Irodov 3.386: Energy Filter Then Momentum Filter – Amit & Deepali Sinha Foundation

Problem Irodov 3.386

A non-relativistic charged particle passes through the electric field between the electrodes of a cylindrical capacitor, of radii $a$ and $b$ with $a < b$ and potential difference $V$ , moving there along an arc of a circle concentric with the electrodes. It then enters a region of uniform magnetic field of induction $B$ , perpendicular to its velocity, where it moves along a semicircle of radius $r$ . Find the speed of the particle and its specific charge $q/m$ .

The charge q enters the gap between the cylindrical electrodes, follows a circular arc through it, and enters the region of uniform field B, where it describes a semicircle of radius r about the centre marked on the region's edge. — The charge $q$ enters the gap between the cylindrical electrodes, follows a circular arc through it, and enters the region of uniform field $\vect B$ , where it describes a semicircle of radius $r$ about the centre marked on the region’s edge.

Given

electrode radii $a < b$
potential difference $V$
field $B$
semicircle radius $r$

Find

the speed

v

· the specific charge

q/m

Solution

Step 1The field inside the cylindrical capacitor

By symmetry the field between the electrodes is radial and the same all round; with charge $\lambda$ per unit length on the inner electrode, Gauss’s law over a coaxial cylinder of radius $\rho$ and length $\ell$ gives

E\cdot 2\pi\rho\,\ell = \frac{\lambda\ell}{\varepsilon_0} \quad\Longrightarrow\quad E = \frac{\lambda}{2\pi\varepsilon_0\,\rho}

The potential difference fixes $\lambda$ . Integrating $E$ across the gap:

V = \int_a^b E\,d\rho = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{b}{a}

Eliminating $\lambda$ between the two lines:

E(\rho) = \frac{V}{\rho\,\ln(b/a)} \tag{1}

Step 2The circular arc in the capacitor

The forces read off the path: in the capacitor q E points to the common axis of the electrodes, in the magnetic region q v· B points to the semicircle's centre. Drawn for q > 0, which needs B into the page. — The forces read off the path: in the capacitor $q\vect E$ points to the common axis of the electrodes, in the magnetic region $q\vect v\times\vect B$ points to the semicircle’s centre. Drawn for $q > 0$ , which needs $\vect B$ into the page.

On its arc, of some radius $\rho_0$ between $a$ and $b$ , the particle has constant speed (the field is perpendicular to the path) and the electric force $qE$ toward the axis is the centripetal force. With $E$ from (1):

\frac{m v^2}{\rho_0} = qE(\rho_0) = \frac{qV}{\rho_0\ln(b/a)}

m v^2 = \frac{qV}{\ln(b/a)}

(2)

Step 3The semicircle in the magnetic field

In the field $B$ the magnetic force $qvB$ , perpendicular to the velocity, is the centripetal force of the semicircle of radius $r$ :

\frac{m v^2}{r} = q v B

m v = qBr

(3)

Step 4The speed and the specific charge

Dividing (2) by (3) removes $m$ and $q$ together:

v = \frac{m v^2}{m v} = \frac{qV/\ln(b/a)}{qBr}

v = \frac{V}{rB\ln(b/a)}

(4)

Then (3) gives the specific charge from this speed:

\frac{q}{m} = \frac{v}{Br} = \frac{1}{Br}\cdot\frac{V}{rB\ln(b/a)}

\frac{q}{m} = \frac{V}{r^2B^2\ln(b/a)}

(5)

Check

Units: $V/(rB)$ is $\text{V}/(\text{m}\cdot\text{T})$ , and since $E = vB$ for balanced electric and magnetic forces, $\text{V/m} = (\text{m/s})\cdot\text{T}$ , so $V/(rB)$ is a speed. For $q/m$ , $V/(r^2B^2)$ is that speed over $rB$ , which is $(\text{m/s})/(\text{m}\cdot\text{T}) = \text{C/kg}$ with $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ . Thin gap: as $b \to a$ , $\ln(b/a) \approx (b - a)/a$ and (1) becomes $V/(b - a)$ , the uniform field of a parallel-plate gap, and (2) becomes $mv^2/a = qV/(b - a)$ , the centripetal balance on a circle of radius $a$ in that field.

Remark

The two devices filter different quantities. The cylindrical capacitor passes particles of one value of $mv^2/q$ , twice the kinetic energy per charge, whatever arc they follow in the gap; the magnet sorts by $mv/q$ , the momentum per charge. Either alone confuses a fast light ion with a slow heavy one; together they fix $v$ and $q/m$ separately, the principle of a double-focusing mass spectrograph.