### Video Transcript

In this lesson, we’re going to
learn how some properties of charged particles determine their expected motion in
the presence of magnetic fields. We’ll then work backwards, starting
with an observation of a particle’s motion, and determine from that properties of
the particle like charge, mass, and charge-to-mass ratio.

Let’s start by considering the
force acting on a particle as it moves in a magnetic field. Here we have a charged particle and
its direction of motion is perfectly horizontal towards the right. We’ll call the charge of the
particle 𝑄 and its speed 𝑣. We’ve represented a magnetic field
in this diagram by an array of circles with x’s in them. The x’s represent that the magnetic
field is pointing away from us. In other words, we’re looking in
the same direction that the magnetic field is pointing. We’ll use the letter 𝐵 to
represent the strength of this magnetic field.

Since the direction of the
particle’s motion and the direction of the magnetic field are perpendicular, the
force on this object when it’s inside the magnetic field is the charge of the object
times its speed times the strength of the magnetic field. The direction of the force is
perpendicular to both the direction of motion and the direction of the magnetic
field, so either directly up or directly down. The particular direction depends on
the sign of the charge of the particle. If the particle has a positive
charge, then in this configuration, the force will be up. And if the particle has a negative
charge, in this configuration, the force will be down. Regardless of whether the charge is
positive or negative, the force is always perpendicular to the direction of
motion.

There is another situation where an
object moves subject to a force that is perpendicular to its direction of
motion. And that’s an object moving in a
circle. For an object moving in a circle,
the force is perpendicular to the motion, and in this case, the force has a special
name. It’s called the centripetal
force. The reason there needs to be a
force acting on an object moving in a circle at all is because the object is
changing direction, and direction changes require a force. For a charged particle in a
magnetic field, the force is perpendicular to the direction of motion, just like for
an object moving in a circle. Therefore, a charged particle in a
uniform magnetic field will also move in a circle with the force on the particle
always pointing towards the center of the circle.

If the particle were negatively
charged, the trajectory would curve downward in keeping with the direction of the
force. Our goal is to relate this circular
motion to the force that the charge experiences because of the magnetic field. To do this, recall that the
centripetal force on an object moving in a circle is the mass of the object times
the square of the object’s speed divided by the radius of the circular path. Since the force on our charged
particle due to the magnetic field is in fact the centripetal force that our
particle is experiencing, we can equate these two expressions. This gives us that the mass of our
charged particle times the square of its speed divided by the radius of its path as
it moves through the magnetic field is equal to its charge times its speed times the
strength of the magnetic field.

In this context, we refer to the
radius of the path as the cyclotron radius because the circular paths made by
charged particles moving in magnetic fields are known as cyclotron motion. Let’s carefully examine the
quantities that we have in this equation. We have the strength of the
magnetic field, which describes the environment of the particle but not the particle
itself. We have the speed of the particle
and the radius of its path, which describe its motion. And we have the mass and the
charge, which are properties of the particle itself. Remember, our goal was to relate
particle properties to a description of its motion. So let’s collect all of the
quantities that describe the motion on one side of the equation and all of the
quantities that are properties of the particle on the other side.

We’ll accomplish this by dividing
both sides by the mass and speed of the particle and also the strength of the
magnetic field. On the right-hand side, 𝑣𝐵
divided by 𝑣𝐵 is just one. And on the left-hand side, 𝑚
divided by 𝑚 is one and one factor of 𝑣 in the denominator cancels one factor of
𝑣 in the numerator. This gives us exactly the sort of
expression we’re looking for. We have information about the
particle’s motion on one side of the equation and information about the particle
itself on the other side.

Before doing anything further with
this equation, let’s observe that the right-hand side 𝑄 divided by 𝑚 is the
charge-to-mass ratio of the particle. The fact that the only appearances
of charge and mass in this equation are as part of the charge-to-mass ratio tells us
several things. Firstly, it tells us that two
particles may have different charges and masses but undergo the same motion in a
magnetic field because the ratio of their charge to their mass is the same. Furthermore, this equation tells us
that we need some kind of independent measurement to establish the particular charge
or particular mass of a particle since knowledge of just its trajectory — that is,
its velocity, the radius, and the strength of the magnetic field — is only enough to
specify the charge-to-mass ratio.

Finally, recall that negatively
charged particles will move in the opposite direction to positively charged
particles. Therefore, since particles and
antiparticles have the same magnitude for their charge-to-mass ratio but opposite
signs, they will follow trajectories of the same shape but opposite direction. Let’s illustrate these ideas by
showing how the motion of a particle, as observed in a particle detector, can be
used to determine information about its charge-to-mass ratio.

This picture shows an example of
what we might observe in a particle detector. The black rectangle is the total
area in which we can observe particles, and the white spiral is the path followed by
one such charged particle. Detectors typically have uniform
magnetic fields, so we expect that the paths followed by charged particles will be
circles. However, instead of the circle that
we expect, our picture clearly shows that the particle is moving in a spiral. This is because when charged
particles accelerate, including by changing direction, they emit electromagnetic
radiation. This radiation carries energy. So, by conservation of energy, as
these particles start to circle around in a magnetic field, they emit radiation,
lose energy, and therefore slow down.

Looking back at our equation that
describes the motion of these particles, the charge-to-mass ratio is constant, as is
the detectors magnetic field. Because those other quantities are
constant, this tells us that the speed of the particle divided by the radius of the
path is also a constant quantity. So as a particle’s speed decreases,
the radius of its path also decreases, and so our particle follows a spiral instead
of a circle. So, charged particles in detectors
with uniform magnetic fields tend to follow spiral not circular paths.

Spirals do not have a single
well-defined radius. Luckily, if we measure from the
center of the spiral to the outer edge, we get approximately the radius that the
particle would have if it was moving in a circle without emitting electromagnetic
radiation. This is not fully accurate for
quantitative calculations, but it is certainly good enough to qualitatively compare
the tracks of two particles. For the particle that we’ve drawn
without any knowledge of its speed or its charge-to-mass ratio, the only thing that
we can really determine is the radius of its path. Luckily, information like speed and
the charge is often available from other measurements.

Furthermore, if we detect two
particles, we can often compare them qualitatively using only a little bit of extra
information. Here, the detector is showing the
paths of two charged particles. We’ll call them one and two. Now, let’s say that we’ve also
determined that these two particles have the same initial speed. So by some external measurement,
the particles have the same initial speed. And since they’re in the same
detector, we know they’re experiencing the same magnetic field. Since these two quantities are the
same, if we can compare the radii of their paths, we can compare their
charge-to-mass ratios.

When we draw the radius of each
path on the diagram, we can clearly see that the two particles actually follow paths
with the same radius. But this means that all three
quantities on the left-hand side of our equation, speed, magnetic field, and radius,
are the same for both particles. This means that at least in
magnitude, the two particles have the same charge-to-mass ratio.

There’s even more information that
we can determine from this picture. The path of particle two and the
path of particle one curve in opposite directions. Since they clearly started in the
same place traveling in the same direction, the force on particle two must have
pointed in the opposite direction to the force on particle one. But for two particles moving in the
same direction in the same magnetic field, they will only experience forces pointing
in opposite directions if they have opposite signs on their electric charge. So from only the detector output,
we know the relationship between the sign of the charges of these two particles.

Furthermore, from a little bit of
information about their initial speed, we’ve also managed to use the detector output
to determine the relationship between the size of their charge-to-mass ratios. Putting these two pieces of
information together, we know that particle one and particle two have charge-to-mass
ratios with the same size but opposite sign, which is characteristic of the
relationship between a particle and its antiparticle. In fact, seeing this particle one
and particle two seem to have spontaneously appeared at the same position in space,
it is almost guaranteed that they’re the particle–antiparticle product of pair
production.

This means that in addition to
charge and charge-to-mass ratio, we’ve also managed to relate the identities of
particle one and particle two. If we later discovered that, for
example, particle two was a positron, we would then conclude that particle one, its
antiparticle, is an electron. So from just a picture of the paths
followed by two particles in a uniform magnetic field and also a relationship
between their initial speeds, we worked out several further relationships between
these two particles and, with an additional piece of information, even managed to
give a likely identification for both particles in the detector.

Great! Now that we’ve seen how to use our
equation and qualitative information from particle detectors, let’s work through a
quantitative example.

A charged particle moves through a
uniform magnetic field. It moves along a circular path with
a radius of 0.0200 meters. Its speed is 3.40 times 10 to the
sixth meters per second. The strength of the magnetic field
is 0.200 teslas. What is the charge-to-mass ratio of
the particle? Give your answer in coulombs per
kilogram to three significant figures.

Since this question is asking us to
find the charge-to-mass ratio of a particle moving through a uniform magnetic field,
it makes sense to consider the equation speed divided by magnetic field strength
times radius is equal to charge-to-mass ratio. This gives us the charge-to-mass
ratio of a particle moving in a uniform magnetic field with motion described by the
speed and the radius of the path. Recall that we derived this
equation by equating the magnetic force on a charged particle moving in a magnetic
field to the centripetal force on any object moving in a circle.

Now that we have this equation, to
calculate charge-to-mass ratio, all we need to do is find appropriate values for the
speed, magnetic field, and radius. In this question, the values are
simply given to us. We have 0.0200 meters for the
radius, 3.40 times 10 to the sixth meters per second for the speed, and 0.200 teslas
for the strength of the magnetic field. Plugging those values into our
formula gives us an expression for the charge-to-mass ratio. Let’s start by evaluating the
numerical portion, 3.40 times 10 to the sixth divided by 0.200 times 0.0200. Plugging into a calculator, this
gives us 850 times 10 to the sixth.

We’re asked to report our answer to
three significant figures. Since this number is written in the
form of three digits, 850 times a power of 10, the three significant figures for
this number are 850. We could either leave the number in
this form, or to be absolutely clear with the significant figures, we could write it
as 8.50 times 10 to the eighth. For the units, meters in the
numerator divided by meters in the denominator is just one. And per seconds in the numerator
divided by teslas in the denominator is just one divided by tesla seconds.

Aside from the fact that these are
somewhat odd units, the question wants us to have our answer in coulombs per
kilogram. So we need to convert one per tesla
seconds to coulombs per kilogram. The most direct way to do this is
to recall that one tesla is exactly one kilogram per coulomb seconds. So one tesla second is one kilogram
second per coulomb second, which is just one kilogram per coulomb. Taking the reciprocal of both
sides, we see that despite one per tesla second looking rather odd, it’s exactly
equivalent to one coulomb per kilogram, which is precisely the units we want for our
answer.

This one-to-one relationship is
actually guaranteed from the definition of SI units and the fact that the tesla is
the basic unit of magnetic field, the second is the basic unit of time, the coulomb
is the basic unit of charge, and the kilogram is the basic unit of mass. Because one per tesla seconds and
coulombs per kilogram are both valid units for charge-to-mass ratio and both only
contain basic SI units, by definition they must be equivalent. So to three significant figures,
the charge-to-mass ratio of our particle is 8.50 times 10 to the eighth coulombs per
kilogram. As it happens, this is almost
exactly the magnitude of the charge-to-mass ratio of both the muon and the
antimuon.

Alright, now that we’ve seen how to
apply this formula both qualitatively and quantitatively, let’s review what we’ve
learned about identifying particles and their properties. We started by deriving the equation
that the charge-to-mass ratio of a particle is equal to its speed divided by the
radius of its path and the strength of the magnetic field in which it is moving. In the context of a particle
detector, this strength is typically fixed so this equation effectively relates the
charge-to-mass ratio of a particle to its motion.

We saw by way of example that this
equation can be used to calculate the charge-to-mass ratio of a particle from
knowledge of its speed, the radius of its path, and the strength of the magnetic
field. And in fact, we can calculate any
quantity appearing in this equation from knowledge of the other quantities. We also saw that particle detectors
can show us the path of a charged particle as it moves through a magnetic field. In this case, the paths are spirals
not circles because as the charged particle changes direction, it loses energy,
which causes its speed to decrease with a corresponding decrease in radius.

Nevertheless, using the output of
such detectors, we can still find the radius at any particular time. We can also determine if two
particles have the same or opposite sign for their charge by looking at whether
their paths curve in the same or opposite directions. Finally, if we include information
like charge and initial speed that are available from other types of measurement, we
can qualitatively compare the charge-to-mass ratio and the mass of two particles in
a detector and sometimes even come up with quantitative values for those
quantities. Sometimes it is even possible to
use all this information to identify a particular particle by its path in the
detector.