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In the last video we learned--
or at least I showed you, I

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don't know if you've learned it
yet, but we'll learn it in

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this video.

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But we learned that the force
on a moving charge from a

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magnetic field, and it's a
vector quantity, is equal to

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the charge-- on the moving
charge-- times the cross

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product of the velocity of the
charge and the magnetic field.

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And we use this to show you that
the units of a magnetic

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field-- this is not a beta, it's
a B-- but the units of a

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magnetic field are the tesla--
which is abbreviated with a

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capital T-- and that
is equal to newton

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seconds per coulomb meters.

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So let's see if we can apply
that to an actual problem.

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So let's say that I have a
magnetic field, and let's say

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it's popping out
of the screen.

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I'm making this up on the
fly, so I hope the

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numbers turn out.

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It's inspired by a problem that
I read in Barron's AP

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calculus book.

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So if I want to draw a bunch of
vectors or a vector field

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that's popping out of the
screen, I could just do the

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top of the arrowheads.

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I'll draw them in magenta.

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So let's say I have
a vector field.

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So you can imagine
a bunch of arrows

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popping out of the screen.

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I'll just draw a couple of them
just so you get the sense

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that it's a field.

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It pervades the space.

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These are a bunch of
arrows popping out.

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And the field is popping out.

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And the magnitude of the field,
let's say it is, I

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don't know, let's say
it is 0.5 teslas.

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Let's say I have some proton
that comes speeding along.

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And it's speeding along at a
velocity-- so the velocity of

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the proton is equal to
6 times 10 to the

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seventh meters per second.

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And that is actually about 1/5
of the velocity or 1/5 of the

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speed of light.

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So we're pretty much in the
relativistic realm, but we

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won't go too much into
relativity because then the

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mass of the proton increases,
et cetera, et cetera.

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We just assume that the mass
hasn't increased significantly

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at this point.

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So we have this proton going at
a 1/5 of the speed of light

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and it's crossing through
this magnetic field.

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So the first question is what is
the magnitude and direction

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of the force on this proton
from this magnetic field?

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Well, let's figure out the
magnitude first. So how can we

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figure out the magnitude?

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Well, first of all, what is
the charge on a proton?

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Well, we don't know it right
now, but my calculator has

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that stored in it.

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And if you have a TI graphing
calculator, your calculator

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would also have it
stored in it.

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So let's just write that down
as a variable right now.

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So the magnitude of the force on
the particle is going to be

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equal to the charge of a
proton-- I'll call it Q sub

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p-- times the magnitude of the
velocity, 6 times 10 to the

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seventh meters per second.

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We're using all the
right units.

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If this was centimeters
we'd probably want to

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convert it to meters.

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6 times 10 to the seventh
meters per second.

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And then times the magnitude of
the magnetic field, which

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is 0.5 soon. teslas-- I didn't
have to write the units there,

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but I'll do it there--
times sine of the

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angle between them.

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I'll write that down
right now.

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But let me ask you a question.

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If the magnetic field is
pointing straight out of the

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screen-- and you're going to
have to do a little bit of

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three-dimensional visualization
now-- and this

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particle is moving in the plane
of the field, what is

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the angle between them?

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If you visualize it in three
dimensions, they're actually

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orthogonal to each other.

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They're at right angles
to each other.

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Because these vectors are
popping out of the screen.

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They are perpendicular to the
plane that defines the screen,

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while this proton is moving
within this plane.

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So the angle between them, if
you can visualize it in three

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dimensions, is 90 degrees.

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Or they're perfectly
perpendicular.

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And when things are perfectly
perpendicular, what is the

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sine of 90 degrees?

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Or the sine of pi over 2?

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Either way, if you want
to deal in radians.

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Well, it's just equal to 1.

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The whole-- hopefully--
intuition you got about the

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cross product is we only want to
multiply the components of

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the two vectors that are
perpendicular to each other.

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And that's why we have
the sine of theta.

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But if the entire vectors are
perpendicular to each other,

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then we just multiply the
magnitude of the vector.

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Or if you even forget to do
that, you say, oh well,

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they're perpendicular.

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They're at 90 degree angles.

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Sine of 90 degrees?

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Well, that's just 1.

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So this is just 1.

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So the magnitude of the force
is actually pretty easy to

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calculate, if we know the
charge on a proton.

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And let's see if we can figure
out the charge on a proton.

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Let me get the trusty
TI-85 out.

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Let me clear there,
just so you can

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appreciate the TI-85 store.

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If you press second and
constant-- that's second and

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then the number 4.

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They have a little constant
above it.

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You get their constant
functions.

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Or their values.

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And you say the built-in--
I care about the built-in

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functions, so let me press F1.

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And they have a bunch of-- you
know, this is Avogadro's

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number, they have a bunch of
interesting-- this is the

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charge of an electron.

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Which is actually the
same thing as

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the charge of a proton.

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So let's use that.

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Electrons-- just remember--
electrons and protons have

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offsetting charges.

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One's positive and
one's negative.

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It's just that a proton
is more massive.

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That's how they're different.

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And of course, it's positive.

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Let's just confirm that that's
the charge of an electron.

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But that's also the charge
of a proton.

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And actually, this positive
value is the

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exact charge of a proton.

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They should have maybe put a
negative number here, but all

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we care about is the value.

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So let's use that again.

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The charge of an electron-- and
it is positive, so that's

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the same thing as the charge for
a proton-- times 6 times

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10 to the seventh-- 6 E 7, you
just press that EE button on

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your calculator-- times
0.5 teslas.

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Make sure all your units are
in teslas, meters, and

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coulombs, and then your result
will be in newtons.

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And you get 4.8 times 10 to
the negative 12 newtons.

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Let me write that down.

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So the magnitude of this force
is equal to 4.8 times 10 to

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the minus 12 newtons.

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So that's the magnitude.

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Now what is the direction?

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What is the direction
of this force?

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Well, this you is where we break
out-- we put our pens

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down if we're right handed, and
we use our right hand rule

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to figure out the direction.

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So what do we have to do?

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So when you take something
crossed something, the first

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thing in the cross product is
your index finger on your

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right hand.

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And then the second thing is
your middle finger pointed at

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a right angle with your
index finger.

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Let's see if I can do this.

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So I want my index finger
on my right hand to

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point to the right.

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But I want my middle finger
to point upwards.

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Let me see if I can
pull that off.

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So my right hand is going to
look something like this.

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And my hand is brown.

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So my right hand is going to
look something like this.

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My index finger is pointing in
the direction of the velocity

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vector, while my middle finger
is pointing the direction of

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the magnetic field.

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So my index finger is going to
point straight up, so all you

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see is the tip of it.

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And then my other fingers are
just going to go like that.

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And then my thumb is
going to do what?

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My thumb is going-- this is the
heel of my thumb-- and so

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my thumb is going to be at a
right angle to both of them.

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So my thumb points
down like this.

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This is often the
hardest part.

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Just making sure you get your
hand visualization right with

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the cross product.

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So just as a review, this
is the direction of v.

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This is the direction of
the magnetic field.

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It's popping out.

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And so if I arrange my right
hand like that, my thumb

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points down.

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So this is the direction
of the force.

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So as this particle moves to the
right with some velocity,

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there's actually going to
be a downward force.

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Downward on this plane.

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So the force is going to
move in this direction.

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So what's going to happen?

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Well, what happens-- if you
remember a little bit about

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your circular motion and your
centripetal acceleration and

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all that-- what happens when you
have a force perpendicular

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to velocity?

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Well, think about it.

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If you have a force here and the
velocity is like that, if

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the particles-- it'll
be deflected a

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little bit to the right.

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And then since the force
is always going to be

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perpendicular to the velocity
vector, the force is going to

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charge like that.

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So the particle is actually
going to go in a circle.

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As long as it's in the magnetic
field, the force

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applied to the particle by the
magnetic field is going to be

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perpendicular to the velocity
of the particle.

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So the velocity of the
particle-- so it's going to

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actually be like a centripetal
force on the particle.

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So the particle is going
to go into a circle.

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And in the next video we'll
actually figure out the radius

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of that circle.

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And just one thing I want
to let you think about.

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It's kind of weird or spooky
to me that the force on a

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moving particle-- it doesn't
matter about

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the particle's mass.

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It just matters the particle's
velocity and charge.

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So it's kind of a strange
phenomenon that the faster you

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move through a magnetic field--
or at least if you're

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charged, if you're a charged
particle-- the faster you move

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through a magnetic field, the
more force that magnetic field

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is going to apply to you.

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It seems a little bit, you know,
how does that magnetic

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field know how fast
you're moving?

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But anyway, I'll leave
you with that.

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In the next video we'll
explore this magnetic

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phenomenon a little bit deeper.

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See

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