Irodov 3.387: The Magnetic Clock Times a Fall – Amit & Deepali Sinha Foundation

Problem Irodov 3.387

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 particle of specific charge $q/m$ starts from the origin $O$ with a non-relativistic velocity $v_0$ along the positive $x$ axis. Find:

the coordinate $y_n$ at which the particle crosses the $y$ axis for the $n$ th time;
the angle $\alpha$ between the particle’s velocity and the $y$ axis at that crossing.

The frame with z toward the viewer. Both uniform fields, E and B, point along +y; the particle leaves 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 particle leaves 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 $O$

Find

(a)

y_n

at the

n

th crossing of the

y

axis · (b) the angle

\alpha

there

Solution

Step 1The motion along the fields

The magnetic force $q\,\vect v\times\vect B$ is perpendicular to $\vect B$ , so it has no $y$ component. Along $y$ the only force is $qE$ , constant, and the particle starts with $v_y = 0$ at $y = 0$ :

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

Step 2The motion across the fields

With $\vect B = B\,\vect j$ the force $q\,\vect v\times\vect B$ involves only $v_x$ and $v_z$ and is perpendicular to them, so the transverse speed stays $v_0$ . The projection of the motion onto the $xz$ plane is therefore a uniform circle of radius $r$ , on which the force $q v_0 B$ is the centripetal force:

\begin{aligned} \frac{m v_0^2}{r} &= q v_0 B \quad\Longrightarrow\quad r = \frac{m v_0}{qB}, \\ T &= \frac{2\pi r}{v_0} \end{aligned}

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

(2)

At the start the force is $q v_0 B\,(\vect i\times\vect j) = q v_0 B\,\vect k$ , so for $q > 0$ the circle’s centre lies at $z = r$ on the $z$ axis. The circle passes through the origin of the $xz$ plane, and the period does not depend on the speed.

Step 3(a) Where the particle crosses the

y

axis

The particle is on the $y$ axis when $x = z = 0$ , which is when its projection is back at the start of the circle. That happens after whole periods, at the times $t_n = nT$ . Putting $t_n$ into (1) with $T$ from (2):

y_n = \frac{qE}{2m}\,(nT)^2 = \frac{qE}{2m}\cdot\frac{4\pi^2 n^2 m^2}{q^2B^2}

y_n = \frac{2\pi^2 m E\, n^2}{qB^2}

(3)

The particle winds round a helix about the field line $x = 0$ , $z = r$ , its pitch growing turn by turn, and touches the $y$ axis once per turn.

The trajectory: a helix about the field line through the circle's centre, with B along it. The pitch grows each turn, and the particle meets the y axis at the dots y_1 and y_2. — The trajectory: a helix about the field line through the circle’s centre, with $\vect B$ along it. The pitch grows each turn, and the particle meets the $y$ axis at the dots $y_1$ and $y_2$ .

Step 4(b) The direction of the velocity at a crossing

After a whole number of turns the transverse velocity is again $v_0$ along $+x$ , while from (1) the longitudinal velocity has grown to $v_y$ at $t_n = nT$ . The velocity therefore lies in the $xy$ plane, with the components shown below.

The velocity v at the nth crossing, resolved along x (the transverse speed v_0) and along y; α is its angle with the y axis. — The velocity $\vect v$ at the $n$ th crossing, resolved along $x$ (the transverse speed $v_0$ ) and along $y$ ; $\alpha$ is its angle with the $y$ axis.

Evaluating $v_y$ at $t_n$ and taking the ratio of the components:

\begin{aligned} v_y &= \frac{qE}{m}\cdot\frac{2\pi n m}{qB} = \frac{2\pi n E}{B}, \\ \tan\alpha &= \frac{v_0}{v_y} \end{aligned}

\tan\alpha = \frac{v_0 B}{2\pi n E}

(4)

Check

Units: $E/B$ is a speed (from $qE = q v B$ for balanced forces), so $v_0B/E$ is a pure number, as a tangent must be; and $mE/(qB^2)$ is $(\text{kg}\cdot\text{V/m})/(\text{C}\cdot\text{T}^2)$ , which with $\text{V}\cdot\text{C} = \text{J}$ and $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ is $\text{J}\cdot\text{s}^2/(\text{kg}\cdot\text{m})$ , a length. Limits: as $E \to 0$ every $y_n$ goes to $0$ and $\tan\alpha$ grows without bound, so $\alpha \to 90^\circ$ : the particle just circles in the $xz$ plane through $O$ , as it must with $B$ alone. For a negative charge both $qE$ and $y_n$ change sign, so the crossings lie below $O$ .

Remark

The crossings are equally spaced in time, so the $y_n$ grow as $n^2$ and the gaps between successive crossings go as $1, 3, 5, \ldots$ , Galileo’s odd-number rule for a uniformly accelerated fall. The magnetic field acts as a clock that strobes the electric acceleration; since $T$ does not depend on $v_0$ , measuring $y_1$ in known fields gives $q/m$ directly, $2\pi^2 E/(y_1 B^2)$ , with no need to know the launch speed.