Electrons start from rest and are accelerated through a potential difference of 1200 V…

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Electrons start from rest and are accelerated through a potential difference of $1200\ \text{V}$ . They then enter a uniform magnetic field of $B = 4.0\ \text{mT}$ at an angle of $60^\circ$ to the field, as in the figure, and move along a helix. Take the charge of an electron as $1.6 \times 10^{-19}\ \text{C}$ and its mass as $9.1 \times 10^{-31}\ \text{kg}$ . Ignore relativistic effects and gravity. A helix is the path of a charged particle whose velocity has a part along the field and a part across it. The field points to the right in the figure. The pitch of the helix is the distance that the electron advances along the field in one turn.

An electron gun, drawn as a small box at the lower left, fires electrons that have been accelerated from rest through 1.2 kV. A bold arrow shows their velocity leaving the gun upwards and to the right. Parallel horizontal arrows pointing right fill the region beyond the gun and show a uniform magnetic field B of 4.0 mT. A dashed horizontal line through the point of entry and an arc mark the angle of 60 degrees between the velocity and the field. No path inside the field is drawn.

What is the speed with which the electrons enter the field?
What is the radius of the helix?
How long does an electron take to complete one turn?
What is the pitch of the helix?
What is the magnitude of the magnetic force on an electron just after it enters the field?

Split the velocity into a part along the field and a part across it before you start.

The worked solution

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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.

Given. $V = 1200\ \text{V}$ , $B = 4.0 \times 10^{-3}\ \text{T}$ , $\theta = 60^\circ$ , $e = 1.6 \times 10^{-19}\ \text{C}$ , $m = 9.1 \times 10^{-31}\ \text{kg}$ .

(a) Speed. $\tfrac{1}{2} m v^2 = e V$ , so $v = \sqrt{\dfrac{2 e V}{m}} = 20.54\ \text{Mm/s}$ . This is $20.54 \times 10^{6}\ \text{m/s}$ . The two parts are $v_\parallel = v\cos 60^\circ = 10.27\ \text{Mm/s}$ along the field and $v_\perp = v\sin 60^\circ = 17.79\ \text{Mm/s}$ across it.

(b) Radius. The force on the electron from $v_\perp$ makes it go round in a circle: $r = \dfrac{m v_\perp}{e B} = 2.53\ \text{cm}$ .

(c) Time for one turn. $T = \dfrac{2\pi m}{e B} = 8.93\ \text{ns}$ .

(d) Pitch. The field does not change $v_\parallel$ , so in one turn the electron advances $v_\parallel T = 10.27 \times 10^{6}\ \text{m/s} \times 8.93 \times 10^{-9}\ \text{s} = 9.18\ \text{cm}$ .

(e) Force. Only $v_\perp$ gives a force: $F = e v_\perp B = 1.6 \times 10^{-19}\ \text{C} \times 17.79 \times 10^{6}\ \text{m/s} \times 4.0 \times 10^{-3}\ \text{T} = 1.14 \times 10^{-14}\ \text{N}$ .

Answer: (a) $20.54\ \text{Mm/s}$ . (b) $2.53\ \text{cm}$ . (c) $8.93\ \text{ns}$ . (d) $9.18\ \text{cm}$ . (e) $1.14 \times 10^{-14}\ \text{N}$ .

Check. The force must equal the centripetal force of the circular part of the motion: $\dfrac{m v_\perp^2}{r} = 1.14 \times 10^{-14}\ \text{N}$ , the same as in (e). Also $v_\perp T = 2\pi r$ gives $15.89\ \text{cm}$ for the circumference.

Watch out. If you use the whole speed $v$ in the radius, you get $2.92\ \text{cm}$ , too large by a factor of $\dfrac{1}{\sin 60^\circ}$ . The part of the velocity along the field gives no force and does not bend the path.

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