Rotating Conductors and the Generator

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Picture a wheel with ten metal spokes, spinning in the earth's magnetic field. Is there an emf between the axle and the rim? And is it ten times the emf of one spoke? Let's find out, and see how the A C generator works. This comes up often in JEE, so let's begin.

0:22A rod rotating about one end

So, here is a metal rod of length small l, turning about one end with angular velocity omega, the Greek letter. It turns in a plane perpendicular to the field, B with a vector arrow; the upside-down T sign means perpendicular. So the rod, its velocity and the field are mutually perpendicular. Okay? Now let x be the distance of a point of the rod from the axis. That point moves with speed v equal to omega times x, from zero at the axis up to omega times l at the free end. Right? For a rod moving in a straight line, the emf is B times l times v. So which speed goes in here? Many students take the end speed, omega times l, and get B times omega times l squared. Is that right? No, careful, that is the trap, and it doubles the answer! The speed grows evenly from zero to omega times l, so the average speed is omega times l, over two. Clear? So the emf, the curly E on screen, equals B times l, times omega times l over two. That is one half times B times omega times l squared, the key result. Got it?

1:37Discs, and two points on a rotating rod

Next, a metal disc of radius small r, rotating about its axis, in an axial field, along that axis. Each radius acts just like our rod, so from the centre to the rim, curly E is one half times B times omega times r squared. With me? Now two points of a rotating rod, at distances r one and r two from the axis, with the small one and two below. Take the emf out to r two, minus the emf out to r one. So curly E is one half times B times omega, times r two squared minus r one squared. Okay? Now try this one. Let our rod spin about its centre instead. What is the emf between its two ends? Each half is a rod of length l over two, turning about the centre, so it gives one half times B times omega, times l over two in brackets, squared. That is one eighth times B times omega times l squared. Right? But both emfs point from the centre outwards, or both inwards, so the two ends are at the same potential, and the emf between them is zero! Between the centre and one end, it is one eighth times B times omega times l squared. Clear?

2:54The rotating coil: an AC generator

Now the big one: a coil of capital N turns, and capital A is the area of each turn. It rotates with angular velocity omega about an axis in its own plane, perpendicular to the field. Okay? The flux through one turn is B times A, times cos of the angle between the field and the normal, and after a time t that angle is omega times t. So the flux linkage, capital N times capital phi, is N times B times A, times cos omega t. Right? By Faraday's law, curly E equals minus N times d phi over d t, and the minus sign is Lenz's law, the induced current opposing the change in flux. On differentiating, the minus signs cancel, so curly E equals N times B times A times omega, times sine omega t. This emf alternates in sign, and that is how the A C generator works! Clear? The largest value of sine omega t is one, so the peak emf, curly E nought, with a small zero below, is N times B times A times omega. Got it? In which position of the coil is the emf the greatest? When the flux through it is the greatest, right? Oh no, careful, it is just the opposite! The emf depends on how fast the flux changes, and the flux changes fastest as it passes through zero. So the emf is greatest when the plane is parallel to the field, the two vertical lines on screen. When the plane is perpendicular, the flux is the greatest, but the emf is zero. With me?

4:29Example 1: emf of a 10-spoke wheel in the earth's field

Now, back to our wheel, with ten metal spokes, each zero point five metres long, rotating at one hundred and twenty revolutions per minute, the rev per min on screen, in a plane normal to the horizontal component of the earth's field, zero point four times ten to the power minus four tesla, capital T for tesla. Find the emf between the axle and the rim. Pause the video and try it first. Done? Okay, let's check. One hundred and twenty revolutions per minute is two revolutions per second, and each revolution is two pi radians. So omega is four pi radians per second, the rad s to the power minus one. Right? Each spoke is a rod turning about one end, so it gives one half times B times omega times l squared. That is one half, times zero point four times ten to the power minus four, times four pi, times zero point two five, which is zero point five squared. Okay? Half of zero point four times ten to the power minus four is two times ten to the power minus five, and four pi times zero point two five is pi. So curly E is two pi times ten to the power minus five, about six point three times ten to the power minus five volts, capital V for volt. The wavy sign means approximately. Clear? Now the second part of our puzzle. With ten spokes, is the emf ten times this? No, it is not, and this is the trap! Every spoke joins the same axle to the same rim, so the spokes are in parallel, and the number of spokes does not matter. Puzzle solved, got it?

6:13Example 2: 100-turn coil at 50 rev/s, emf and its peak

Now the generator itself: a coil of one hundred turns and area zero point zero five square metres, the m squared, rotating at fifty revolutions per second in a field of zero point two tesla. Write the emf as a function of time, and find its peak value. Pause the video and try it first. Done? Okay, let's check. The peak is N times B times A times omega. So can we put omega equal to fifty, and get fifty volts? No, careful, because fifty is in revolutions per second. In radians per second, omega is two pi times fifty, which is one hundred pi, about three hundred and fourteen radians per second. Right? Now curly E nought is N times B times A times omega, which is one hundred, times zero point two, times zero point zero five, times three hundred and fourteen. The first three numbers multiply to exactly one, so the peak emf is three hundred and fourteen volts. Clear? At any time t, the emf is the peak times sine omega t. So curly E equals three hundred and fourteen times sine of one hundred pi t, in volts. With me?

7:27Quick revision: rod, disc and AC generator emf formulas

So, quick revision. First, a rod turning about one end gives one half times B times omega times l squared. Second, a disc gives one half times B times omega times r squared, and between r one and r two, one half times B times omega, times r two squared minus r one squared. Third, the generator gives N times B times A times omega, times sine omega t, with peak N times B times A times omega. Now go and try the practice problems on this page. You have got this. All the best!