Irodov 3.382: Pitch from the Speed-Blind Period – Amit & Deepali Sinha Foundation

Problem Irodov 3.382

An electron, starting from rest, is accelerated through a potential difference $V = 1.0$ kV. It then moves through a uniform magnetic field of induction $B = 29$ mT, its velocity making the angle $\alpha = 30^\circ$ with the direction of $\vect{B}$ . Its path is a helix. Find the pitch of this helix, the distance it advances along $\vect{B}$ in one turn.

The electron in the uniform field B (a royal field line), moving with velocity v at the angle α to the field direction. — The electron in the uniform field $\vect B$ (a royal field line), moving with velocity $\vect v$ at the angle $\alpha$ to the field direction.

Given

$V = 1.0$ kV
$\alpha = 30^\circ$
$B = 29$ mT

Find

the pitch

\Delta l

of the helix

Solution

Step 1The speed the voltage gives

The field of the accelerating gap does work $eV$ on the electron, which starts from rest, so its kinetic energy at entry is $eV$ . This energy is $1.0$ keV, far below the rest energy $m c^2 = 511$ keV, so the non-relativistic form applies.

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

Step 2Split the motion along and across the field

The magnetic force $-e\,\vect v\times\vect B$ is perpendicular to $\vect B$ , so it never changes the velocity component along the field. Resolving $\vect v$ along $\vect B$ and across it:

v_\parallel = v\cos\alpha, \qquad v_\perp = v\sin\alpha \tag{2}

The component $v_\parallel$ stays constant. The force is also perpendicular to $\vect v$ , so it does no work: $v_\perp$ keeps its magnitude, and the projection of the motion onto a plane normal to $\vect B$ is a uniform circle.

Step 3The period of the circling

The force $e v_\perp B$ supplies the centripetal force of that circle, which fixes the radius $r$ ; one turn takes the circumference over the speed.

\begin{aligned} \frac{m v_\perp^2}{r} &= e v_\perp B \quad\Longrightarrow\quad r = \frac{m v_\perp}{eB}, \\ T &= \frac{2\pi r}{v_\perp} \end{aligned}

T = \frac{2\pi m}{eB}

(3)

Combining the uniform drift along $\vect B$ with this circle gives the helix below.

The electron's helix about a field line B, with v at the start and its dashed components along and across the field; Δ l is the advance along B in one turn. — The electron’s helix about a field line $\vect B$ , with $\vect v$ at the start and its dashed components along and across the field; $\Delta l$ is the advance along $\vect B$ in one turn.

Step 4The advance in one turn

During one period the electron drifts along the field at the constant speed $v_\parallel$ . Using (2) for $v_\parallel$ , (3) for $T$ and (1) for $v$ :

\begin{aligned} \Delta l &= v_\parallel T = v\cos\alpha\cdot\frac{2\pi m}{eB} \\ &= \frac{2\pi m\cos\alpha}{eB}\sqrt{\frac{2eV}{m}} \\ &= 2\pi\sqrt{\frac{m^2}{e^2B^2}\cdot\frac{2eV}{m}}\,\cos\alpha \end{aligned}

Numerically, with $m = 0.911\times10^{-30}$ kg and $e = 1.60\times10^{-19}$ C,

\Delta l = 2\pi\sqrt{\frac{2(0.911\times10^{-30})(1.0\times10^{3})}{(1.60\times10^{-19})(0.029)^2}}\;(0.866) \text{ m}

\Delta l = 2\pi\sqrt{\frac{2mV}{eB^2}}\,\cos\alpha \approx 2.0 \text{ cm}

(4)

Check

Units: $mV/(eB^2)$ is $\text{kg}\cdot\text{V}/(\text{C}\cdot\text{T}^2)$ ; with $\text{V}\cdot\text{C} = \text{J}$ and $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ it becomes $\text{J}\cdot\text{s}^2/\text{kg} = \text{m}^2$ , so the root is a length. Step by step: $v = 1.87\times10^{7}$ m/s ( $v/c = 0.06$ , comfortably non-relativistic), $T = 1.23$ ns, $v_\parallel = 1.62\times10^{7}$ m/s, and $v_\parallel T = 2.0$ cm. The radius is $r = 1.84$ mm, and the ratio $\Delta l / r$ is $2\pi\cot\alpha = 10.9$ , the stretched helix of the figure.

Remark

Because $T$ does not depend on the speed, electrons leaving one point with slightly different directions all return to the same field line after one period. For small $\alpha$ the cosine is nearly 1, so they meet again at nearly the same distance $2\pi m v/(eB)$ : a longitudinal magnetic field focuses a diverging beam, and measuring the focal field and distance gives $e/m$ .