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Current time:0:00Total duration:10:47

In the last video we learned--
or at least I showed you, I don't know if you've learned it
yet, but we'll learn it in this video. But we learned that the force
on a moving charge from a magnetic field, and it's a
vector quantity, is equal to the charge-- on the moving
charge-- times the cross product of the velocity of the
charge and the magnetic field. And we use this to show you that
the units of a magnetic field-- this is not a beta, it's
a B-- but the units of a magnetic field are the tesla--
which is abbreviated with a capital T-- and that
is equal to newton seconds per coulomb meters. So let's see if we can apply
that to an actual problem. So let's say that I have a
magnetic field, and let's say it's popping out
of the screen. I'm making this up on the
fly, so I hope the numbers turn out. It's inspired by a problem that
I read in Barron's AP calculus book. So if I want to draw a bunch of
vectors or a vector field that's popping out of the
screen, I could just do the top of the arrowheads. I'll draw them in magenta. So let's say I have
a vector field. So you can imagine
a bunch of arrows popping out of the screen. I'll just draw a couple of them
just so you get the sense that it's a field. It pervades the space. These are a bunch of
arrows popping out. And the field is popping out. And the magnitude of the field,
let's say it is, I don't know, let's say
it is 0.5 teslas. Let's say I have some proton
that comes speeding along. And it's speeding along at a
velocity-- so the velocity of the proton is equal to
6 times 10 to the seventh meters per second. And that is actually about 1/5
of the velocity or 1/5 of the speed of light. So we're pretty much in the
relativistic realm, but we won't go too much into
relativity because then the mass of the proton increases,
et cetera, et cetera. We just assume that the mass
hasn't increased significantly at this point. So we have this proton going at
a 1/5 of the speed of light and it's crossing through
this magnetic field. So the first question is what is
the magnitude and direction of the force on this proton
from this magnetic field? Well, let's figure out the
magnitude first. So how can we figure out the magnitude? Well, first of all, what is
the charge on a proton? Well, we don't know it right
now, but my calculator has that stored in it. And if you have a TI graphing
calculator, your calculator would also have it
stored in it. So let's just write that down
as a variable right now. So the magnitude of the force on
the particle is going to be equal to the charge of a
proton-- I'll call it Q sub p-- times the magnitude of the
velocity, 6 times 10 to the seventh meters per second. We're using all the
right units. If this was centimeters
we'd probably want to convert it to meters. 6 times 10 to the seventh
meters per second. And then times the magnitude of
the magnetic field, which is 0.5 soon. teslas-- I didn't
have to write the units there, but I'll do it there--
times sine of the angle between them. I'll write that down
right now. But let me ask you a question. If the magnetic field is
pointing straight out of the screen-- and you're going to
have to do a little bit of three-dimensional visualization
now-- and this particle is moving in the plane
of the field, what is the angle between them? If you visualize it in three
dimensions, they're actually orthogonal to each other. They're at right angles
to each other. Because these vectors are
popping out of the screen. They are perpendicular to the
plane that defines the screen, while this proton is moving
within this plane. So the angle between them, if
you can visualize it in three dimensions, is 90 degrees. Or they're perfectly
perpendicular. And when things are perfectly
perpendicular, what is the sine of 90 degrees? Or the sine of pi over 2? Either way, if you want
to deal in radians. Well, it's just equal to 1. The whole-- hopefully--
intuition you got about the cross product is we only want to
multiply the components of the two vectors that are
perpendicular to each other. And that's why we have
the sine of theta. But if the entire vectors are
perpendicular to each other, then we just multiply the
magnitude of the vector. Or if you even forget to do
that, you say, oh well, they're perpendicular. They're at 90 degree angles. Sine of 90 degrees? Well, that's just 1. So this is just 1. So the magnitude of the force
is actually pretty easy to calculate, if we know the
charge on a proton. And let's see if we can figure
out the charge on a proton. Let me get the trusty
TI-85 out. Let me clear there,
just so you can appreciate the TI-85 store. If you press second and
constant-- that's second and then the number 4. They have a little constant
above it. You get their constant
functions. Or their values. And you say the built-in--
I care about the built-in functions, so let me press F1. And they have a bunch of-- you
know, this is Avogadro's number, they have a bunch of
interesting-- this is the charge of an electron. Which is actually the
same thing as the charge of a proton. So let's use that. Electrons-- just remember--
electrons and protons have offsetting charges. One's positive and
one's negative. It's just that a proton
is more massive. That's how they're different. And of course, it's positive. Let's just confirm that that's
the charge of an electron. But that's also the charge
of a proton. And actually, this positive
value is the exact charge of a proton. They should have maybe put a
negative number here, but all we care about is the value. So let's use that again. The charge of an electron-- and
it is positive, so that's the same thing as the charge for
a proton-- times 6 times 10 to the seventh-- 6 E 7, you
just press that EE button on your calculator-- times
0.5 teslas. Make sure all your units are
in teslas, meters, and coulombs, and then your result
will be in newtons. And you get 4.8 times 10 to
the negative 12 newtons. Let me write that down. So the magnitude of this force
is equal to 4.8 times 10 to the minus 12 newtons. So that's the magnitude. Now what is the direction? What is the direction
of this force? Well, this you is where we break
out-- we put our pens down if we're right handed, and
we use our right hand rule to figure out the direction. So what do we have to do? So when you take something
crossed something, the first thing in the cross product is
your index finger on your right hand. And then the second thing is
your middle finger pointed at a right angle with your
index finger. Let's see if I can do this. So I want my index finger
on my right hand to point to the right. But I want my middle finger
to point upwards. Let me see if I can
pull that off. So my right hand is going to
look something like this. And my hand is brown. So my right hand is going to
look something like this. My index finger is pointing in
the direction of the velocity vector, while my middle finger
is pointing the direction of the magnetic field. So my index finger is going to
point straight up, so all you see is the tip of it. And then my other fingers are
just going to go like that. And then my thumb is
going to do what? My thumb is going-- this is the
heel of my thumb-- and so my thumb is going to be at a
right angle to both of them. So my thumb points
down like this. This is often the
hardest part. Just making sure you get your
hand visualization right with the cross product. So just as a review, this
is the direction of v. This is the direction of
the magnetic field. It's popping out. And so if I arrange my right
hand like that, my thumb points down. So this is the direction
of the force. So as this particle moves to the
right with some velocity, there's actually going to
be a downward force. Downward on this plane. So the force is going to
move in this direction. So what's going to happen? Well, what happens-- if you
remember a little bit about your circular motion and your
centripetal acceleration and all that-- what happens when you
have a force perpendicular to velocity? Well, think about it. If you have a force here and the
velocity is like that, if the particles-- it'll
be deflected a little bit to the right. And then since the force
is always going to be perpendicular to the velocity
vector, the force is going to charge like that. So the particle is actually
going to go in a circle. As long as it's in the magnetic
field, the force applied to the particle by the
magnetic field is going to be perpendicular to the velocity
of the particle. So the velocity of the
particle-- so it's going to actually be like a centripetal
force on the particle. So the particle is going
to go into a circle. And in the next video we'll
actually figure out the radius of that circle. And just one thing I want
to let you think about. It's kind of weird or spooky
to me that the force on a moving particle-- it doesn't
matter about the particle's mass. It just matters the particle's
velocity and charge. So it's kind of a strange
phenomenon that the faster you move through a magnetic field--
or at least if you're charged, if you're a charged
particle-- the faster you move through a magnetic field, the
more force that magnetic field is going to apply to you. It seems a little bit, you know,
how does that magnetic field know how fast
you're moving? But anyway, I'll leave
you with that. In the next video we'll
explore this magnetic phenomenon a little bit deeper. See