30. Induction and Inductance
Motional EMF
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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 have 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 sine 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 for with the magnetic force. So these things will feel a sort of electric force in this direction that's gonna be FE, 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 v b. We can sort of drop this sine 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 is equal to 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 times 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 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. So you might be wondering what this has to do with Faraday's law. Well, what happens is now things get interesting if we're if we attach this conducting rod here to a conducting wire like this and make a loop. Because now what happens 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 circuits throughout the circuit is constantly changing. So we have to use Faraday's law on the circuit that this thing creates. And 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 it the magnetic field? Is it the area? Or is it the angle? We'll see. The magnetic field is uniform and it points into the page and it never changes. The angle of this thing makes, right, their velocity is to the right. Magnetic field is out or is into the page, never changes. So what's really going on is that the area is changing. That's our changing variable. So we can write our Delta phi over Delta T as B times Delta A times cosine of theta over delta t. Now what happens is the or sorry. That's, yeah. That's cosine of theta, where this cosine of the angle here is, remember, we're 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 point in the same exact direction. So that means that the cosine of the angle is equals to 0. 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 then 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 rectangle that we've made is actually equal to L times Delta X, and that's gonna be over Delta T. So remember we've gotten rid of that cosine of term here. Okay. So we have 2 constants over here, the magnetic field and the length of the rod, and we have 2 variables, delta x over delta t. But if you remember back from way back in, like, early physics 1, 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 the conducting rod is moving through the field. So what basically happens is that the induced EMF, which remember is equal to delta phi over delta t, is just equal to VBL, which is the exact same equation that we got from before. It's no coincidence that we got the same exact equation because remember, motional EMF is just an applied Faraday's law. So, really, this is the equation that we're gonna be using to solve these problems. Alright? So let's get to it. So we have a circuit that's below here, and the wire has resistance of 10 milliohms. And we have to do is figure out what the induced current is if the length of the bar is 10 centimeters. And we're given some other conditions right here. So let's write out our variables. We're supposed to be figuring out what the strength of the induced, current. Sorry. That's gonna be the induced current. So our induced current is always gonna be related to our induced EMF. That's epsilon induced divided by the resistance. That's just Ohm's law. Now remember, what happens is that this induced EMF, when we're talking about motional EMF, is actually given a special form. That's gonna be vb times l. That's the new equation that we have divided by r. So let's take a look here. I've got my I induced is gonna be the magnetic field or, sorry, the velocity first. The velocity is 25 meters per second, which is 0.25. Now we have 0.2 Tesla. And now we have the length of the bar, which is 10 centimeters. So it's gonna be 0 point 1. Now we have the resistance is equal to 10 milli ohms, which is actually point this is 0.010. Right? So if you work all this out, you're gonna get an induced current that's equal to 0.5 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? Well, let's see. How do we get power? Remember that power actually has 3 equations. Again, this is sort of an equation that we used a little while ago. So we have three forms of it. We have the voltage times the current. 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 Epsilon I equals I squared R equals Epsilon squared over r. Remember that these epsilons 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 0.5 squared times the resistance, which is 0.010. So what I get when I plug that in is I get 2.5 times 10 to the minus 3, and that's in watts. So that's gonna be my power output. This is actually equal to 2.5 milliwatts as my final answer. Alright, guys. So that's it. We're gonna get a couple more practice problems, and I'll see you guys in the next one.
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