Crossed Fields, the Cyclotron and the Hall Effect

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Push a proton round and round in a magnetic field, kicking it with energy again and again. It gets faster, and its circle keeps growing. So must the timing of the kicks keep changing? Most students say yes. Keep that answer in mind, we will test it today. This topic comes up often in JEE, so let's begin.

0:26Crossed fields: the velocity selector

So, crossed fields. Take an electric field E and a magnetic field B, at right angles to each other and to the velocity v of a charge. The arrows on top mean vectors, and the upside-down T sign means perpendicular. Okay? The charge q feels two forces: the electric force q E one way, and the magnetic force q v B the other way. So when will it go straight through, without bending? When the two forces balance, exactly! q E equals q v B, so v equals E over B, nice and simple. The charge cancels and the mass never appears, so this works whatever the charge or mass. That is a velocity selector. Clear? Now, which way does a faster particle bend? Many students guess the electric way, and that is the trap. For a faster particle, q v B is bigger than q E, so the magnetic force wins and it bends the magnetic way. Slower ones bend the electric way. Got it?

1:28The mass spectrometer

Now let us put the selector to work, in a mass spectrometer. Every ion leaving the selector has the same speed, E over B, right? Then a second field bends the ions into semicircles. It is B with a little dash, read as B prime, and the radius is m v over q B prime. So with the same speed and charge, which ions make the bigger semicircle, heavy or light? The heavy ones, exactly! The curly sign means proportional to, so r is proportional to m, and the different masses separate. With me? Measure r, put in v equals E over B, and q over m equals E, over B times B prime times r. Okay? Without a selector, ions accelerated from rest through a voltage, capital V, and circling with radius r in a field B, have mass q B squared r squared, over two V, where V is a voltage, not a speed. Clear?

2:28Released from rest, and parallel fields

Now release a charge from rest in crossed fields. At first, only the electric field pulls it. So does it keep speeding up along E forever? It does not, because once it moves, the magnetic force starts bending it. It follows a looping curve called a cycloid, drifting at speed E over B, perpendicular to both fields. Right? The farthest it gets along E is two times m times E, over q B squared, and its greatest speed is two E over B, double the drift. Got it? Now a different case: E parallel to B, shown by the two vertical lines, and a charge that has some velocity across the fields. The velocity along the fields grows uniformly, while the part across them goes round in a circle. That gives a helix whose pitch keeps increasing. Careful: released from rest here, it simply moves in a straight line along the fields. With me?

3:25The cyclotron: one fixed frequency

Now, back to our puzzle, the cyclotron. The ion circles inside two hollow D-shaped boxes called dees, and a voltage across the gap kicks it at each crossing. Remember: does the period of a circling charge depend on its speed? Exactly, it does not! The circle grows, but each half round takes the same time, so one fixed rhythm works. The oscillator frequency must be q B over two pi m, where pi is the Greek letter, the circle constant. So our puzzle's answer is no. Okay? Crossing the gap twice per round, with a gap voltage capital V, the ion gains two q V of energy per revolution. Right? It leaves at the dee radius, capital R, with its greatest kinetic energy: K max, the max written small below, equals q squared B squared R squared, over two m. So can this machine speed up electrons? The answer is no, and that is a common trap. An electron's mass changes quickly with speed, so its frequency changes and the kicks fall out of step. And neutral particles feel no force at all. Clear?

4:36The Hall effect

Last idea, the Hall effect. Take a strip carrying a current, capital I, with a field B along its thickness, small t, okay? The carriers move along the strip. So what does the field do to them? It pushes them sideways onto one face, yes! So a voltage appears across the strip, at right angles to the current. This Hall voltage, V with a small H below it, is I B over n q t, with n the number of carriers per unit volume. Got it? For the same current, positive and negative carriers pile up on the same face, so the sign of the Hall voltage gives the sign of the carriers, and its size gives n, the carrier density. With me?

5:20Example 1: velocity selector speed, E 3x10^4 V/m, B 0.02 T

Okay, the examples. In a velocity selector, E is three times ten to the power four volts per metre, written V m to the power minus one, and B is zero point zero two tesla, capital T on screen. Which particles pass straight through? Pause the video and try it first. Done? Okay, let's check. Straight through means v equals E over B, whatever the charge or mass. So, what is three times ten to the power four, over zero point zero two? Zero point zero two is two times ten to the power minus two. So v is one point five times ten to the power six metres per second, written m s to the power minus one. Right? So every particle with that speed goes straight through, whatever it is. If you tried to work out which particle, that was the trap. Clear?

6:17Example 2: proton cyclotron frequency and KE, R 0.5 m, 1 T

Now a harder one: a cyclotron has dees of radius zero point five metre, and a field of one tesla. Find the oscillator frequency for protons, and their final kinetic energy. Pause the video and try it first. Done? Okay, let's check. In f equals q B over two pi m, put one point six times ten to the power minus nineteen, times one, on top, and two pi times one point six seven times ten to the power minus twenty seven at the bottom. That is about fifteen point two megahertz, and the wavy sign means approximately. Got it? Now the final energy. The proton circles at many radii on its way out. So which radius goes into the formula? The dee radius, exactly, since the proton leaves at the edge with its greatest energy. R squared is zero point five squared, which is zero point two five metre squared. Right? Now substitute into K max. On top, one point six times ten to the power minus nineteen, squared, times zero point two five; at the bottom, two times one point six seven times ten to the power minus twenty seven. That is about one point nine two times ten to the power minus twelve joule. How many mega electron volts, M e V on screen, is that? One mega electron volt is one point six times ten to the power minus thirteen joule, so one point nine two over zero point one six gives twelve mega electron volts. And one fixed frequency took the proton all the way, which settles our puzzle. Clear?

8:01Quick revision: v = E/B, cyclotron f and K max, Hall voltage

So, quick revision. The selector passes speed E over B, and m v over q B prime sorts ions by mass. In a cyclotron, f equals q B over two pi m stays fixed, and K max is q squared B squared R squared, over two m. And the sign of the Hall voltage, I B over n q t, gives the sign of the carriers. Now go and try the practice problems on this page. All the best!