Motional EMF - Online Tutor, Practice Problems & Exam Prep

Motional EMF

Hey, guys. So for this video, we're gonna be taking a look at a special kind of EMF called motional EMF. Let's check it out. So remember the theme of these last couple videos has been about is that the changing magnetic flux through a loop or something like that produces an induced EMF. 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 motional EMF. So what I want you guys to take away from this video is that really motional EMF is just a special case of Faraday's law. Alright? 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, the 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 rod 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 going to do is we're going to take our fingers and we're going to point our fingers in the direction of the magnetic field, which points into the page. And we also want our thumb to be pointing off towards the right, and then 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 build up 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 sin of theta , 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 we have an electric field that points sort of through this conductor like this, and what ends up 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 it that they experience due to this magnetic field is actually 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 × v b . We can sort of drop this sin of theta 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 you or sorry, straight into the page. So that means that the angle is equal to 90 there. So what that means is that so we can cancel out the charges involved, and that means we have a relationship between the electric field e and the velocity with the magnetic field, so e = v b . Now what that's basically means is that we've now have an EMF that is induced on the charges or in this conducting rod. The EMF is just equal to Vb × l . So where do we actually come up with this equation? Well, remember, way back from a couple chapters ago, that the voltage, which is really just a, which is really what the EMF 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, which you have a relationship with the velocity and the magnetic field, that's just v b , 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. It's like a special kind of EMF that's due to motion.

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

Forces on Loops Exiting Magnetic Fields

Hey, guys. Let's take a look at this example problem. We're going to 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 of the induced current is when the loop is halfway out of the field. So for this first part right here, what we need to figure out is the magnitude and the direction of some induced currents like this. So, we're going to have to use Faraday's law for motional EMF. Right? So we know that if we want the induced currents, we're going to have to relate that to the ε_induced divided by the resistance. Now this ε_induced is from motional EMF because this loop is moving outside of this field. So what we can do is say that the induced current here is just going to be the motional EMF. So instead of I IND, I'll write mot for motional, which means we're going to replace this with VBL divided by r. That means that the induced current is going to be VBLr. 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 uniform. I have the length of the loop here, L, and I have what the resistance is. So this tells me everything I need to know about the induced currents. That's just going to be 12 meters 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. The induced current is equal to 3 amps. Now how do we figure out what the direction is? Well, remember whenever we're trying to find out what the direction of an induced current is, we're going to have to use Lenz's law. So the direction of the induced current is going to be given to us using Lenz's law. First, what we have to do is we have to figure out what is the magnetic field, and then we have to figure out what is the change in the magnetic flux, and that will tell us the direction of the induced b field. Right? So, we're going to use both of these things here to figure out what the induced b field looks like. And then from here, we can figure out what the induced current direction is. So that's always how we do Lenz's 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 point into the page sort of away from you. Now the change in the magnetic flux, what's happening here? Well, remember that the change in the magnetic flux depends on 3 variables, b a and cosine of theta. Now what's happening here is that just as we saw what the equation for motional EMF comes from, it comes from the fact that this area is changing. Same thing's happening here. So the magnetic field isn't changing, and the cosine of the angle isn't changing. What's happening here is that the area that this loop goes through or 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. That's our change in magnetic flux is going to be going downwards. It's going to be decreasing. So Lenz's 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 b field. So what happens is it 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 going to find out what the direction of that induced field is by using our right hands. So what we're going to 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 fingers should be curling in the direction of that induced current. So if our fingers do 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 going to be going this way, like this, and that's going to be the direction of my induced currents. So that means here, it's going to be going this way, this way, that way, and 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 of the external force that is needed to keep this loop exiting at a constant velocity. Okay. So 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 0. So what they're actually asking us is what is the external force required so that the acceleration is equal to 0? So we're actually going to be talking about magnetic forces and so we're going to 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 going to be mass times acceleration. We know this a is equal to 0, 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 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 going to 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 IND, that is in the presence of a magnetic field. So that means that there's actually going to be a magnetic force here, and we want the sum of those forces to equal 0. So remember that the strength of this magnetic field on a current-carrying wire, so that the strength of this magnetic force is ILB. So we're going to have ILB, and this is actually going to be our induced current. So what happens is our F external, the magnitude of that force is going to be equal to this force because these things are going to perfectly balance 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 LB. 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 they can kinda do that backwards. So the f external is going to be the induced current, which is 3 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 newtons, and that is our answer. Alright? So that was kinda tricky, but hopefully, you guys, hopefully, you guys don't have any questions. If you do, let me know. And thanks for watching. I'll see you in the next one.