Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two long parallel wires, six centimetres apart, carry three amperes and six amperes in the same direction. Where is the magnetic field zero? At the midpoint, three centimetres from each wire, right? Hmm, keep your guess; Example one will settle it. Today we see how this chapter works, collect the tools for its three sets, and solve two of its problems. Let's begin.

0:32How this chapter works

So, how does this chapter work? Magnetism often gives two or three questions in a JEE Main shift. Okay? Set one is on the fields of currents, and Set two is on charged particles. Right? And Set three is on forces, torques, magnets and magnetic materials. With me? These problems are original, about one to two minutes each. The real ones, from twenty twenty one to twenty twenty six, are at the end of Chapters one to five. Got it?

1:02Set 1 tools: fields of currents

Now, Set one tools. For a long straight wire, B equals mu naught I over two pi d, with pi the Greek letter, capital I the current and d the distance. Mu naught, the Greek mu with a small zero, is the permeability of free space: four pi times ten to the power minus seven tesla metre per ampere. Okay? For an arc of angle phi, the Greek letter, in radians, the centre field is mu naught I phi over four pi R, with capital R the radius. Put phi equal to two pi for one full turn, and multiply by capital N: a coil of N turns gives mu naught N I over two R at its centre. Right? The direction comes from the right-hand thumb rule: thumb along the current, fingers curling along B; the arrow on top marks a vector. With me? Now, at a point, two wires give fields in opposite directions. Do you add their sizes, or subtract them? You subtract them, exactly! Fields add as vectors, and our puzzle needs exactly this. Clear?

2:05Set 2 tools: charged particles

Now Set two. Here, a charge q with velocity v in a field B feels the force q v cross B, the cross being the cross product. Its size is q v B sine theta, where theta, the Greek letter, is the angle between the two vectors. Okay? This force is always perpendicular to the velocity. So does it change the speed? It does not, and that is the key point! The upside-down T sign means perpendicular: such a force does no work, so the speed stays constant. Right? So when v is perpendicular to B, the path is a circle of radius r equal to m v over q B, with m the mass. With me? Now, double the speed, and the circle gets bigger. Does the time for one round change? It does not, and this is a common trap! The period, capital T, is two pi m over q B, with no speed in it. Clear?

3:00Set 3 tools: forces, torques and materials

Now Set three. Here, a straight wire of length L, with the L vector along the current I, in a field B feels the force I L cross B, of size ILB sine theta. Okay? For a loop of capital N turns, each of area capital A, the magnetic moment, mu vector, is N I times the A vector, normal to the loop. The torque, tau, another Greek letter, is mu vector cross B, of size mu B sine theta. Clear? Now, the plane of a coil makes thirty degrees with the field; the small circle means degrees. So theta is thirty degrees, right? No, no, wait, that is exactly the trap! Theta is the angle between mu vector, along the normal, and B, and not the plane. The normal is at ninety minus thirty, so here theta is sixty degrees. Got it? And for materials, look at chi, the Greek letter, the susceptibility. Small and less than zero means diamagnetic. Small and greater than zero means paramagnetic, with Curie's law, chi equals C over T, where C is a constant and capital T is the absolute temperature. Large and greater than zero means ferromagnetic. With me?

4:13Example 1: zero field point between 3 A and 6 A wires

Now, Example one, from Set one: our puzzle from the start. Two long parallel wires, six centimetres apart, carry three amperes and six amperes in the same direction. Where is the field zero? Here, c m is centimetres and capital A is ampere. Pause the video and try it first. Done? Okay, let's check. Where can the two fields cancel at all: between the wires, or outside them? They can cancel only between the wires, exactly! By the right-hand thumb rule, currents in the same direction give opposite fields between the wires; outside, the fields add. Okay? Let x be the distance from the three ampere wire; the other wire is six minus x away. Equal fields: mu naught times three, over two pi times x, equals mu naught times six, over two pi times, in brackets, six minus x; here the cross simply means times. Right? Mu naught and two pi cancel, so three over x equals six over, in brackets, six minus x, all in centimetres. The double arrow means implies. Cross multiply: eighteen minus three x equals six x, so x equals two. Clear? So the field is zero between the wires, two centimetres from the three ampere wire. It is not the midpoint; the zero point sits closer to the weaker current. And that settles our puzzle. Got it?

5:40Example 2: pitch of an electron's helix at 60 degrees

Now Example two, from Set two, a bit harder. An electron moves at ten to the power six metres per second, at sixty degrees to a field of one millitesla. Find the pitch of its helix. Here, m s to the power minus one means metres per second, and m T is millitesla. Pause the video and try it first. Done? Okay, let's check. In the figure, the velocity is at an angle theta to the field, so the path is a helix. The pitch is the distance moved along the field in one round. So which part carries it along: v sine sixty, or v cos sixty? It is v cos theta, the part along the field, exactly! So the pitch, small p, is v cos theta times the period T, the dot meaning times: two pi m v cos theta, over e B, with e the electron's charge. Okay? Now the period, which is two pi m over e B: two pi times nine point one times ten to the power minus thirty one, the electron's mass, over one point six times ten to the power minus nineteen, times ten to the power minus three tesla, the field. That gives about three point five seven times ten to the power minus eight seconds; the wavy sign means approximately. Right? The speed along the field is ten to the power six times cos sixty, which is zero point five, so five times ten to the power five metres per second. With me? So the pitch is five times ten to the power five, times three point five seven times ten to the power minus eight. That gives about one point eight times ten to the power minus two metres, or one point eight centimetres. Cosine for the pitch, not sine: mixing them up is the classic trap. Clear?

7:26Quick revision: wire and arc fields, radius, pitch, torque

So, quick revision. Set one: a wire gives mu naught I over two pi d, an arc gives mu naught I phi over four pi R, and fields add as vectors. Set two: r equals m v over q B, the period, two pi m over q B, does not depend on speed, and the pitch of a helix uses v cos theta, not v sine theta. Set three: tau equals mu B sine theta, with theta from the normal. Now open Set one, and give each problem one to two minutes. All the best!