Forces on Loops Exiting Magnetic Fields

by Patrick Ford
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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.