Irodov 3.394: Angular Momentum Sets the Cutoff Field

Problem Irodov 3.394

A magnetron consists of a filament of radius $a$ and a coaxial cylindrical anode of radius $b$ , placed in a uniform magnetic field parallel to the filament. An accelerating potential difference $V$ is applied between the filament and the anode. Electrons leave the filament with zero velocity. Find the values of the magnetic induction at which they reach the anode.

The magnetron end-on: the filament of radius a inside the coaxial anode of radius b, with the uniform field B along the axis, out of the page. — The magnetron end-on: the filament of radius $a$ inside the coaxial anode of radius $b$ , with the uniform field $\vect B$ along the axis, out of the page.

Given

$a$
$b$
$V$
$\vect B$ along the axis
zero initial velocity

Find

the induction

B

at which electrons reach the anode

Solution

Step 1The magnetic torque and the angular momentum

Use polar coordinates $r$ , $\varphi$ about the axis, with $\vect B$ along $+z$ . For the electron the force $-e\,\vect v\times\vect B$ has the azimuthal part $eB\,\dot r$ , so its torque about the axis changes the angular momentum $m r^2\dot\varphi$ as

\frac{d}{dt}\left(m r^2\dot\varphi\right) = r\,eB\,\dot r = \frac{eB}{2}\,\frac{d(r^2)}{dt}

Integrating from the filament, where $r = a$ and $\dot\varphi = 0$ , and dividing by $mr$ for the azimuthal speed $v_\varphi = r\dot\varphi$ :

m r^2\dot\varphi = \frac{eB}{2}\left(r^2 - a^2\right)

v_\varphi = \frac{eB\left(r^2 - a^2\right)}{2mr}

(1)

Step 2The energy available

The magnetic force does no work. With $U(r)$ the potential at radius $r$ measured from the filament, so that $U(b) = V$ , energy conservation from rest reads

\frac{m}{2}\left(\dot r^2 + v_\varphi^2\right) = eU(r) \tag{2}

Step 3The condition at the anode

The electron reaches the anode if $\dot r^2$ is still not negative at $r = b$ , that is, if the azimuthal motion there takes no more than the whole energy $eV$ . Using (1) at $r = b$ in (2):

\begin{aligned} \frac{m}{2}\left(\frac{eB\left(b^2 - a^2\right)}{2mb}\right)^2 &\leq eV, \\ \frac{eB\left(b^2 - a^2\right)}{2mb} &\leq \sqrt{\frac{2eV}{m}} \end{aligned}

Solving for $B$ :

B \leq \frac{2b}{b^2 - a^2}\sqrt{\frac{2mV}{e}}

(3)

At the equality the electron arrives with $\dot r = 0$ , grazing the anode; for stronger fields it turns back to the filament.

Step 4No earlier turning point

Between coaxial cylinders the potential grows as the logarithm of the radius. Measuring the radius by $s = \ln(r/a)$ , the potential and the azimuthal speed (1) become

\begin{aligned} U &= V\,\frac{s}{S}, \qquad S = \ln\frac{b}{a}, \\ v_\varphi &= \frac{eB}{m}\cdot\frac{r^2 - a^2}{2r} = \frac{eBa}{m}\,\sinh s \end{aligned}

At the critical field, the equality in the answer, (2) then gives for every radius

\dot r^2 = \frac{2eV}{m}\left(\frac{s}{S} - \frac{\sinh^2 s}{\sinh^2 S}\right)

Because $\sinh^2 s / s$ increases with $s$ , the bracket is positive for every $s$ below $S$ ; a weaker field only lowers the second term. So whenever the electron can reach $r = b$ at all, it gets there without turning back on the way.

At the critical field: the electron leaves the filament from rest and curls in B while the radial E (crimson) accelerates it, touching the anode tangentially. — At the critical field: the electron leaves the filament from rest and curls in $\vect B$ while the radial $\vect E$ (crimson) accelerates it, touching the anode tangentially.

Check

Units: $\sqrt{mV/e}$ is $\sqrt{\text{kg}\cdot\text{J}}/\text{C} = \text{kg}\cdot\text{m}/(\text{C}\cdot\text{s})$ , and dividing by a length gives $\text{kg}/(\text{C}\cdot\text{s}) = \text{T}$ . Narrow gap, $d = b - a$ small against $a$ : then $b^2 - a^2 \approx 2bd$ and the critical field becomes

B \approx \frac{1}{d}\sqrt{\frac{2mV}{e}}

which is the cutoff of a plane magnetron. There an electron released from rest in crossed uniform fields $E = V/d$ and $B$ moves on a cycloid of height $2mE/(eB^2)$ , and setting that height equal to $d$ gives the same field:

\frac{2mV}{eB^2d} = d \quad\Longrightarrow\quad B = \frac{1}{d}\sqrt{\frac{2mV}{e}}

Remark

The condition at the anode does not depend on how the potential varies between the electrodes, only on $V$ and the radii: any radial electric field leaves the angular momentum argument untouched. That nothing turns the electron back earlier was shown above for the vacuum field; a potential that lagged near the filament (space charge, say) would need that step redone. This sharp cutoff, the Hull criterion, is where a magnetron is run: just above it the electrons circulate near the anode instead of landing on it, and their swirling cloud is what drives the resonator’s oscillations.