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Electrons leave a hot cathode almost at rest and are sped up through a potential difference of 600…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Electrons leave a hot cathode almost at rest and are sped up through a potential difference of $600\ \text{V}$ . They then pass through a tiny hole on the axis of a long solenoid. Beyond the hole they spread out in a narrow cone, all at small angles to the axis. A screen across the solenoid stands $25\ \text{cm}$ beyond the hole, as in the figure. The solenoid has $2000$ turns per metre, and inside it the field is uniform and along the axis. The current is set so that the electrons first come back together on the axis exactly at the screen, giving a sharp spot. Take $m_e = 9.11 \times 10^{-31}\ \text{kg}$ , $e = 1.60 \times 10^{-19}\ \text{C}$ and $\mu_0 = 4\pi \times 10^{-7}\ \text{T m/A}$ . Leave out relativity and the forces between the electrons.

Side view of a long solenoid with 2000 turns per metre, its field B along its axis. On the left, electrons leave a hot cathode and are sped up through 600 V towards a plate with a tiny hole on the axis. Just beyond the hole a few short dashed lines spread out in a narrow cone. A screen stands across the solenoid 25 cm beyond the hole.

What is the speed of the electrons?
What field is needed in the solenoid?
What current is needed in the solenoid?
One electron leaves the hole at $3.0^\circ$ to the axis. What is the greatest distance between it and the axis on its way?
That electron does not return to the axis exactly at the screen. How far in front of the screen does it cross the axis?
The current is then raised slowly. At what field is the spot sharp again for the first time?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Each electron moves on a helix that starts and ends on the axis. All the helices take the same time, $2\pi m/(eB)$ , for one turn. For small angles they all move almost the same distance along the axis in that time, so they meet again: the beam is focused.

Step 1. Speed (a). $\tfrac12 m_ev^2 = eV$ gives $v = \sqrt{\dfrac{2eV}{m_e}} = 1.45 \times 10^{7}\ \text{m/s}$ .

Step 2. Field (b). In one turn an electron moves $vT = \dfrac{2\pi m_ev}{eB}$ along the axis. Setting this to $25\ \text{cm}$ :

\displaystyle B = \frac{2\pi m_ev}{eL} = \frac{2\pi \times 9.11 \times 10^{-31}\ \text{kg} \times 1.45 \times 10^{7}\ \text{m/s}}{1.60 \times 10^{-19}\ \text{C} \times 0.25\ \text{m}} = 2.077\ \text{mT}

Step 3. Current (c). $B = \mu_0 nI$ , so $I = \dfrac{B}{\mu_0 n} = 0.827\ \text{A}$ .

Step 4. Greatest distance (d). The part across the field is $v\sin3.0^\circ$ , so the circle has radius $r = \dfrac{m_ev\sin\alpha}{eB} = 2.082\ \text{mm}$ . The circle starts on the axis, so the electron gets as far as one diameter away: $2r = 4.16\ \text{mm}$ .

Step 5. The crossing point (e). In one turn this electron moves $v\cos\alpha\,T = L\cos\alpha$ along the axis. It crosses the axis $L(1 - \cos\alpha) = 0.343\ \text{mm}$ in front of the screen.

Step 6. The next sharp spot (f). As $B$ grows, the turns get shorter. The spot is sharp again when two whole turns fit into $25\ \text{cm}$ : twice the field, $4.155\ \text{mT}$ .

Answer: (a) $1.45 \times 10^{7}\ \text{m/s}$ . (b) $2.077\ \text{mT}$ . (c) $0.827\ \text{A}$ . (d) $4.16\ \text{mm}$ . (e) $0.343\ \text{mm}$ . (f) $4.155\ \text{mT}$ .

Watch out. The radius $2.082\ \text{mm}$ is measured from the centre of the circle, which is off the axis; the greatest distance from the axis is twice that. The tiny error in (e) is why the cone must be narrow: it grows as $\alpha^2$ .

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