We have an example of a cyclotron problem here
in which we ask what happens with an electric charge moving
at a velocity of 1 meter per second
through a field whose strength is 1 tesla,
so this is a magnetic field.
The question is, what is the radius of the circle
created by the motion of the electron
if we're given that the mass-to-charge ratio
of an electron is this very small number?
So go ahead and use this cyclotron material that we've just discussed
and see if you can solve for the radius of this path
as the electron moves through a magnetic field.
Hopefully, what you did looks something like this.
What we have is again that we have a particle moving in some circle
because it has a force towards the center of that circle.
This force is the force from the magnetic field
and from our prior chapters we know that the force from the magnetic field,
which we said is q times v times strength of the magnetic field
has to be equal to mv squared over r,
if it's gonna stay in this uniform circular motion.
But we're interested in, in this problem, is the radius of the path.
So let's solve for the radius
rather than what we did earlier, the velocity.
You should get that the radius
is equal to mv squared over q times v times B.
Do some quick simplifying
cancelling one of these units of velocity.
We have mass times velocity over
the charge times the magnetic field.
So how are we supposed to solve for this radius in terms of the givens?
As what we know that we are given a velocity of 1.
Let's rewrite that 1 meter per second.
We know that we have a magnetic field of 1 tesla
and we also know the mass-to-charge ratio of the electron.
So what is that?
This is simply, this expression here, the mass-to-charge ratio, m over q.
Be aware that sometimes you're given the charge-to-mass ratio
which will be a very, very different number
so be careful to make sure you know what you're being given in the problem.
In this case, we have the mass-to-charge ratio.
We're told that this is, this a very small number,
5.7 times 10 to the minus 12 kilograms per coulomb.
And now we are ready to plug these values in
by rearranging this equation here in terms of our knowns.
So we have m over q in this equation here on the left.
We also have a v over a B, so we'll have to plug those in as well.
So these are now our known quantities.
We now have the mass-to-charge ratio
which is 5.7 times 10 to the minus 12th kilograms per coulomb
times the velocity over the strength of the magnetic field, B,
where the velocity is 1 and the strength of the magnetic field is also 1.
Since we've written everything here in terms of our simple SI units,
our standard units, we don't have to worry about the units of the radius.
We know that they'll come out to be meters,
so since these are just one
the magnitude of this quantity stays exactly what it is
5.7 times 10 to the minus 12th
and again we know all these units
will boil down to our SI units
since we used our standard units throughout the entire problem.
So this is a simple example of a cyclotron equation problem.
You can simply use the equations that we talked about for a cyclotron.
And again, always remember that the force of the magnetic field
is going to work together with the equation we have for uniform circular motion
in order to create the cyclotron equation which gets used very often.
This wraps up our discussion of the magnetic field and how it works.
Both its definition, how we talked about in measuring and defining it
as well as the kinds of forces that charged particles experience
when they are in the presence of a magnetic field.
Thanks for watching.