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## Physics library

### Course: Physics library>Unit 13

Lesson 3: Electric motors

# Electric motors (part 3)

Sal finishes the explanation of how a commutator will allow a loop of wire to continue spinning in a magnetic field, thereby allowing it to work as an electric motor. Created by Sal Khan.

## Want to join the conversation?

• in the generator the coil is moved,this induces current in it.on the other hand this current may also be responsible for generating a force on the coil.according to me , one situation will use right hand rule another the will use left hand rule.
if the magnetic field and current remains constant the the direction of force will be opposite in the two cases.is this the effect of the lenz's law which states that the induced current opposes the cause that produces it • Aren't the rotational arrows supposed to be going in the opposite direction? Sal corrected himself in the first Electric motor video but he forgot to do it here. • can any one tell me hoe to measure magnetic flux??
basically wats a proper definition and explanation of magnetic fulx..??
thnk u • Richard is right. The units are Webers. 1 Weber= 1 Tesla times 1 square meter. It tells you how much of a magnetic field passes directly through a surface (i.e. what component is perpendicular). Conceptually, this isn't difficult, but it changes how we calculate magnetic flux. Instead of just taking the product of area and field strength, we need to take the product of area and the component of field strength normal to that area. For this we use the dot product. Here are some ways of finding magnetic flux mathematically (in decreasing difficulty).

If you have been through a multivariable calculus class then you know how to take the surface integral. Magnetic flux is defined mathematically as the surface integral of the dot product between the magnetic field vector and the differential area vector, over the whole surface. In physics and calculus the area vector has a direction normal to the surface (which means it points straight out). It looks like this (\Phi)=int(B dot dA). The math can get tricky, but this definition will work no matter how ugly your problem looks.

Another way, which is what you see in most examples, assumes a uniform magnetic field, and flat surface. If this is true, then you can pull the magnetic field vector outside of the integral as a constant. Then you have (\Phi)=B dot int(dA). The integral of dA is then just A (the area vector), and (\Phi)=B dot A. If you do not know the dot product, this is equivalent to (\Phi)=B*A*cos(/theta), where we treat B and A as scalars (values without direction), and /theta is the angle between the direction normal to the area and direction of the magnetic field. This is the first formula I learned for magnetic flux, and I always found it easy to remember because the equation looks like "Oh: BAcon"
• i knw this could be silly but when you are using the right hand rule for the right hand side of the circuit, do u have to invert your hand? • I'm a little confused. Does the commutator switch the direction of the current? If so, does it make the DC motor a AC motor? • Why is the rotation counter-clockwise. If the force on the left is acting into the screen, and the force on the right is acting out of the screen, shouldn't it be rotating clock-wise? I mean, in the previous video, it was rotating clockwise. • How do a AC and DC generator work? • Hello Suvan,

At a fundamental level there is no difference between the AC and DC machine. In both cases we find a wire moving in the presence of a magnetic field. A current is induced in this wire.

The magic is in the commutator. See: https://en.wikipedia.org/wiki/Commutator_(electric)

In a DC machine the commutator acts as a mechanical switch reversing the connection to the rotating coil every 180°. This way the current leaving the machine is always going the same direction. There is no such mechanism in an AC machine consequently the current changes direction every 180 mechanical degrees.

Note - as technology moves forward the commutator has been replaced with semiconductor switches. So no you have a machine that has the internal construction of an AC machine but has the characteristics of a DC machine.

This is a fascinating field of study. I hope you get the chance to take an electromechanical machines course in the future.

Regards,

APD
• hence what is the difference between dc motor and ac motor. Is commutator arrangement is needed for the working of ac motor. • Can I get to know what a Cyclotron is, how does it works and what it does? • A cyclotron is a way to accelerate particles with a strong static magnetic field and a changing electric field
It goes like this:
The cyclotron is composed of two semicircles(a and b) with a distance d separating both of them, and there's an electric field between them that imposes a potential difference of V. The proton starts at a's surface facing b, with a potential V and its accelerated by the electric field until b, gaining a kinetic energy of V.q
When it gets to B, the magnetic field of intensity B takes over and the proton realizes a circular motion of radius r
magnetic force F = q.v.B = mv²/r , r = mv/Bq
Since mv²/2 = Vq , then v = sqrt(2Vq/m) , so r = m.sqrt(2Vq/m)/Bq = sqrt(2Vm/q)/B
So, r = sqrt(2Vm/q)/B , and the time it takes for the proton to come to b until it realizes half of a circle and reaches b again is f = pi.r/v = pi.m/Bq
Now, when the proton reaches b, the electric field must have changed (it needs to be set up to do so), in such a way that now b is at a potential V and a at a potential 0. This way, the proton, once again, gains a kinetic energy Vq from the electric field. Only this time, the radius increased, according to the equation written above
r = mv/Bq , where v is, now, sqrt(4Vq/m)
The ratio r/v, however, remains constant and equals to m/Bq , so f is constant and always equal to pi.m/Bq
So, the electric field must be always changing at this rate, in order for the proton to keep drawing semicircles with a growing radius and speed. At some point, the particles may get so fast you need to account for relativistic effects, too. 