Mixed Problems: Set 2 (Charged Particles)

Transcript

0:00Introduction

Hello everyone! Okay, a quick question before we start. Say a charged particle is going round in a circle, in a magnetic field. Now you double its kinetic energy. Does the circle become twice as big? Many students say yes. Keep your answer; we will check it soon. Today is Set two of our mixed JEE Main problems: charged particles in magnetic fields. Questions like these come up often in JEE Main. So, let's begin.

0:31The force on a moving charge

So, first the tool kit. Take a charge q, moving with velocity v in a magnetic field B; the small arrows on top mark vectors. It feels the force q v cross B, where the cross sign is the cross product. Its size is q v B sine theta, and theta, the Greek letter, is the angle between v and B, okay? This force is always perpendicular to both the velocity and the field. So let me ask you this: does it change the kinetic energy of the charge? It does not change it at all, exactly! The upside-down T sign means perpendicular. The force is perpendicular to the velocity, so it does no work. The speed and the kinetic energy, capital K, stay constant; only the direction turns. Right? And for a negative charge, like an electron, the force points opposite to v cross B, so always watch the sign of the charge. With me?

1:28The radius of the circle

Now, when v is perpendicular to B, the force has a constant size, q v B, and it keeps turning the charge, so the path is a circle. The magnetic force supplies the centripetal force: q v B equals m v squared over r, where m is the mass and r is the radius. The double arrow means implies, so r equals m v over q B, clear? Now, m v is the momentum, small p, so r is also p over q B, and the momentum equals the square root of two m K, so r equals root two m K, over q B; the tick-shaped sign is the square root. Got it? Now, back to our opening question. Look at this last form. If the kinetic energy doubles, what happens to the radius? Oh no, it does not double, and that is the trap! The radius goes as the square root of K, so it becomes root two times r, about one point four one times r; the wavy equals sign means approximately. So the circle grows by root two, not by two. Okay? And if the charge starts from rest and is accelerated through a voltage, capital V, it gains kinetic energy q V, and then r equals one over B, times the square root of two m V over q. Careful here, this capital V is the voltage, not the speed. Right?

2:46Period, frequency and the velocity selector

Next, the timing. The period, capital T, is two pi m over q B, and pi is the Greek letter, the circle constant. The cyclotron frequency, small f, is q B over two pi m, and omega, the Greek letter like a curly w, is the angular frequency, q B over m. Okay? Now, a faster charge moves on a bigger circle. So does it take longer to go round once? No, it takes exactly the same time! There is no v in any of these formulas, so they are independent of the speed. Bigger circle, faster run, same period. With me? Now the velocity selector. Here an electric field, capital E, is perpendicular to B, and both are perpendicular to v. The charge goes straight when the electric force q E balances the magnetic force q v B, right? Now tell me, does the speed it selects depend on the charge of the particle? It does not, very good! Cancel q from both sides, and v equals E over B, whatever the charge or mass. Clear?

3:52Example 1: radius of a proton at 5x10^6 m/s in 0.5 T

Now, Example one, an easy one. Here, a proton moves at five times ten to the power six metres per second, perpendicular to a field of zero point five tesla. Find the radius. On screen, m s to the power minus one means metres per second, and capital T is tesla. Pause the video and try it first. Done? Okay, let's check. The velocity is perpendicular to the field, so the path is a circle, with r equal to m v over q B, and for a proton, m is one point six seven times ten to the power minus twenty seven kilogram, and q is one point six times ten to the power minus nineteen coulomb. Okay? First the top, m v; in these lines the cross sign simply means times. One point six seven times ten to the power minus twenty seven, times five times ten to the power six, gives eight point three five times ten to the power minus twenty one. Right? Now the bottom: q B is one point six times ten to the power minus nineteen, times zero point five, which gives eight times ten to the power minus twenty. Got it? Now divide. Eight point three five times ten to the power minus twenty one, over eight times ten to the power minus twenty, gives about zero point one zero four metre. So the radius is about ten point four centimetres, written c m on screen. Clear?

5:18Example 2: mass in u of an ion accelerated through 500 V

Now Example two, a little harder. Read it with me: a singly charged ion is accelerated through five hundred volts, and then moves in a circle of radius ten centimetres, in a field of zero point one tesla. Find its mass in u, the atomic mass unit. Singly charged means it carries one electronic charge, and capital V on screen is volt. Pause the video and try it first. Done? Okay, let's check. The speed of the ion is not given. So where does it come from? From the voltage, exactly! The ion gains kinetic energy q V, so we use our radius formula: r equals one over B, times the square root of two m V over q. Square both sides, and r squared equals two m V over q B squared. So m equals q B squared r squared, over two V, okay? Now put in the numbers. The radius is ten, so r squared is one hundred, right? No, no, wait, the radius must be in metres! Ten centimetres is zero point one metre, so r squared is zero point zero one. Also, q is one point six times ten to the power minus nineteen coulomb, and B squared is zero point one squared, which is zero point zero one. With me? Now substitute. One point six times ten to the power minus nineteen, times zero point zero one, times zero point zero one, all over two times five hundred, which is one thousand. That gives one point six times ten to the power minus twenty six kilogram. Right? Last step, we change the mass into atomic mass units. One u is about one point six six times ten to the power minus twenty seven kilogram. Divide, and the mass is about nine point six u. So, energy from the voltage, then the radius formula: two tools, one problem. Got it?

7:16Quick revision: no work, r = mv/qB, double K gives root 2 r

So, quick revision. The magnetic force, q v B sine theta, does no work, so the speed and the kinetic energy stay constant. The radius is m v over q B, or root two m K over q B, so doubling K makes the radius grow by root two. The period, two pi m over q B, does not depend on the speed, and the velocity selector picks v equals E over B, okay? Now go to the practice problems on this page, give each problem one to two minutes, and begin. All the best!