in this video we're introducing the idea of cyclotron motion so this is the uniform
circular motion of a charged particle in a magnetic field so in the diagram i have a
particle with a positive charge on it and it's moving with a speed of v perfectly to the right
when it enters this magnetic field that's pointing inward that's what the x's mean for the direction
and i just want to work things out qualitatively real quick so let's look at the particle after
it's gone into the magnetic field so at this moment i still have a rightward velocity and
i have this inward magnetic field and we can find the direction of the magnetic force on this
particle by pointing our fingers in the direction of v and then sweeping out to the direction of b
and your thumb will point in the direction of the magnetic force so that looks something like this
and i see that my thumb is pointing up so the magnetic force is upward here i'll call that fb
and it's important to note that the magnetic force is always perpendicular to the velocity
well that means that force can't be speeding up the particle or slowing it down a force
perpendicular to the motion does no work so my speed's always going to stay the same and all this
force can do is turn the velocity vector so it's going to rotate it a little bit counterclockwise
now i can visualize my particle a moment later again the speed is going to be exactly the same
because forces that are perpendicular to the direction of motion can't do any work
i run through the right hand rule again and i put my fingers in the direction of v sweep out
to the direction of b and now my thumb is still pointing at the center of curvature for this path
again that's exactly perpendicular to the velocity and now i'm starting to get the idea the force
always points to the center of curvature for the path of this object and what i'm looking at is
an example of uniform circular motion so now i can get into the physics analysis and just a
quick reminder to begin with the magnitude of the magnetic force is given by the lorentz force law
and i'll post a link real quick to where this was first introduced and the sine of theta there the
theta is the angle between the velocity vector and the magnetic field and that's 90 degrees in
this case because my magnetic field is perfectly into the page and my velocity is within the plane
of the page well with theta equals 90 degrees my magnetic force is maxed out sine theta is one and
i get a magnetic force magnitude of q v b now i simply apply newton's second law to this particle
f net equals m a and there's only one force acting on it it's the magnetic force
in the acceleration now that i know that i'm looking at uniform circular motion
the acceleration is v squared over r where r is the radius of curvature
i plug in my magnetic force cancel one factor of v and i'm going to say
my motivation here is to solve for the radius of curvature so i get mv over qb
next we're going to apply our new formula to a simple example
in the example i have a proton and i'm giving you the proton mass here as 1.0073 atomic mass units
an atomic mass unit converts to kilograms like this and i have a speed of 2.2 times 10 to the
fifth meters per second if you're curious how we can control the speed of a proton entering
a magnetic field you do it by using a velocity selector device and i'll post a link real quick
to the video where we worked this out and then i'm told that my magnetic field has a
strength of 2.0 teslas our goal is to get the radius of curvature for this proton
so just as a side note let's get the mass of the proton in kilograms 1.0073 atomic mass units
multiplied by 1.66 times 10 to the negative 27 kilograms for each atomic mass unit and my
atomic mass units cancel out and i've got 1.67 just keeping three sig figs times 10 to the 20
10 to the negative 27 kilograms it looks like everything else is ready to go into
our radius of curvature formula so i have mv over qb and i need to remember the elementary
charge on a proton real quick that's 1.60 times 10 to the negative 19 coulombs
and then finally my magnetic field was 2.0 tesla i get .0011 just keeping two sig figs here meters
and i could write that as 1.1 millimeters radius of curvature
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