\cdot\frac{B^2a\,(a+2b)}{2E}

Solving for $q/m$ :

\frac{q}{m} = \frac{2E\,\Delta x}{B^2a\,(a + 2b)}

(4)

The side to which the spot moves gives the sign of $q$ ; the formula gives its magnitude per unit mass.

Check

Units: with $\text{T} = \text{V}\cdot\text{s}/\text{m}^2$ , $E/B^2$ is $(\text{V}/\text{m})\cdot\text{m}^4/(\text{V}^2\text{s}^2) = \text{m}^3/(\text{V}\cdot\text{s}^2)$ , and multiplying by $\Delta x/a^2$ gives $(\text{m}/\text{s})^2/\text{V} = \text{J}/(\text{kg}\cdot\text{V}) = \text{C}/\text{kg}$ , a specific charge. Limit: for a screen right at the exit edge, $b = 0$ , (3) reduces to $qEa^2/(2mv^2)$ , the parabola’s own displacement $y_1$ from (2), as it must.

Remark

The exit direction makes $\tan\theta = v_\perp/v = qEa/(mv^2)$ with the axis, and (3) can be written $\Delta x = (a/2 + b)\tan\theta$ : the straight exit line, run backwards, crosses the axis exactly at the middle of the field region, so the deflected particles seem to come from its centre. Selecting the speed with crossed fields and then deflecting with $E$ alone is how J. J. Thomson measured the specific charge of the electron in 1897.