Irodov 3.393: The Filament Field Turns Electrons Back

Problem Irodov 3.393

A long cylindrical anode of radius $a$ surrounds a coaxial cylindrical cathode of radius $b$ , with $b < a$ . A filament along the common axis carries a heating current $I$ , which sets up a magnetic field in the space around it. Electrons leave the cathode with zero initial velocity. Find the least potential difference between the cathode and the anode at which these electrons begin to reach the anode.

A section through the axis: the filament carrying the current I, the cathode wall at radius b and the anode wall at radius a, both cylinders coaxial with the filament. — A section through the axis: the filament carrying the current $I$ , the cathode wall at radius $b$ and the anode wall at radius $a$ , both cylinders coaxial with the filament.

Given

$a$
$b < a$
$I$
zero initial velocity

Find

the least potential difference

V

at which electrons reach the anode

Solution

Step 1The filament’s magnetic field

Take the $z$ axis along the current. By symmetry $\vect B$ circles the axis and has the same magnitude all round a circle of radius $r$ , so Amp`ere’s law on that circle gives

B\cdot 2\pi r = \mu_0 I \quad\Longrightarrow\quad B = \frac{\mu_0 I}{2\pi r} \tag{1}

Step 2The magnetic force builds up axial momentum

Write the electron’s velocity as $v_r$ outward plus $v_z$ along the axis. With $\vect B$ along the circles, the force $-e\,\vect v\times\vect B$ has an axial part set by $v_r$ alone:

F_z = -e\,v_r B = -\frac{e\mu_0 I}{2\pi r}\,\frac{dr}{dt}

An electron moving outward with v_r: the filament's B there points out of the page, and the magnetic force F on the electron points against the current. — An electron moving outward with $\vect v_r$ : the filament’s $\vect B$ there points out of the page, and the magnetic force $\vect F$ on the electron points against the current.

Newton’s law along $z$ , $m\,dv_z/dt = F_z$ , integrates directly in $r$ , from the cathode where $v_z = 0$ at $r = b$ :

m v_z = -\frac{e\mu_0 I}{2\pi}\int_b^r \frac{dr'}{r'}

m v_z = -\frac{e\mu_0 I}{2\pi}\,\ln\frac{r}{b}

(2)

The electron is turned to run against the current, more strongly the further out it gets.

Step 3Energy decides whether the anode is reached

The magnetic force does no work, so all the kinetic energy comes from the electric field. Let $U(r)$ be the potential at radius $r$ measured from the cathode; between coaxial cylinders it grows as $\ln(r/b)$ , reaching $V$ at the anode. Energy conservation from rest at the cathode reads

\frac{m}{2}\left(v_r^2 + v_z^2\right) = eU(r), \qquad U(r) = V\,\frac{\ln(r/b)}{\ln(a/b)} \tag{3}

At the anode $U = V$ . The electron arrives there only if $v_r^2 \geq 0$ at $r = a$ , that is, if the axial motion of (2) has not used up more than the available energy:

eV \geq \frac{m}{2}\,v_z^2(a) = \frac{m}{2}\left(\frac{e\mu_0 I}{2\pi m}\right)^2 \ln^2\frac{a}{b}

Step 4No turning point before the anode

It remains to see that $v_r^2$ stays positive at every radius in between. With the shorthand

u = \ln\frac{r}{b}, \qquad L = \ln\frac{a}{b}, \qquad k = \frac{e\mu_0 I}{2\pi m}

(3) with (2) gives

v_r^2 = \frac{2eV}{mL}\,u - k^2u^2 = k^2u\,\left(u_\ast - u\right), \qquad u_\ast = \frac{2eV}{mk^2L}

This vanishes only at the cathode and at $u_\ast$ , and is positive between them. The condition of the previous step says precisely that $u_\ast$ lies at or beyond $L$ , so when it holds the electron moves outward all the way to the anode, and when it fails it turns back short of it. The least voltage puts the turning point exactly on the anode:

V = \frac{m}{2e}\left(\frac{e\mu_0 I}{2\pi m}\right)^2 \ln^2\frac{a}{b}

V_{\mathrm{min}} = \frac{e\mu_0^2 I^2}{8\pi^2 m}\,\ln^2\frac{a}{b}

(4)

Check

Units: $\mu_0 I$ is $\text{T}\cdot\text{m}$ , so $e\mu_0^2I^2/m$ is $\text{C}\cdot\text{T}^2\text{m}^2/\text{kg}$ ; with $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ this is $\text{kg}\cdot\text{m}^2/(\text{C}\cdot\text{s}^2) = \text{J}/\text{C} = \text{V}$ . Limits: as $I \to 0$ the threshold goes to zero, since without a magnetic field any accelerating voltage brings the electrons over; and as $a \to b$ it goes to zero too, since the electrons have no room to be turned. Narrow gap: for $d = a - b$ small against $b$ the logarithm is close to $d/b$ , and with $B_0$ the filament’s field at the cathode the threshold becomes

V_{\mathrm{min}} \approx \frac{e\mu_0^2I^2}{8\pi^2 m}\cdot\frac{d^2}{b^2} = \frac{eB_0^2d^2}{2m}, \qquad B_0 = \frac{\mu_0 I}{2\pi b}

the cutoff of a plane diode in a uniform field $B_0$ parallel to its plates, where the sideways speed $eB_0x/m$ reaches the full speed $\sqrt{2eV/m}$ exactly at $x = d$ . A first power of the logarithm would leave $V$ proportional to $d$ there and fail this test. The threshold grows as $I^2$ , because $v_z$ at the anode is proportional to $I$ and the energy it takes to the square of $v_z$ .

Remark

Result (2) is the conservation of the axial canonical momentum $m v_z - eA_z$ in a field whose vector potential, $A_z = -(\mu_0 I/2\pi)\ln r$ , depends on $r$ only. The anode current therefore switches on sharply at $V_{\mathrm{min}}$ : below it every electron turns back, above it every one arrives, so a large enough heating current can cut off the anode current altogether, the same magnetic cutoff that a magnetron uses with an external field.