Circular and Helical Motion

Transcript

0:00Introduction

Hello everyone! Okay, a small race first. Two protons enter the same uniform magnetic field, at right angles to it. One of them is slow, and the other is twice as fast, so its circle is twice as big. Now, which proton comes back to its starting point first? The slow one, with the smaller circle, right? Let us see who really wins. This idea comes up often in JEE problems, so let's begin.

0:28Why the path is a circle

So, take a charge q with velocity v, at right angles to a uniform magnetic field B, okay? The small arrows on top mark vectors, and the upside-down T sign means perpendicular. The magnetic force has size q v B, always at right angles to the velocity. The force is always perpendicular to the motion. So what does it do to the speed of the charge? It does nothing to the speed, exactly! Since the force is perpendicular to the velocity, it does no work, so the speed stays constant. It only turns the direction, and that gives uniform circular motion. Right? So the magnetic force supplies the centripetal force: q v B equals m v squared over r, where m is the mass and r the radius. With me? Cancel one v, and r equals m v over q B, okay? Write this one down. Clear?

1:23One radius, written three more ways

Now, m v is just the momentum, small p. So r equals p over q B, okay? But often a question gives you the kinetic energy, capital K, not the speed. So how do you get the momentum from K? Yes, exactly, p equals the square root of two m K; the tick-shaped sign is the square root. So r equals root two m K, over q B, got it? And if the charge was accelerated from rest through a voltage, capital V, it gains kinetic energy q V, right? Then r equals one over B, times the square root of two m V over q. But careful here: this capital V is the voltage, not the speed. Clear?

2:08The period does not depend on speed

Alright, back to our race. The time for one round is the period, capital T: the distance two pi r, over the speed v. Pi is the Greek letter, the circle constant. Okay? Now, if the speed doubles, the radius also doubles. So what happens to the period? Put in r equals m v over q B, and look, the v cancels! The dot just means times. T equals two pi m over q B, with no speed in it at all. Double the speed, double the circle, same time. So both protons come back together! Got it? Turn it upside down, and the frequency f equals q B over two pi m, the cyclotron frequency. And omega, the Greek letter like a curly w, is the angular frequency, q B over m. No speed in any of them. With me?

3:01Which way does it go round?

Now let us find the direction of turning. Say B points out of the page, towards you, and a positive charge, plus q, moves to the right. Does it go round clockwise or anticlockwise? Use q v cross B; the cross sign is the cross product. Velocity to the right, crossed with B out of the page, gives a force pointing downwards. So the charge bends down and goes round clockwise. With the field into the page, it goes anticlockwise. Okay?

3:33Velocity at an angle: the helix

Now, what if the velocity makes an angle theta, the Greek letter, with B? Split v into two parts, as the figure on the right shows. v parallel, with two small vertical lines as its subscript, is v cos theta, along the field. Clear? Now tell me, what magnetic force does this parallel part feel? The force is zero, exactly! Motion along B feels no magnetic force, so the charge drifts along the field at a steady v cos theta. Right? And v perpendicular, with a small upside-down T as its subscript, is v sin theta. It turns in a circle of radius m v sin theta, over q B, okay? Circle plus drift gives a spring-like path, a helix, as in the figure. With me?

4:23The pitch of the helix

Now the pitch, small p, and careful, here p is not the momentum! It is the distance the charge moves along B in one full round, that is, in one period. Okay? So which velocity carries it along the field, v sin theta or v cos theta? It is v cos theta, the parallel part, exactly! So p equals v cos theta times T, which is two pi m v cos theta, over q B, right? Radius with sine, pitch with cosine; mixing them up is a classic trap. And divide the pitch by the radius: m, v, q and B all cancel. So p over r equals two pi cot theta, where cot is the cotangent. Got it?

5:08Example 1: radius of a proton at right angles to 0.1 T

Read the first example with me: a proton moves at ten to the power six metres per second, perpendicular to a field of zero point one tesla. Find the radius of its path. Here m s to the power minus one means metres per second, and capital T is tesla, not the period. Pause the video and try it first. Done? Okay, let's check. The velocity is perpendicular to the field, so r equals m v over q B, and the cross sign here simply means times. 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. Now substitute the values. On top, one point six seven times ten to the power minus twenty seven, times ten to the power six. At the bottom, one point six times ten to the power minus nineteen, times zero point one. Which powers of ten are left? On top, minus twenty seven plus six gives ten to the power minus twenty one. At the bottom, zero point one is ten to the power minus one, so ten to the power minus twenty. One point six seven over sixteen, so r is about zero point one zero four metre; the wavy sign means approximately. Right?

6:31Example 2: radius and pitch, proton at 60 degrees to 0.2 T

Now a harder one: a proton moves at two times ten to the power six metres per second, at sixty degrees to a field of zero point two tesla; the small circle means degrees. Find the radius and the pitch of its helix. Pause the video and try it first. Done? Okay, let's check. Both answers need q B, so let us find it first. One point six times ten to the power minus nineteen, times zero point two, gives three point two times ten to the power minus twenty. Okay? Now the radius. Which one goes in, sine sixty or cos sixty? It is sine sixty, about zero point eight six six, the perpendicular part! So r is m v times zero point eight six six, over q B, okay? That gives about zero point zero nine zero metre, or nine point zero centimetres. Got it? Now the pitch uses cos sixty, which is zero point five. Two pi, times the same m v, times zero point five, over the same q B, clear? That gives about zero point three three metre, so the pitch is thirty three centimetres. Radius with sine, pitch with cosine. Right?

7:47Quick revision: r = mv/qB, T = 2 pi m/qB, helix sin and cos

So, quick revision. Velocity perpendicular to the field gives a circle, with r equal to m v over q B, or root two m K over q B, okay? The period, two pi m over q B, is the same for every speed. At an angle theta, you get a helix: sine in the radius, cosine in the pitch. Now try the practice problems on this page. You can do them, all the best!