Drift Velocity and Mobility

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You press a switch, and the lamp lights up at once. But the electrons in the wire crawl along at only a fraction of a millimetre per second. So how does the lamp light so fast? Hold on to that question. Today we learn drift velocity and mobility, an idea that comes up often in JEE current electricity. So, let's begin.

0:29Random motion, then a slow drift

So, let us look inside a metal. Each atom gives one or more free electrons, which move randomly at speeds of the order of ten to the power five metres per second, colliding with the lattice ions, the grey dots in the figure. The wavy sign means of the order of, and m s to the power minus one is metres per second, okay? Now let me ask you this. With electrons this fast, the wire must carry a huge current, correct? No, there is no current at all. The motion is random, in every direction, so the average velocity is zero, and there is no current. Right? Now switch on a field, E, with a little arrow on top because it is a vector; here it points to the right. Between collisions, every electron gets an acceleration e times E, over m, where e is the magnitude of the electron's charge and m its mass. Clear? The electron is negative, so it is pushed against the field. See the orange path of the electron, e with a small minus sign: it zigzags, and between collisions the field bends each piece slightly, too little to see in the drawing, so the path slowly creeps to the left. Now average over tau, the Greek letter tau, the mean time between collisions, called the relaxation time. You get a small average velocity opposite to the field, the green arrow: the drift velocity. Got it?

1:51Drift velocity and mobility

So, how big is this drift? The acceleration is e times E, over m, and on average it acts for a time tau. So what velocity does the electron pick up? Yes, exactly, acceleration times time! v d, the v with a small subscript d, equals e times E times tau, over m. The green label in the figure has vector arrows and a minus sign; the minus says the drift is opposite to the field. With me? Now, the drift speed per unit field is the mobility, mu, the Greek letter mu. Mu equals v d over E, which is e tau over m. Okay? Its unit is metre per second, over volt per metre, which gives metre squared per volt per second: m squared, V to the power minus one, s to the power minus one. And on the right of the figure, I equals n e A v d is what we do next. The other line, J equals sigma E, where sigma, the Greek letter sigma, is the conductivity, together with the formula for sigma, belongs to the next concept on Ohm's law. Right?

2:55From slow drift to current

Now, how does this drift give a current? I, the current, equals n times e times A times v d. And J, the current density, equals n e v d. Clear? Here n is the number of free electrons per unit volume, and capital A is the area of cross-section of the wire. Okay? Now the surprise. Typical drift speeds are only a fraction of a millimetre per second, m m s to the power minus one on screen. Got it? So, back to our puzzle. With such a slow drift, why does the lamp light the moment you press the switch? Here is the answer. The electric field is set up along the whole wire at nearly the speed of light. So the electrons all along the wire, even inside the lamp, start drifting almost together. And that solves our puzzle. With me?

3:49A fixed potential difference across a wire

Alright, one more result that comes up often. Put a potential difference V across a wire of length l, small l. The field inside is E equals V over l. Right? Put this field into our drift formula. What do you get? Yes, exactly: e times E times tau, over m, becomes e times tau times V, over m times l. Okay? Now look: there is no capital A in it. So, for a given potential difference, the drift speed does not depend on the area of cross-section. Clear?

4:23Example 1: drift speed in a 1 mm2 copper wire carrying 1.6 A

Now, Example one. Here is a copper wire of cross-section one square millimetre, m m squared, carrying one point six ampere. Copper has eight point five times ten to the power twenty eight free electrons per cubic metre, m cubed. Find the drift speed. Pause the video and try it first. Done? Okay, let's check. From I equals n e A v d, the drift speed v d equals I, over n times e times A, got it? Now the area. One millimetre is ten to the power minus three metre, so one square millimetre is ten to the power minus three square metre, correct? No, no, wait, that is the trap! The whole bracket, ten to the power minus three metre, is squared, giving ten to the power minus six square metre. Right? Now substitute: one point six, over eight point five times ten to the power twenty eight, times one point six times ten to the power minus nineteen, times ten to the power minus six. Here is a small trick: the one point six of the current cancels the one point six in the electron's charge! The powers give twenty eight minus nineteen minus six, which is three, so we get one over eight point five times ten to the power three. With me? That is about one point two times ten to the power minus four metre per second; the wavy equals sign means approximately. So about zero point one two millimetre per second, just as slow as in our puzzle. Clear?

5:53Example 2: drift speed when V, length or radius is doubled

Now Example two, a little harder. How does the drift speed change if, part a, the potential difference across a wire is doubled; part b, the length is doubled at the same potential difference; part c, the radius is doubled at the same potential difference? Hint: the potential difference is given, not the current. Pause the video and try it first. Done? Okay, let's check. The potential difference is given, so use v d equals e times tau times V, over m times l. Okay? Part a: V becomes two V; the arrow means becomes. Since v d is proportional to V, the drift speed doubles. Right? Part b: l becomes two l, at the same potential difference. The field, V over l, halves, so the drift speed halves. Got it? Now part c, the radius doubles, so the area becomes four times. And from I equals n e A v d, more area means less drift speed, so it becomes one quarter, correct? Oh no, careful, this is a common JEE trap! That works only when the current is fixed, and here the potential difference is fixed. Yes, r becomes two r, and the area, pi r squared, becomes four A, but our formula has no capital A in it at all. With me? So the drift speed is unchanged. The current changes instead: in I equals n e A v d, the area is four times and v d is the same, so the current is four times. Clear?

7:26Quick revision: vd, mobility, I = neAvd, area at fixed V

So, quick revision. Drift speed is e times E times tau, over m, opposite to the field, and mobility is v d over E, which is e tau over m. The current is n e A v d; the drift is slow, but the field is set up almost at once. And for a given potential difference, v d equals e times tau times V, over m times l, with no area in it. Now go and try the practice problems on this page. All the best!