A Velocity Selector A group of physicists at Argonne National Laboratory in
Illinois wants to bombard metals with monoenergetic beams of alpha particles
to study radiation damage. (Alpha particles are helium nuclei, which consist
of two neutrons and two protons and thus have a net charge of \(+2 e\) where \(e\)
is the amount of the charge on the electron.) They have managed to create a
beam of alpha particles from the decay of radioactive elements, but some of
the alpha particles lose energy as they collide with other atoms in the
source. As a new physicist assigned to the group you have been asked to use a
velocity selector to select only the alpha particles in the beam that are
close to one velocity and get rid of the others. The velocity selector
consists of: (1) a power supply capable of de-
livering large potential differences between capacitor plates and (2) a large
permanent magnet that has a uniform magnetic field perpendicular to the beam.
The setup for the velocity selector is shown in Fig. 29-38. The direction of
the \(\mathrm{B}\) -field is out of the paper. Your magnet has a field of \(0.22
\mathrm{~T}\) and the capacitor plates have a spacing of \(2.5 \mathrm{~cm}\).
You are asked to figure out how the velocity selector works and then tell your
group what voltage to put across the capacitor plates to select a velocity of
\(4.2 \times 10^{6} \mathrm{~m} / \mathrm{s} .\)
This is your first job and you feel overwhelmed by the assignment, but you
calm down and begin to analyze the situation one step at a time. You come up
with the following:
(a) The magnet is oriented so its magnetic field is out of the paper in the
diagram you are given, so you use the right-hand rule to determine the
direction of the magnetic force on an alpha particle passing from left to
right into the magnetic field. What direction did you come up with for the
force?
(b) You realize that by using \(\vec{F}^{\prime} \mathrm{mag}=q \vec{v} \times
\vec{B}\), you can calculate the magnitude of force on an alpha particle moving
at speed \(v\) just as it enters the uniform magnetic field as a function of the
charge on the alpha particle and the magnitude of the magnetic field \(B\). What
is the expression for the magnitude of the force in terms of \(e\), \(v\), and \(B
?\)
(c) You realize that you might be able to put just the right voltage across
the two capacitor plates so that the electrical force on a given alpha
particle will be equal in magnitude and opposite in direction to the Lorentz
magnetic force. Then any alpha particles with just the right velocity will
pass straight through the poles of the magnet without being deflected. First
you think about whether the voltage on the upper capacitor plate should be
positive or negative to give a canceling force. What do you decide?
(d) Next you realize that if you know the electric field between the plates
and the charge on the alpha particle then you can compute the electrical force
on it. What is the relationship between the electrical force \(\vec{F}^{\text
{elec }}\), charge, \(q\), and electric field \(\vec{E} ?\)
(e) Finally, you use the fact that the magnitude of the electric field between
capacitor plates is given by \(E=|\Delta V| / d\) where \(d\) is the spacing
between the plates. Show that the voltage needed to have the electrical force
and the magnetic force be "equal and opposite" can be calculated using the
equation \(|\Delta V|=v B d\). Calculate the voltage needed.
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