A Rod Sliding on Rails

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Two smooth metal rails lie in a magnetic field, and they are joined at one end by a resistor. We lay a metal rod across them, give it one push, and let go. With no friction anywhere, will the rod slide on forever? Many students would say yes, but that magnetic field changes everything. By the end of this video, you will find exactly how far the rod goes. This set-up comes up often in JEE, so let's begin.

0:30The set-up: emf, current and its direction

Here is our set-up. Two rails, a distance small l apart, are joined by a resistance R on the left. The rod slides right with speed v, and the little crosses mean the field B, with its vector arrow, points into the page. The rod, its velocity and the field are mutually perpendicular, so the emf, the curly E on screen, is B times l times v. Okay? The rod, the rails and the resistor make a closed circuit, so a current I flows. By Ohm's law, I equals B times l times v, over R, clear? But which way does it flow? Think about the flux through the loop to the left of the rod. Let's see. As the rod moves right, the loop grows, so the flux into the page grows. By Lenz's law, the induced current opposes this growth, so inside the loop it makes its own field out of the page. That needs an anticlockwise current, like the orange arrows, going up through the rod. With me? Now, this current flows in the rod, inside a magnetic field. So what does the rod feel? It feels a magnetic force, yes, exactly! This is F, with the short word mag below it, for magnetic, and it equals B times I times l. Put in the current, and it becomes B squared times l squared times v, over R, opposing the velocity, as Lenz's law demands. So it acts like a brake. Right?

1:59Constant speed: work in, heat out

Now, to keep the speed constant, we must push forward with a force F equal to this brake: B squared times l squared times v, over R, okay? But we keep doing work on the rod, and its speed never changes. So where does that energy go? Let's track it. The power we supply is F times v, which is B squared times l squared times v squared, over R, right? And the heat produced every second is I squared times R, okay? Put in the current, and look, it is exactly the same! So all our work turns into heat. That is the energy balance at constant speed. Got it?

2:41A constant force: terminal velocity

Now, suppose instead we pull the rod with a constant force F, and let it move freely. Does it keep speeding up forever? No, it does not, and here is why. As the rod speeds up, the drag, B squared times l squared times v, over R, grows as v grows. Clear? So the speed creeps up towards the value where the drag equals F, okay? That is the terminal velocity, the v with a small T below it: v T equals F times R, over B squared times l squared. Got it? The time constant tau, the Greek letter, sets how quickly it gets there. Tau equals m times R, over B squared times l squared, with small m the mass of the rod. Clear?

3:28No push at all: the rod slows down

Now, back to our puzzle. The rod starts with speed v nought, the v with a small zero below it, and then nothing pushes it. Only the brake acts, so the speed dies away exponentially: v equals v nought times e to the power minus t over tau. Right? But this exponential never quite reaches zero. So does the rod travel an infinite distance? No, it does not, and this surprises many students! The distance adds up to a finite total, v nought times tau. So even with no friction, the rod does not slide forever. With me?

4:05Example 1: rod at 5 m/s, emf, current, force and power

Now our first example, an easy one. Here, a rod slides at five metres per second, the m s to the power minus one, on rails zero point five metres apart. The field is zero point four tesla, capital T for tesla, and the rails are joined by two ohms, the Greek capital omega. Find the emf, the current, the force needed to keep the speed constant, and the power dissipated. Pause the video and try it first. Done? Okay, let's check. As in our set-up, everything is mutually perpendicular, so curly E equals B times l times v. Zero point four times zero point five times five gives one volt, capital V for volt. Okay? The current is one volt over two ohms, which gives zero point five amperes, capital A for ampere. Right? Now, the speed is constant, so the net force on the rod is zero. Does that mean we need no push at all? Oh no, careful, that is the trap! The net force is zero only because our push balances the magnetic brake. F equals B times I times l: zero point four times zero point five times zero point five, which gives zero point one newton, capital N for newton. Clear? Next comes the power dissipated. I squared times R is zero point two five times two, which gives zero point five watts, capital W for watt. Got it? And here is a quick check from our energy balance. F times v is zero point one times five, again zero point five watts, and the tick mark says it matches. That is a perfect match between power in and heat out!

5:52Example 2: rod left alone, time constant, distance, heat

Now, remember our rod from the start? Let's finish it. Here, a rod of mass zero point one kilograms, the k g on screen, lies on rails zero point four metres apart, in a field of zero point five tesla, with a circuit resistance of two ohms. It is given a speed of two metres per second and left alone. Find the time constant, the total distance it travels, and the total heat produced. Pause the video and try it first. Done? Okay, let's check. Tau equals m times R, over B squared times l squared. On top, zero point one times two is zero point two. Below, B squared times l squared is zero point two five times zero point one six, which is zero point zero four. Zero point two over zero point zero four gives five seconds, small s for second. Okay? So the total distance is v nought times tau, two times five, which gives ten metres. There is the answer to our opening puzzle: even with no friction, the rod covers just ten metres. Right? Now, the current keeps falling as the rod slows down. So how do we get the total heat quickly? Use energy, yes, exactly! That is the topper's trick. All the kinetic energy of the rod ends up as heat. So the heat is one half times m times v nought squared: one half times zero point one times two squared, which gives zero point two joules, capital J for joule. See how fast that was?

7:30Quick revision: Blv, Fv = I2R, terminal velocity, v0 tau

So, quick revision. First, curly E is B times l times v, the current is that over R, and the force, B squared times l squared times v, over R, opposes the motion. Second, at constant speed, power in, F times v, equals heat out, I squared times R, okay? Third, the terminal velocity is F times R, over B squared times l squared, and with no push the rod covers v nought times tau, where tau is m times R, over B squared times l squared. Now go and try the practice problems on this page. All the best!