(II) A conducting rod rests on two long frictionless parallel ...

Question

(II) A conducting rod rests on two long frictionless parallel rails in a magnetic field $\vec{B}$ (⊥ to the rails and rod) as in Fig. 29–53. (a) If the rails are horizontal and the rod is given an initial push, will the rod travel at constant speed even though a magnetic field is present? (b) Suppose at t = 0, when the rod has speed v = v₀, the two rails are connected electrically by a wire from point a to point b. Assuming the rod has resistance R and the rails have negligible resistance, determine the speed of the rod as a function of time. Discuss your answer.

Accepted Answer

Welcome back everyone in its laboratory, a physicist constructs a square of mass M and a side length L. It has three sides made of non conducting materials and a conducting wire as its remaining side. If it is put on top of two isolated frictionless metal rods, P and Q and given a push with an initial velocity of V I in the presence of a magnetic field B into the screen as shown in the figure below. First, will it keep moving with the same velocity? And second, if the rods are electrically connected and the resistance of the wire is R determine the expression for the velocity of the square. In terms of time T here we have a diagram showing our setup. And for our answer choices, A says it will move with the same velocity and the velocity of the square in terms of time T equals Vie to the negative bl square divided by Mr multiplied by T B says it's yes. And Vie raised to the power of negative BLT divided by Mr C says no and that VT equals vie to the negative bl squared T divided by Mr R and D says no and Vie to the negative BLT divided by Mr now we have two situations here. So let's try to think both of them through. So first, we want to figure out that or we want to figure out if rather our rod will be, will keep moving with the same velocity. OK? If it is pushed along or square. Now, to help us figure this out, there's a very important part in our problem statement. Our problem statement tells us that initially our frictionless metal rods, P and Q are isolated. What does that mean? Well, that means that the rods P and Q are disconnected. And in this case, there would be no current in the wire as it moves to the right. Hence, no magnetic force would act on the wire as the rods are frictionless, then the square keeps moving with the same velocity. So for the first part of our problem, yes, it will keep moving. OK with the same velocity. Now, let's move on to the next part of our problem. If the rods are electrically connected and the resistance of the wires are what's an expression for the velocity of the square in terms of the time T, now this one is a bit different because now we are specifically told that they are connected. OK. In other words, we can imagine here that we have our two rods, P and Q. OK? And no P and Q are connected. No, we have our rod, I think it's in blue here. We have, sorry, we have our conducting wire rather, it's in blue here. OK. And we know that our conducting wire has a length of L and you know what, it should actually be the same length as the distance between P and Q. So let me adjust my diagram. OK. And now we want to figure out the velocity, OK, of the square as a function of our time. T So the question is how can we figure out what that velocity is? Well as the wire moves to the right? OK. Then the area enclosed gets larger. No, this results in an, in an EMF as per Lindsay's law recall that Lindsay's law tells us that an EMF is induced. OK? As the flux changes over time, what's causing this change in flux? Well, our area enclosed is getting larger and recall that flux is equal to the strength of the magnetic field multiplied by the area. So this is really then OK, the derivative of B with respect to time where area is what's changing. So it's really be multiplied by D AD T. OK. Which we know that our change in area is going to be equal to, OK. The product of the velocity and the time multiplied by the length. OK. With respect to T. So here this is how we can represent our induced EMF no, for a uniform velocity case, OK. For a uniform velocity case, what do we know about our EMF? Well, our EMF is going to here then be equal to BVL because our velocity is uniform. OK. We also know that the EMF equals the current multiplied by the resistance. In this case, we are told the resistance of our wire is R. So that means the current equals the EMF divided by the resistance. So that's BVL divided by R. OK. So we know about our current, we know some stuff about our EMF but the question is how can we relate this to help us express our velocity in terms of our time? T well, what can help if we is if we start to think about the force that's involved because the magnetic force on the wire F is going to be equal to the current multiplied by the length multiplied by B. In that case, that's going to be negative BVL divided by R. OK? Multiplied by lb. And we take the opposite direction to which the wire is moving in as negative. So now we get that force to be equal to negative B squared L squared V divided by R. What else do we know about the force? Well, we know that force is equal to mass multiplied by acceleration and really acceleration is defined as the change in velocity over time. OK. So now if we equate this, we create both of these expressions for force, this tells us then that M DVD T is equal to negative V squared L squared V divided by R no. In this case, OK, we can get the V terms on one side and the T terms on the other. Let me make some space to the right. No, we can rewrite this as DV, divided by V equal to negative B squared L squared divided by Mr multiplied by DT No, this is pretty good. OK? Because no, OK. This should have been a capital R as well. No, we can integrate both of our sides. OK. We can integrate the left hand side between our initial velocity V and our F sorry, initial velocity V I and our final velocity V of DV prime. OK. Divided by V prime. And on the right hand side, we can see negative B squared L squared divided by Mr. All of those are constants multiplied by the derivative starting from zero or initial time of zero to the time T of the integral of BT prime. Now when we integrate both sides and let me put our result in red, OK. Then on the left hand side, we'll get the natural log of V divided by V I, OK. And on the right hand side, we will get negative B squared L squared divided by Mr multiplied by T. In that case, if we use the definition of the log, then we can say that we as a function of time is going to be equal to V I multiplied by E raised to the power of negative B squared L squared divided by Mr multiplied by T this would be the expression for our velocity for the velocity of the square as a function of the time. Therefore, this tells us that or if the rods are isolated, then it will keep moving with the same velocity. But if the rods are electrically connected, then the velocity is going to be equal. So sorry, the velocity of the square as a function of the time equals vie to the negative B squared L squared. OK. Negative B squared L squared divided by Mr multiplied by T OK. And when we look back at our answer, choices that tells us a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Electromagnetic Induction

Lorentz Force

Resistance and Ohm's Law

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