Irodov 3.390: Only the Cross-Field Speed Is Fixed – Amit & Deepali Sinha Foundation

Problem Irodov 3.390

In a right-handed $xyz$ frame a uniform electric field of strength $E = 4.0$ kV/m points along the positive $y$ axis and a uniform magnetic field of induction $B = 50$ mT along the positive $z$ axis. Non-relativistic protons move along a straight line through these fields; their path lies in the $xz$ plane, between the positive $x$ and $z$ axes, at the angle $\varphi = 30^\circ$ to the $x$ axis. The electric field is then switched off and the protons move along a helix. Find the pitch of this helix.

The frame with z toward the viewer: E along +y, B along +z, and the protons' velocity v in the xz plane at the angle φ to the x axis, with its parallelogram on the axes. — The frame with $z$ toward the viewer: $\vect E$ along $+y$ , $\vect B$ along $+z$ , and the protons’ velocity $\vect v$ in the $xz$ plane at the angle $\varphi$ to the $x$ axis, with its parallelogram on the axes.

Given

$E = 4.0$ kV/m along $y$
$B = 50$ mT along $z$
straight path in the $xz$ plane at $\varphi = 30^\circ$ to $x$

Find

the pitch

\Delta l

of the helix once

E

is off

Solution

Step 1Straight motion needs zero net force

Resolving the velocity along the axes, $\vect v = v\cos\varphi\,\vect i + v\sin\varphi\,\vect k$ , the magnetic force picks out the component across $\vect B$ :

q\,\vect v\times\vect B = q B\,(v\cos\varphi\,\vect i\times\vect k + v\sin\varphi\,\vect k\times\vect k) = -q v B\cos\varphi\,\vect j

Adding $qE\,\vect j$ and setting the sum to zero:

qE - q v B\cos\varphi = 0

v\cos\varphi = \frac{E}{B}

(1)

Only the velocity component across $\vect B$ is fixed; the component along $\vect B$ , $v\sin\varphi$ , feels no force at all.

Step 2The helix once the electric field is off

With $\vect B$ alone, the component along $\vect B$ stays constant, and the component across it turns round a circle in the $xy$ plane. With $v\cos\varphi$ from (1):

v_\parallel = v\sin\varphi = \frac{E}{B}\tan\varphi, \qquad v_\perp = v\cos\varphi = \frac{E}{B} \tag{2}

The circling has the period that does not depend on the speed, since $e v_\perp B$ is its centripetal force:

\frac{m v_\perp^2}{R} = e v_\perp B \quad\Longrightarrow\quad T = \frac{2\pi R}{v_\perp} = \frac{2\pi m}{eB}

The proton's helix about the field direction B once E is off, with v at the start and its dashed components along and across B; Δ l is the pitch. — The proton’s helix about the field direction $\vect B$ once $\vect E$ is off, with $\vect v$ at the start and its dashed components along and across $\vect B$ ; $\Delta l$ is the pitch.

Step 3The pitch

In one period the proton advances along $\vect B$ at the constant speed $v_\parallel$ from (2):

\Delta l = v_\parallel T = \frac{E}{B}\tan\varphi\cdot\frac{2\pi m}{eB}

Numerically, with $m = 1.672\times10^{-27}$ kg and $e = 1.60\times10^{-19}$ C,

\Delta l = \frac{2\pi\,(1.672\times10^{-27})(4.0\times10^{3})(0.577)}{(1.60\times10^{-19})(0.050)^2} \text{ m}

\Delta l = \frac{2\pi m E}{eB^2}\tan\varphi \approx 6.1 \text{ cm}

(3)

Check

Units: $mE/(eB^2)$ is $\text{kg}\cdot(\text{V/m})/(\text{C}\cdot\text{T}^2)$ ; with $E/B$ a speed and $m/(eB)$ a time ( $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ ) it is a length. Limit: for $\varphi = 0$ the protons move across $\vect B$ only, and the pitch is zero, a closed circle, as the formula gives. Size: the circle’s radius is $R = m v_\perp/(eB) = mE/(eB^2) = 1.67$ cm, and $\Delta l/R = 2\pi\tan\varphi = 3.6$ , matching $6.1/1.67$ .

Remark

The straight-line condition is the $\vect E\times\vect B$ drift: across $\vect B$ the proton must move at exactly $E/B$ along $\vect E\times\vect B$ , here the $+x$ axis, while along $\vect B$ any speed is allowed. Switching off $E$ turns that drift into gyration of the same speed, and the free motion along $\vect B$ becomes the helix’s pitch; so the pitch measures the hidden parallel speed, $\tan\varphi$ times the drift speed.