Slide – Exercise 43 – Motors and Generators


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Direction and Magnitude of the Force

A small metal ball is placed on a pair of electrified rails, as seen in this picture, it's in the presence of a magnetic field so that it is free to roll back and forth. So it's free to sort of move forward and backwards depending on the forces acting on it.
So if the current passing through the metal bowl is 90 A finds the direction and magnitude of the force on it. Now, at first, this might seem a bit large, but remember that we actually have quite a short circuit over here. So we, in reality, we might even get less than 90 A of resistance. But that's what the questions asking us. So these that.
Now, the first thing to find is the direction and that should be fairly easy. If we have the current flowing in this direction around the circuit, which we can see from the power supply, then we can just use the right-hand rule in order to figure out the direction of the force. Current moves that way, magnetic fields into the board. So the force will be upward and will be propelled toward the end of the rails.
So what's the magnitude of the force? Well, we have the strength of the magnetic field here 2T. This is a fairly strong field, but not the strongest you can get. The distance between the Rails is 1 cm. So that will be the length of the conductor inside the magnetic field. So if we have our equation, F=B I L sin θ, we can substitute in some of these numbers. We know that B is 2T, I is 90 A, and L is 1 cm, we can see from the direction of the magnetic field that is going to be perpendicular to the direction of the cart.
So we can ignore the sin θ, it'll just take on its maximum value. So shooting in those numbers, we have 2T x 90 A, x 1 cm, which gives us 1.8 N, and it will be away from the voltage source and toward the end of the rails.
So there's a force 1.8 N. Now if the ball weighs 5 g, find its speed offered as travelled along 70 cm of the rails. And that seems fairly accurate. I think, if we have a 1 cm diameter sphere, and it's made of metal, then it will tend to have a mass of between 5 and 10 g. So we can assume that the rails are 70 cm long. And during all of this, the ball will experience the acceleration due to this force.
So now we can use our equations of motion to figure out exactly how fast the ball ends up going. So in this question, we're given a displacement, r. and acceleration, we can figure out from a force and our initial velocity is of course zero, or in trying to find is V, the velocity of the ball after it's been accelerated over 70 cm.
So we know that F = ma, we have the mass, 5g or 5 times 10 to the minus three kilogrammes [sysco]. And we have the force so we can calculate acceleration.
Substituting in our numbers and rearranging the equation, we have v^2 = 2ar, or taking the square root, v^2 = 2ar. Of course, a is equal to f/m. And you can see I've already made that substitution here.
So now we're taking our numbers, F is going to be 1.8, M is going to be 5g. R is going to be 0.7 because we have to measure that in meters.
So now our equation looks like this. And of course, we can use a calculator to figure out exactly what that value will be. It turns out to be about 22.45 m/s, which is roughly the speed of a car.
So we've actually propelled this little pellet quite fast off the rails. If we used our equations of motion to figure out the time that it would take, it turns out that it would take about a 20th of a second for it to travel this 70 cm, obviously, we could increase the speed of the pellet going off the rails. Simply by increasing the strength of the magnetic field, increasing the current throw it or increasing the distance of the rails. In fact, if we get it fast enough, then we could be firing this little pellet-like a bullet from a gun.

Describe the amount of torque on the coil of an electric motor

Which equation can we use to describe the amount of torque on the coil of an electric motor?
Well, by now you should know this. F equals N-B-I-A cosine theta. Of course, that’s the torque, not the force.
So it would be force times distance where the distance is the radius of the coil.
So the equation in part A is sometimes zero because cosine theta will sometimes be zero, right? So why is it that the coil keeps spinning even though the torque on it is sometimes zero?
Well, the answer is momentum or inertia. When the torque is zero, the angular momentum of the coil is usually sufficient to make it keep spinning.
If the force or the torque suddenly shuts off, it’s not going to stop moving instantly. It’s going to keep moving a bit and it will move far enough that the torque will actually stop being zero and become a positive number again.
So that means that it will keep spinning.
If the motor isn’t spinning, then it doesn’t have any of that angular momentum. So, in fact, it won’t actually start and you will need a little push to get it going.