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Electric motors (part 1)
Sal shows that there will be a net torque on a loop of current in a wire. Sal shows that this net torque will cause the loop to rotate. Created by Sal Khan.
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sooo confused on the hand rule. Wow I'm like beyond confused on the hand rule I don't even know where the middle finger is supposed to point(50 votes)
I'm learning the right hand rule like this:
Thumb in the direction of the current
Fingers in the direction of the field
Magnetic force is now coming out of your palm
:)(134 votes)
at7:20if the one on the left is going in and the one on the right is going out, shouldnt it spin the other direction as the blue arrow?(9 votes)
Can we actually somehow use the magnetic field of the Earth and its induction as a source for any kind of engine?(7 votes)- It is a very weak field. There is no practical way to use such a weak field as an engine.(6 votes)
- Why is the current a scalar and not a vector? Is current analogous to speed?(6 votes)
current does not obey the laws of vector addition,or in other words, it does not depend on the angle in which the current carrying conductor is bent...
for example,if we connect two conductors in parallel(and suppose they are inclined to each other by 90 degrees,just for simplicity),and through each,a current of 'i' passes..
then, the net current through the system is '2i' ,not squareroot ( (i)2 + (i)2 ) ,i.e. not i root 2
so since it does not behave like a vector and does not follow its rules,current got kicked out of the vector club :)(4 votes)
I don't understand why the circuit rotates. Can anyone help me?(4 votes)- I think it may also help to look at the problem as follows:
To determine why the circuit rotates, we can look at each section of wire individually and determine the force exerted on it using the right hand rule. Let's look at the bottom, horizontal chunk of wire first. The current is traveling antiparallel to the magnetic field, so no force will be exerted on that section of wire. The other two small horizontal sections of wires closer to the top also have current running parallel to the magnetic field, so no force will be exerted on those either. This is due to the fact that in our formula:
F = I (L x B), the cross product of parallel or antiparallel vectors (meaning going in the same or opposite directions) is 0.
Now we can look at the interesting stuff that does have forces applied to it. First let's look at the left side wherein the current travels downwards. Using our right hand rule, where the current travels downwards and the magnetic field travels left, we can determine the force to be into the page. For the right side, wherein the current travels upwards, we can use the right hand rule to determine the force on that chunk of wire is out of the page. This means the loop of wire as a whole has a force into the page on the left and out of the page on the right, meaning it will rotate about its central vertical axis. I hope this helps.(6 votes)
- I don't understand how the torque decreases as the circuit rotates around the axis.. can anyone explain?(5 votes)
- Imagine flipping the circuit around. The part of the circuit that was initially on the right is now on the left and vice versa. Initially current on the left flowed towards the bottom of the screen, but now because we've flipped the circuit around, the current on the new left side of the circuit (the original right) is flowing the towards the top of the screen. Reversing the direction of the current reverses the direction of the force acting on the wire, so the circuit now begins to spin in the opposite direction.(2 votes)
Wait. His index finger is pointing down for L, so how does his thumb also point down?(3 votes)- the left hand rule is used to find the force acting on the conductor in the magnetic field and not the right hand rule..!(2 votes)
the right hand rule is for the generator effect.
you got it the wrong way round which must have confused so many viewers.
I think it is because you confused the middle and index fingers and labelled them wrongly. The index should be magnetic field/B and the middle should be current/I(1 vote)
I didn't understand how the torque changed due to the movement of the circuit. Can anyone explain please?(2 votes)
Hello Prakhar,
Let’s explore this with a bar magnet.
We need to simplify this setup. Picture a clock. Let a constant magnetic field flow from the 3 o’clock position across to the 9 o’clock. And from the 2 to the 10 and 1 to 11 and so forth. If we drew the magnetic vectors the clock face would be full of lines running right to left.
Now let’s replace the loop of wire with a bar magnet. We can do this because they do the same thing. In Sal’s video the loop had a constant current. It follows that the loop also had a constant magnetic field. Therefore we can replace it with a bar magnet.
Don’t confuse the bar magnet with the magnetic field on the face of the clock!
Now that we have the setup, install your bar magnet on a pivot in the center of the clock. Let N be at 1 o’clock and S point to 7 o’clock. Let the bar go and describe what happens. In fact your question will be answered as you consider the torque at each position on the clock face.
Be sure to read up on commutation (DC machines) and AC motors to see how we can make the torque rotate smoothly.
Regards,
APD(1 vote)
- When you have the current going in the wire loop, it will also generate a magnetic field pointing outside the screen right?? Will it affect the external magnetic field that points to the left? Or it is too small and can be ignored?(2 votes)
- It does generate a field, but it is small in comparison to B, so we can ignore it(1 vote)
7:20Video transcript
Let's say we have a magnetic
field that's coming out of the right side of the screen. And it's not just along the
screen, it's actually three dimensional. So it's going above the
screen, below, but the direction of the magnetic
field is from the right to the left. Let me just draw that. And I'm not going to draw a
bunch of the field vector arrows because that'll just
take up a lot of valuable space on our-- so that's the
vector magnetic field. It's down here, too. If I could, I would draw it
above your screen and below your screen. But it's coming from the
right to the left. And then in that magnetic
field I have an electric circuit. I won't draw the whole
circuit right now. I'll do that in a second. But let's say I have--
part of the electric circuit is a loop. And the loop looks like this. I'm trying to draw
it carefully. So that we can-- because I think
a careful drawing will be more useful in
a second than an uncarefully drawn drawing. So let's see. So it's a loop. You can almost-- you could
imagine taking a paper clip and putting it into
this shape. Oh, that's good enough,
I think. And I have a current going
in this direction in this paper clip. So this is the positive,
that's the negative. So the current is
going like that. Current is going in
a loop like that. The current's coming out
of this end, it's coming into this end. And let's say that loop of--
it could be a paper clip or anything-- let's say
that it can rotate. And that's important. What's going to happen? Well, my magnetic field is
coming in this direction. The current is going
down here, up here. What's going to be the net
force of the magnetic field on this loop? Well, let's try it out. And it turns out it's going to
be a different magnitude at different points
of the current. So here. And we don't worry about-- all
we're worried about right now is direction. And then maybe a little
intuition of the magnitude. So we know that the force of the
magnetic field is equal to the current times the
length vector cross the magnetic field. Well, what would be the force of
the magnetic field on this segment of wire? We could call this L. And that L goes in the same
direction as the current. Well, let's see. Current is just a scalar,
but L is going down. Magnetic field going
to the left. Cross product. Cross product, I take my right
hand, put my index finger in the direction of the current,
or in the direction of L, because that's the first term
of the cross product. So that's the index finger. So my index finger goes down,
because that's the direction of the current. And then my middle finger-- and
remember, you have to do this with the right hand. If you do it with the left hand,
you're going to get the opposite result. And now my middle finger is
going to go in the direction of the field. So let me point my
middle finger. My middle finger is
going to go in the direction of the field. I keep having to look
at my own hand. And then my other two fingers
are just going to do what they need to do. So that's my third finger. That's my pinky. And then what is my
thumb going to do? What is my thumb going to do? Well, my hand-- that's
my hand. This is what my hand is doing. I'm pointing downward. And my palm is kind of
pointing at my body. So what is my thumb doing? I know it's hard to see. This is my middle finger
right here. So my thumb is on the other
side of this drawing. And my thumb is pointing
downwards. I hope you see that. And you try it with
your own hand. So my thumb is pointing
downwards. So the direction of the force
created by the magnetic field on this current is going
to go downward. So let me draw that. So the force vector--
I'll do it in this orangey brown color. The force vector on this segment
of the wire is going to be going down. Now what about this segment
of the wire? Well, think about it. This segment of the wire, the
L vector-- this L vector-- It's parallel to the magnetic
field just in the opposite direction. And so when you take the cross
product-- remember, the cross product is you're multiplying
the magnitude of the vectors that are perpendicular
to each other. But if this is the L vector
right here, there's no component of it that it's
perpendicular to the magnetic field. So the magnetic field and the
current are in the same plane. They're parallel. They're not orthogonal at all. There's no components of them
that are at 90 degrees. So when you take the cross
product, you're going to see that the net force on this
segment of the wire is 0. And likewise on this segment of
the wire and this segment of the wire. Because they aren't in any
way perpendicular. No components of them are
even perpendicular. So fair enough. So all we know right now is the
magnetic field is exerting a downward force on this
side of our paper clip or of our circuit. Now what about this side? Well, same thing. Take the cross product. If this is our L, L cross B. So take your index finger
in the direction--. So index finger goes
like that. Your middle finger will go in
the direction of the field. So your middle finger is going
to look something like that. And then your other
two fingers are going to be like that. And what is your thumb
going to do? And this has to be your
right hand to work. Your thumb is going to
point straight up. This is like the heel
of your thumb. Your thumb is going to--. I don't know if that's a good
drawing of a thumb. But your thumb is essentially
pointing out of the page. Middle finger in the direction
of the current, or in the direction of our length. Sorry, index finger in the
direction of the current. Middle finger in the direction
of the field. Thumb points out of the page. Do that with your own right hand
and you'll see that the net force of the magnetic field
on this segment of the wire is going to be upwards. Let me do it in a different
color just to get some contrast. So what's going to happen? Assuming that this circuit
can rotate, what's going to happen? On this side there's
a downward force. On this side there's
an upward force. So the magnetic field is
actually exerting a torque on this wire. If you viewed this little dotted
line as our axis of rotation, the whole coil is
actually going to rotate around that line. And so there's some force over
here, along this whole line being applied downwards. And it's actually perpendicular to our moment arm. If you remember what we had
learned about torque. So it will actually exert all
of that force-- that force times this distance
will be the torque applied on this side. And then likewise there's a
torque-- it's really the same sign, in the same direction,
because here on the other side of the arm it's pushing
upwards. So they're not going
to cancel out. They're both going
to reinforce. And this whole coil
is going to be turning in this direction. Here it's going to be moving up
out of your video screen. Here it's going to be moving
down into your video screen. Now what happens? I'm going to try to not run
out of time either. So it's going to
start rotating. So the left hand side's going
to go below the page. The right hand side's going
to be above the page. I want to draw some perspective,
that's why I'm just drawing it bigger. Maybe it looks like that. Maybe my circuit starts
to look like that. And I'll redraw my
axis of rotation. So this is my axis
of rotation. And on the way I drew it--
this part, the axis of rotation is still in the
plane of our video. But this part of the
coil is, you could imagine it popping out. I wish you had 3D glasses on. It's popping out
of your screen. This part is going
into your screen. And the current is still going
in the same direction. Current is going in that
direction there. So using the same right hand
rule, on this side of the wire the magnetic field is going
to be exerting a net downward force. But the torque is actually going
to be less because our moment arm distance is going to
be like-- I want to draw it with some perspective. It's going to look something
like that. So it's going to be going to the
left and behind the page while the torque is still
just into the page. So you would actually take the
component of the torque that's perpendicular. So there's some component of
the torque that's actually perpendicular. I don't want to confuse
you too much. But you could imagine
the torque lessens. Even though the net magnetic
force is the same, the component of that force that's
perpendicular to your moment arm, that lessens. So there's still going to be
some torque that's going to be causing it to rotate downwards
in that direction. Sorry. You know what? I drew this wrong here. We're pushing up on the right
hand side, we're pushing down on the left hand side. So the direction is going
to be like that. Pushing up on the right hand,
down on the left hand. So you're still going to be
doing the same thing here. You are going to be
pushing up here. But you're going to be pushing
up directly out of the page. But that's not completely
perpendicular to the moment arm. So the component that is
perpendicular, that's actually creating rotational torque,
that's going to be a little less. And then you could imagine that
all the way to the point, the coil's going to keep
rotating with smaller torque. At some point you'll be
looking at it head on. So I can just draw it like
a straight line, right? You can imagine. This arm is on top and this
arm is behind it. And at this point, what's
going to happen? All the magnetic force on
this top arm is going to be popping up. It's going to be popping up out
of your page but it's not going to be providing
any torque. Because it's not perpendicular
anymore to your moment arm. And likewise, on the bottom
behind this, if you could visualize this, it would be
exerting a net downward force. And that's also not going
to be helpful. But maybe they have some angular
momentum so the wire will still rotate. But then when it still rotates, what's going to happen? And this is where I'll
leave you with a little bit of a conundrum. Actually, I don't want to go
over the Youtube limit, so I'm going to continue this in the
next video and I'll show you the conundrum. See you soon.