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### Course: Physics library>Unit 13

Lesson 3: Electric motors

# 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.

## Want to join the conversation?

• 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
• 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
:)
• at if 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?
• Can we actually somehow use the magnetic field of the Earth and its induction as a source for any kind of engine?
• It is a very weak field. There is no practical way to use such a weak field as an engine.
• Why is the current a scalar and not a vector? Is current analogous to speed?
• 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 :)
• I don't understand why the circuit rotates. Can anyone help me?
• 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.
• I don't understand how the torque decreases as the circuit rotates around the axis.. can anyone explain?
• 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.
• Wait. His index finger is pointing down for L, so how does his thumb also point down?
• the left hand rule is used to find the force acting on the conductor in the magnetic field and not the right hand rule..!
• 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?
• 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?