A Conductor Moving in a Magnetic Field

Transcript

0:00Introduction

Hello everyone! Let's start with a strange question. An aeroplane is flying high in the sky, and its metal wings stretch thirty metres from tip to tip. There is no battery connected between the wing tips. So, is there a voltage between them? Most students say no. But the earth has a magnetic field, and those wings move through it. By the end of this video, you will calculate it yourself. This idea is called motional emf, and it comes up often in JEE, so let's begin.

0:31Why a moving rod develops an emf

So, take a straight metal rod of length l, small l. It moves with velocity v, perpendicular to its length, and the magnetic field B is perpendicular to both. The arrows on v and B mean they are vectors, and the upside-down T sign means perpendicular. Okay? Now, the rod is full of free charges, and they move with it. So tell me, what does a charge moving in a magnetic field feel? It feels a magnetic force, yes, exactly! Each charge q feels q times v cross B; the cross means the cross product. In this set-up, that force points along the rod. Right? In a metal, the free charges are electrons, with negative q, so they are pushed opposite to v cross B, towards one end. That end turns negative, and the other end, which loses electrons, turns positive. Clear? Now, do these charges keep piling up at the ends forever? No, they stop, and here is why. The separated charges create an electric field, capital E, inside the rod, which pulls them back. With no closed circuit, the charges settle when q times E balances q times v times B, the magnetic force. The double arrow means which gives, so E equals v times B, okay? And an electric field E along the length l gives a potential difference of E times l. So the emf, the curly E on screen, equals B times v times l. This is motional emf: an emf produced by motion alone. With me?

2:11The key result and its conditions

So, curly E equals B times v times l, but only when the rod, the velocity and the field are all mutually perpendicular. Okay? In general, the emf is the integral of v cross B, dot d l, along the rod. Here d l, with its arrow, is a tiny piece of the rod, and the dot is the dot product. Clear? So only the part of v cross B along the rod counts. And the rule is this: if the three are not mutually perpendicular, take the components that are. Got it? For instance, the field is perpendicular to the plane, but the velocity makes an angle theta, the Greek letter, with the rod. Which part of the velocity counts? Only the component perpendicular to the rod, which is v sine theta, that is correct! The part along the rod does nothing. So curly E equals B times l times v, times sine theta. Right?

3:07Which end is at the higher potential?

Now, let us find which end becomes positive. Positive charges are pushed along v cross B, so the end that v cross B points to is at the higher potential. Okay? Let's try one. The rod lies along the y axis, it moves in the plus x direction, and the field is along plus z. Clear? Now you tell me: which way does v cross B point? Let's see. The letters i, j and k with the little hats are the unit vectors along x, y and z. So v cross B is along i hat cross k hat, which is minus j hat. Did you get plus j? Careful, because k cross i is plus j, so i cross k is minus j. With me? So positive charge is pushed towards minus y, and the end at lower y is at the higher potential. Okay?

3:57Bent or curved wires: the effective length

Now, what if the wire is bent, or curved? In a uniform field, replace the wire by the straight line joining its two ends, and use that as the length. Right? So, take a semicircular wire of radius R, moving perpendicular to its diameter, with the field normal to its plane. Which length do you use? The diameter, two R, that is right! The l with a small e f f below it means the effective length, and here it equals two R, okay? Many students use the arc length, pi times R, and that is the trap, so don't fall for it. Got it?

4:35Trap: when is there no emf at all?

Now, a quick trap. Suppose a rod slides along its own length, in a field perpendicular to it. Is there an emf? There is no emf at all, and many students get this wrong! Here the velocity is along the rod, so v cross B is perpendicular to the rod, and nothing pushes the charges along it. Clear? And the second case is when the velocity is parallel to the field. Then v cross B is zero, so again there is no emf. Okay? In both cases, v cross B, dot l, is zero. So always check the directions first. Right?

5:12Example 1: emf of a 0.5 m rod at 4 m/s in a 0.2 T field

Now our first example, an easy one. Here is a rod of length zero point five metres, written m, moving at four metres per second, the m s to the power minus one on screen. It moves perpendicular to its length, and to a field of zero point two tesla, and the capital T stands for tesla. Find the emf across it. Pause the video and try it first. Done? Okay, let's check. Before using the formula, what must you check first? The directions, exactly! The rod, the velocity and the field are all mutually perpendicular, so curly E equals B times v times l directly. Right? So, zero point two times four times zero point five, which gives zero point four volts, capital V for volt. That's it, just one line! Got it?

6:08Example 2: emf between aeroplane wing tips, vertical B

Now, remember our aeroplane? Let's finish it. An aeroplane with a wingspan of thirty metres flies horizontally at seven hundred and twenty kilometres per hour, the k m h to the power minus one. The vertical component of the earth's field there is four times ten to the power minus five tesla. Find the emf between the wing tips. Pause the video and try it first. Done? Okay, let's check. Can you put seven hundred and twenty straight into the formula? No, you cannot, and that is the first trap! The formula needs metres per second. Seven hundred and twenty times one thousand, over three thousand six hundred, gives two hundred metres per second. Clear? Now, the earth's field has a horizontal part and a vertical part. Which one do we use here? The vertical one, and here is why. The wings and the velocity are both horizontal, so the horizontal part of the field lies in the same plane. Then v cross B either points straight up or down, across the wings, or is zero, so it never acts along the wings and gives no emf. With me? So we use the vertical component, the B with a small V below it. It is perpendicular to both the wings and the velocity, so curly E equals B vertical times l times v. That is four times ten to the power minus five, times thirty, times two hundred. Okay? Thirty times two hundred is six thousand, and six thousand times four times ten to the power minus five gives zero point two four volts. So yes, there really is a voltage between the wing tips, made by motion alone. Puzzle solved, right?

7:54Quick revision: Bvl, higher potential end, when emf is zero

So, quick revision. First, curly E equals B times v times l when the rod, the velocity and the field are mutually perpendicular; otherwise, use components. Second, the end that v cross B points to is at the higher potential, and for a bent wire, join the ends. Third, there is no emf if the rod moves along itself, or if the velocity is parallel to the field. Now go and try the practice problems on this page. All the best!