1

concept

## Motional EMF

9m

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Hey, guys, over this video, we're gonna be taking a look at a special kind of MF called emotional MF. Let's check it out. So remember the theme of these last couple videos have been about Is that the changing magnetic flux through a loop or something like that produces an induced E M F. That's what Faraday's Law tells us. Well, sometimes what happens is that this change in magnetic flux can happen through something moving. So it happens through motion. And when that happens, we call it emotional e M f. So what I want you guys to take away from this video Is that really emotional? E m f is just a special case of Faraday's law. All right, so let's go ahead and check it out. So we have this bar or this conducting rod that is moving through a magnetic field. We have the velocity to the right magnetic field that points into the page so away from you. And what happens is as this conducting rod moves through a magnetic field with some velocity. The charges on this magnetic rods So if I have a charge like this, they feel a magnetic force, remember that charges inside of a moving inside of a magnetic field will produce will feel a magnetic force if they have some velocity. And to sort of figure out what the direction is, we have to fuse our right hand rule. So remember, what we're gonna do is we're gonna take our fingers and we're gonna point their fingers in the direction of the magnetic field, which points into the page. We also want our thumb to be pointing off towards the right. And what happens is that the palm of your hand should face in the direction of the magnetic force. So what I've drawn here on the diagram above is that the magnetic force actually points upwards like this. And so what ends up happening is that positive charge will end up moving to the top of the rod like this, so positive charges will feel a force upwards. And what that means is that negative charges will feel a force in the opposite direction. So negative charges like this. So I have a negative charge over here. Those will actually start moving in the opposite direction so you'll start to get negative charges that buildup on the bottom. Now what happens here is we've actually separated these charges because we have a magnetic force. F B is equal to Q V b sign of data, and it depends on the Q that's involved. So what ends up happening is these charges that have now separated to the ends of the Rod have now actually produced an electric field. So he have an electric field. That point sort of through this conductor like this. And when is it happening is that these charges will eventually sort of work themselves out to balance the magnetic field. So what that means is that the force that is that that they experienced due to this magnetic field is actually perfectly balanced with the magnetic with a magnetic force. So these things will feel a sort of electric force in this direction. That's gonna be F E. But that's gonna be perfectly balanced with the magnetic force that's keeping them upwards. Now all that's really happening here is if we sort of write out some equations for this, remember that the electric force on a charge Q is Q times the electric field, and that's gonna be equal to Q times. VB we could sort of drop this sign of fate a term because we know theta is equal to 90 degrees for this particular example. The velocity goes to the right and the magnetic field points sort of straight out towards your story straight into the page. So that means that the angle is equal to 90 there. So what that means is that so you can cancel out the charges involved And that means we have a relationship between the electric field, he and the velocity with the magnetic field so easy equal to V B. Now what that basically means is that we have now have an e m f that is induced on the charges or in this conducting rod, the E M f is just equal to V b times L. So what do we actually come up with? This equation will remember way back from a couple of chapters ago that the voltage which is really just a which is really what the IMF is, is equal to the electric field times a distance. This is an equation that we used a few chapters ago. Well, the electric field Would you have a relationship with the velocity and the magnetic field. That's just VB and the distance is actually just equal to L. The length of the conducting rod. So, really, this is the equation. We're gonna have to use its like a special kind of IMF that's due to motion. So you might be wondering what this has to do with Faraday's law or what happens is now Things get interesting if we if we attach this conducting rod here to a conducting wire like this and make a loop. Because now what happens is we've created a circuit in this little region right here. So we have a circuit of charges that's moving through either this way or this way. We have no idea. But as this bar is moving through in this direction, what happens is that the magnetic flux through that the circus throughout the circuit is constantly changing. So we have to use Faraday's law on the circuit that this thing creates. Remember, when we're using Faraday's law, we're gonna be looking at the change in the magnetic flux. So let's think for a second as this rod is moving to the right sort of outside of this circuit, which of the three variables is changing. Is that the magnetic field? Is that the area or is it the angle? We'll see? The magnetic field is uniformed. It points into the page and it never changes the angle. This thing makes right there. Velocities to the right magnetic field is out there is into the page, never changes. So what's really going on is that the area is changing. That's are changing variable. So we can write our Delta Phi over Delta T as b times Delta a times cosine of fada over Delta T Now what happens is the art. Sorry that Z that's cosine theta where the cosine of the angle here is remember gonna have to take the normal of this sort of surface that we've made and the normal points in the same direction as the magnetic field. So this is the normal, or this is the area sort of vector like this, and these things points in the same exact direction. So that means that the coastline of the angles equals zero. Right? So that means that we can write this Delta A. So let's see what happens here. The rod is gonna be moving to the right, and in some time it's gonna go from here and it's gonna end up over here. So it's changed a distance of Delta X. So if we have the length, which is L and the width of this rectangle, which is little X, then the new area that we've made as this rod has been moving. So in other words, we've made a new area as this thing is moving the length or the area of this sort of new Breck tangle that we've made is actually equal to L. A times Delta X And that's gonna be over Delta T. So remember, we've gotten rid of that coastline of term here. Okay, so we have two constants over here, the magnetic field and the length of the rod. And we have two variables Delta X over Delta T. But if you remember back from way back in like early physics one this Delta X over Delta T the change in the position and the change in time is really just the velocity. So this velocity here is that the velocity which they're conducting Rod is moving through the field. So what basically happens is that the induced E M f. Which remember, is equal to Delta. Fire of a Delta T is just equal to V B l, which is the exact same equation that we got from before. It's no coincidence that we got the same exact equation because remember, emotional MF is just an applied Faraday's law. So, really, this is the equation that we're gonna be using to solve these problems. All right, so let's get to it. So we have a circuit that's below here, and the wire has a resistance of 10 Mila homes, and we have to do is figure out what the induced current is. If the length of the bar is 10 centimeters were given some other conditions right here. So let's write. Our variables were supposed to be figuring out what the strength of the induced current Sorry, that's gonna be the induced current. So they are induced currents is always gonna be related to our induced e M F. That's absolute induced, divided by the resistance that's just owns law. Now remember, what happens is that this induced e m f. When we're talking about emotional E. M f is actually given a special form that's gonna bvb times l. That's the new equation that we have divided by our. So let's take a look here. I've got my eye induced is gonna be the magnetic field or some of the velocity. First, the velocity is 25 m per second, which is 0.25. Now we have the strength of the magnetic field, which is 0.2 Tesla. And now we have the length of the bar, which is 10 centimeters. So it's gonna be 0.1. Now we have the resistance is equal to 10 million homes, which is actually point. This is 0.10 Right? So if you work all this out, you're gonna get an induced current that's equal to 05 amps, and that's actually the answer. So that's the induced current as a result of this bar moving through this field. And that's again just using Faraday's law. So how about the power? Let's see, How do we get power? Remember, that power actually has three equations again, this sort of an equation that we use a little while ago, so we have three forms of it. We have the voltage times, the currents We also have the current squared times, the resistance, and we have the voltage squared, divided by the resistance. Now, remember, in these equations you can always replace a V with an epsilon. So we have excellent. I equals I squared. R equals Epsilon squared over r, remember that these EP salons and voltages are the exact same thing. So let's see, which one of these things do we have? Well, I just figured out what my current is, and I know what the resistance of the wire is. So that means that I'm gonna be using not this equation, Not this equation. But I have the equation that relates the current and the resistance. So what I have is that the power is gonna be equal to 0. squared times the resistance, which is 0.10 So what I get when I plug that in is I get 2.5 times 10 to the minus three, and that's in Watts. So that's gonna be my power output. This is actually equal to 2.5 million watts as my final answer. All right, guys. So that's it. We're gonna get a couple more practice problems, and I'll see you guys the next one

2

Problem

A thin rod moves in a perpendicular, unknown magnetic field. If the length of the rod is 10 cm and the induced EMF is 1 V when it moves at 5 m/s, what is the magnitude of the magnetic field?

A

0.02 T

B

0.5 T

C

2 T

D

50 T

3

example

## Forces on Loops Exiting Magnetic Fields

7m

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Hey, guys, let's take a look at this example. Problem. We're gonna be working this one out together. So we have this rectangular loop here. We're told some dimensions, and we're told that this thing is basically being pulled outside of this magnetic field with some constant velocity. And we're supposed to figure out what the magnitude and the direction is off the induced current when the loop is halfway out of the field. So for this first part right here, we need to figure out is the magnitude and the direction off some induced currents like this. So we're gonna have to use Faraday's law for emotional e M f. Right. So we know that if we want the induced currents, we're gonna have to relate that to the Epsilon induced divided by the resistance. Now, this Epsilon induced is for emotional effort because this loop is moving outside of this field. So we can do is say that the induced current here is just going to be the emotional e m f. So instead of I m d. All right, m o t for emotional. Which means we're gonna replace this with vbl divided by our so that means that the induced current is gonna be the velocity. Let's see. Do I have everything I need to solve this problem? Well, let's see. I have the velocity right here. I have the strength of the magnetic field and I'm told it's uniforms. I have the length of the loop here. L and I have what the resistance is. So that tells me everything I need to know about the induced currents. That's just gonna be 12 m per second times the strength of the field, which is 0.5 times the length, which is 0.2. And then we divide that by 0.4, which is the resistance, right? That's 0.4 homes. So if you work this out, you're actually gonna find that this is equal to three amps. So the magnitude of the induced current is equal to three amps. Now, how do we figure out what the direction is? Well, remember, whenever we're trying to find out what the direction often induced current is, we're gonna have to use lenses law. So the direction of the induced currents is gonna be given to us using lenses law. So first, what we have to do is we have to figure out what is the magnetic fields. And then we have to figure out what is the change in the magnetic flux. And then that will tell us the direction of the induced Byfield. Right. So we're gonna use both of these things here to figure out what the induced be field looks like. And then from here, we can figure out what the induced current direction is. So that's always how we do lenses law. Okay, so the magnetic field, let's see the magnetic field points into the page like that, right, Because all of our field lines basically points into the page, sort of away from you. Um, now, the change in the magnetic flux What's happening here? Well, remember that the change in magnetic flux depends on three variables B A and CO. Sign of Fada. Now, what's happening here? Is that just how we saw what the equation for emotional MF comes from? It comes from the fact that this area is changing same things happening here, so the magnetic field isn't changing on the coast and of the angle isn't changing. What's happening here is that the area that this loop goes through, that this magnetic field goes through is changing. So let's see, As the loop is being pulled away from this uniform magnetic field, the area is decreasing. So that means that the Delta Phi is actually going to be negative. Let's get our change in magnetic flux is gonna be going downwards is gonna be decreasing. So lenses law tells us that the induced magnetic field wants to oppose that change. So if we have a B field that points away from you and it's getting weaker, the induced field actually wants to reinforce that weakening be fields. So what happens is sort of wants to bring it back to the way it was before by inducing a field that also goes into the page to basically reinforce it. So now what we do is we're gonna find out what the direction of that induced field is by using our right hands. So what we're gonna do is take your right hand and now I want you to point your finger. I want you to point your finger into the page, sort of away from you. So on your paper, point your finger downwards or your thumb and your finger should be wrap Your fingers should be curling in the direction off that induced currents. So far, fingers does this. My thumb points into the page, our fingers will be curling in the clockwise direction. Okay, so going over here to my diagram, what happens is that my direction is actually going to be this way. So this is going to be clockwise. That is the direction of my induced current. So if you think about this, this is actually gonna be going this way like this, and that's gonna be the direction of my induced currents. So that means here, it's gonna be going this way. This way, that way, This way. Okay, so that's the magnitude and the direction of our induced currents. Cool. That's part a Part B is now asking us what is the magnitude off the external force that is needed to keep this loop exiting at constant velocity. Okay, so let's let's see what's going on here. We're being asked to find out what is the external force so that this thing is exiting at constant velocity. Now, remember, when you see constant velocity, that means that the acceleration is equal to zero. So what they're actually asking us is, what is the external force required so that the acceleration is equal to zero. So we're actually gonna be talking about magnetic forces and sort of we're gonna have to go back to f equals m a. Right. So whenever we have forces and we're trying to relate that to acceleration, we're going to say that the sum of all the forces is gonna be mass times acceleration. We know this A is equal to zero. So now we just have to figure out what all the forces are. Well, what's happening is that you are pulling this thing out with some external magnetic or sorry, you're you're pulling this loop outwards with your hand, and that's the external Force. And that's actually what we're trying to solve for So what's the other force that's acting on this? Well, remember, so we have this sort of external here, so that's gonna be in the positive direction. So let's say let's say this is the positive direction. So now what other force is there? Well, let's see. We have an induced current. That's I I n D. That is in the presence of a magnetic field. So that means that there's actually going to be a magnetic force here. We want thes some of those horses to equal zero. So remember that the strength of this magnetic field on a current carrying wire So the strength of this magnetic force is I l b So we're gonna have I'll be and this is actually gonna be our induced currents. So what happens is our f external. The magnitude of that force is gonna be equal to this force because these things are gonna perfectly balanced each other out. Right? So if we set these two things equal to each other, the external force is just going to be the induced current times L b So do we have everything we need to solve the problem? We have the induced current. We actually figured that out in the first part. We have the magnetic field is and we also have the length of the loop. So I know that kind of kind of the backwards. So the f external is going to be the induced current, which is three amps times the length of the rod, which is 0.2 and Now we have the strength of the field, which is 0.5. So that means that the force that we need to produce on this loop in order to keep it pulling in at a constant velocity or keep it exiting is equal to 0.3 Newton's. And that is our answer. All right, so that was kind of tricky, but hopefully you guys, hopefully I don't have any questions. If you do, let me know. And thanks for watching. We'll see you the next one.