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Example-Terminal velocity
Let’s consider an interesting example associated with motional emf. Let’s assume that we have an inclined rail system which is making an angle of Theta with the horizontal. And along this rail we have a moveable conducting bar. Something like this. And we apply a magnetic field pointing out in this direction. In other words, it is perpendicular to the floor. And this moveable rail of course, when we release it, assuming that the rail is frictionless, it slides down along this rail. Now, as it slides down it will eventually reach to a terminal velocity. And let’s try to determine this terminal velocity.
Let’s say that the length of this rail is l. And, first of all, as soon as we release this rail of course, it’s going to be sliding down along the influence of the gravitational force acting along the incline. Well, as it slides down then this conducting loop area is going to get smaller and smaller. It means that the magnetic flux through this area will get smaller and smaller. So, we are going to end up with a change in magnetic flux through the area surrounding the bidus [sp] conducting loop. Therefore from [??] we’re going to end up with an induced emf and hence induced current. And that current is going to flow in a direction such that it will oppose it’s cause. And the cause is the the decrease in flux due to the decrease in the area surrounding the bidus conducting loop as the rod slides down. Well, in that case, the induced current will show up such that it will try to oppose its cause or it will try to compensate that decrease and it can do that only by generating a magnetic field. Which is going to be in a way, adding to this field. Well, of course it’s not going to be in exactly the same direction in this field because this field is not really perpendicular to this surface; the surface surrounded by the conducting loop. But in order to try to compensate again to that magnetic field, it should be somehow in this direction. And if we apply right hand rule holding the thumb so in the directional flow of current and we want it’s magnetic field is coming out of the plane surrounded by this loop. Therefore, that b has to be, let’s say the b prime, the induced current’s magnetic field through this loop should be pointing something like this in order to have that, the associated current has to be flowing in counter clockwise direction along that loop from Land’s Law. Well, once this current is induced, then we’re going to end up with a situation that a current carrying conductor in an external magnetic field of this b. Which will generate a force and that force is going to be from il cross b holding the right hand thumb in the flow of directional current and curling toward the external magnetic field b. il cross b is going to generate a force something like this, in this direction. And that force, therefore, a component of this force is going to balance the gravitational force acting along the incline plane and as a result of that balance, the force on the rod is going to be equal to 0 as it slides down. So, it will reach that terminal velocity. In other words, at that instant, in that moment, it’s acceleration is going to be equal to 0.
To be able to see the orientation of these forces, let’s introduce a coordinate system. And also look at the problem from this front view. And, therefore, if we look at the problem from the front view, here is the incline plane with an angle Theta. And, let’s see, here is the rod sitting over here and now the current i is coming out of plane at this point. And, the forces acting on this system, we have mg, due to the gravitational field’s weight of this rod. And, we are going to introduce a coordinate system such that it is perpendicular to the incline plane. And as well as parallel to the incline plane. And if you do that, parallel to the incline plane like this, and this is one force due to the gravity acting in downward direction. Of course we are going to have force acting on this sliding bar, current carrying conducting bar, due to the external magnetic field B. And from this point of view, i l is coming out, b is on the plane pointing up. i l cross b is therefor going to give us a force. A magnetic force acting in this direction. Let’s call this one f sub b. Relative to this coordinate system, we are going to have two components of these forces. By taking the projection of mg along the X direction along the incline plane. Another component of this, if this angle is Theta, therefore, this angle will also be Theta because they will have mutual perpendicular sides. This side is perpendicular to this side and this side is perpendicular to this side. So, these are congruent angles and we are going to have, therefore, these two angles to be equal to one another. So, the force along the incline plane, the component of mg along the incline plane using this right triangle will be equal to mg sin Theta. And of course, we have its component along the vertical direction, which is this one. And for the other force, again, if this angle is Theta we can easily see that this angle and that angle is equal to one another. This too will be Theta. So, if we take the projection of this force, again, along the direction of incline plane we are going to end up that component aligning like this and the other component will be in the same direction. With the component of the mg perpendicular to the incline plane. So, whenever these two forces become equal to one another, then the net force along the incline plane will be equal to 0. It means that the acceleration will be equal to 0. So, the velocity of the object as it slides down will be constant and therefore, that’s what we call as the terminal velocity. And that is what we would like to calculate. V terminal as the conducting bar slides down is the question mark. So, this is conducting rail and this is a conducting bar with length l. Alright, so if we write down of course, the vertical forces, or the perpendicular forces relative to the incline plane are going to be balanced by the normal force generated due to the con by the surfaces. In other words, due to the contact forces between the bar and incline plane surface. And, since we don’t have any friction, we are assuming that the friction is 0 between the bar and the rail, therefore we are not going to end up with a frictional force. Then these are the only forces acting on the system. Since there is no motion in this perpendicular direction, naturally, this normal force is going to be equal to sum of these two components. And, again, the system is now moving at a constant velocity along the incline plane. Therefore, the component of the magnetic force along the incline plane should be equal to component of the weight of this bar down along the incline plane.
If we write that down over here, or if we just write down the net forces acting on the system, sum of the forces along, let’s say, calling this direction as x and the other direction as y, along x is going to give us mg sin Theta, this is in positive x direction minus the x component of the magnetic force and that is using this right triangle over here, cosine of Theta, should add up to mass times acceleration in x direction from Newton’s Law. And, we want that acceleration to be 0. That’s what’s going to happen for the terminal velocity. Therefore, this whole equation is going to be equal to 0. Similarly, some of the forces along y direction is going to give us normal force, which is in positive y direction and minus mg cos Theta. The component of ipsa b along y direction is, or say negative y direction, and that will be minus fb sin Theta and that will be equal to mass times acceleration along y direction. There is no motion in that direction , so a sub Y is also equal to 0. And, these are the net forces acting on this rod as it slides down. Let’s call this equation as number 1 and this one as equation number 2. From equation number 1 we can say that mg sin Theta is going to be equal to fb cos Theta. fb is equal to idl cross b integrated over the length over this current carrying wire. Now, of course, there will be magnetic force acting on these segments of this rail, but obviously they are going to be in opposite directions with equal magnitudes. So, they will basically cancel, and also it’s a fixed rail and the same thing is over here then those forces will be balanced by the reaction forces. Well, if we look over here, the magnitude of this force will be equal to idl magnitude, B magnitude times cosine of the angle between these two vectors. Well, again from our front view, idl is coming out of plane and b there is the magnetic field is in this direction. Direction is up and idl is coming out of plane, therefore, the angle between these two vectors is 90 degrees. So, we have sin 90 over here and sin 90 is equal to just 1 and we are going to integrate this whole thing along the length of that moving bar. Therefore, the boundaries are going to go from 0 to l. f sub b magnitude then will be equal to i’s constant b is constant take it outside of the inter-glow. Inter-glow from 0 to l will give us ibl.
Now, let’s try to obtain the value of i. Let’s say r represents the resistance of the sliding bar. And, the rail in that case i is going to be equal to, from Ohm’s Law, induced emf divided by that total resistance. Induced emf is negative of the change in flux with respect to time. Of course this negative sign is associated with the direction of flow of the current and so what we are interested in over here is the absolute value of this rate of change of flux. Phi sub b at that instant of time, as the rail slides down, is going to be equal to, let’s say this distance, in other words if you just take the flash photography for example as it slows down it’s going to be equal to b dot a. In other words, the area surrounded by this conducting loop. And that’s going to be equal to the width for an incremental displacement of this. l times dx, in other words, d pi b incremental change in flux is going to be equal to b dot da and that’s going to be equal to d magnitude, da magnitude times cos of the angle between these two vectors. In order to see that angle let’s go back to our diagram again. So what we are seeing is as the rod slides down, it will experience a change of let’s say, some dx distance as it slides down along this x direction, and at that moment, the change in flux is going to be as much as this swept area. Which is going to be equal to b from b dot da and that da is going to be equal to l times dx and the area of vector of this region will be perpendicular to this so it’s going to be a vector something like this. Alright, perpendicular to this region. And, the magnetic field vector at that point is pointing this way so we’re talking about this angle as the angle between b and da. In our three dimensional, I mean front view picture b is in this direction, da is going to be perpendicular to the area surrounded by this loop. In other words, it’s going to be perpendicular to this incline plane. So, da is going to be pointing like this, and the angle between these two vectors will be again, equal to Theta. Because this angle is equal to that angle. Moving down, then we have b magnitude da magnitude times cosine of Theta. Or, b times da is ldx times cosine of Theta. If we look at the rate of change of flux, d Phi b over dt then will be equal to b times l times dx over dt times cosine of Theta. So, only change in quantity is dx with respect to time. b and length of the rod and the angle between b and da, Theta remains constant. But dx over dt is the velocity of this bar as it slides down. So, this becomes equal to b times l times v times cosine of Theta. So, once we determine the d Phi over dt, then the induced current is going to be equal to, from Ohm’s Law, Epsilon over the resistance, blv cosine Theta over r. Now, if we go back to equation 1 which eventually gave us this expression over here, mg sin Theta is equal to fb cos Theta. Let’s say from 1, mg sin Theta is equal to f sub b times cos Theta and f sub b in explicit form is ibl. Now we have cos Theta. And, therefore mg sin Theta is equal to 4i. We can write down its equivalent blv cos Theta divided by r and we have another b, another l, another cos Theta in the equation. In other words, this part is the explicit form of the current. Therefore, this equation takes the form of mg sin Theta is equal to b square l square, v times cos square Theta divided by r. If you solve this expression for the velocity or terminal velocity, that will be equal to mgr sin Theta divided by b square l square and cos square of Theta. So, these are all the given quantities: mass of the rod, gravitational acceleration, resistance of the loop, the angle of the incline plane, external magnetic field and length of the moving rod. Therefore, the terminal velocity of this bar as it moves down along the incline plane is going to be equal to this quantity.