Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle first. At three metres from a point source of light, the electric field amplitude is twelve volts per metre. Now go three times farther, to nine metres. Intensity falls as one over r squared, so the field becomes twelve over nine, right? Hmm, hold that thought; Example one will settle it. Today we see how this chapter of mixed JEE Main problems works, collect its tools, and solve two problems. Let's begin.

0:32How this chapter works

So, how does this chapter work? It has three timed sets, each with twelve original problems, in the style of recent JEE Main papers. Okay? Give yourself twenty minutes per set, and only then check the solutions. Now tell me: how much time is that for each problem? One hundred seconds, very good! Twenty times sixty gives twelve hundred seconds, the small s on screen. And twelve hundred over twelve gives one hundred seconds per problem. Right? And every set mixes all four chapters of this unit. So now the tools, chapter by chapter. With me?

1:12Chapter 1 tools: displacement current

Now chapter one. i d, the i with a small d below, is the displacement current: epsilon naught, the Greek epsilon with a small zero, times d phi E by d t, where capital phi with a small E is the electric flux. For a capacitor, it is also capital C, the capacitance, times d V by d t, with V the voltage. Now, a fully charged capacitor has a strong field between its plates. Is there a displacement current? No, because the field is not changing! It exists only where the electric flux changes. While charging, i d in the gap equals i c, the conduction current in the wires; fully charged, i d is zero. Clear?

1:58Chapter 2 tools: the wave

Now chapter two. In vacuum, c equals one over the root of mu naught epsilon naught, mu naught being the Greek mu with a small zero. It also equals omega over k, the angular frequency over the wave number, and nu times lambda, the Greek letters for frequency and wavelength. Now the amplitudes, E naught and B naught, with a small zero: is E naught equal to c times B naught, or is B naught equal to c times E naught? E naught equals c times B naught, yes, exactly! Swapping the two is a common trap. Okay? In the figure, the wave travels along x, the red E swings along y, and the blue B along z; arrows on top mean vectors, and the upside down T means perpendicular. So E and B are perpendicular to each other and to the travel, in phase, and the wave travels along E cross B, the cross product. With me? In a medium, the speed v is c over the root of mu r epsilon r, the relative permeability and permittivity. The frequency nu stays the same, and lambda becomes lambda naught over n, where n is the refractive index. Got it?

3:12Chapter 3 tools: energy, intensity, pressure

Now chapter three. u is the energy density, and the angle brackets mean its average. The average u is one half epsilon naught E naught squared, where E naught is the peak value, the amplitude. And on average it is shared equally between the electric and magnetic fields. Okay? The intensity, capital I, is the average u times c, so it is one half epsilon naught E naught squared, times c. Right? For a point source of power capital P, radiating uniformly, at a distance r, the intensity is P over four pi r squared. Clear? In the figure, light is absorbed on the left, and in the middle a perfect reflector sends it straight back; the small p under each is the pressure. So why should the reflector feel twice the pressure? Yes, exactly: reflection reverses the momentum, so the change in momentum is double! At normal incidence, the pressure is I over c for a perfect absorber, and two I over c for a perfect reflector. And for a perfect reflector with the beam at an angle theta to the normal, on the right, a cos squared theta comes in. Got it?

4:26Chapter 4 tools: the spectrum

And chapter four, the spectrum. From the longest wavelength to the shortest: radio, microwave, infrared, visible, U V or ultraviolet, X-rays and gamma rays; the sign that opens to the left means greater than. So which region has the highest frequency? Gamma rays, that is correct! Frequency and photon energy follow the reverse order of wavelength. With me? And to place a photon: its energy E in electron volts, e V, equals one thousand two hundred and forty over lambda in nanometres, n m. Okay?

5:05Example 1: E0 at 9 m when it is 12 V/m at 3 m from a source

Example one, an easy one from Set three: our opening puzzle. The field amplitude is twelve volts per metre, written capital V, small m, to the power minus one, where capital V is volt and small m is metre. That is at three metres from a point source. Find it at nine metres. Pause the video and try it first. Done? Okay, let's check. Intensity falls as one over r squared, so the field amplitude is twelve over nine, right? Oh no, careful, that is the trap! The sign like an open sideways eight means proportional to. I is proportional to one over r squared, but I is also proportional to E naught squared. So E naught is proportional to one over r, not one over r squared. Clear? So E naught equals twelve times three over nine, which gives four volts per metre. Puzzle solved: three times farther, the intensity falls nine times, but the field only three times. Right?

6:09Example 2: direction of travel and E field from a given Bz

Example two, a hard one from Set two. Here, in vacuum, the magnetic field is B z, the B with a small z below, equal to five times ten to the power minus eight, times sine of, two times ten to the power seven, times t, plus x over fifteen; the capital T is tesla. Find the direction of travel, and write the electric field. Pause the video and try it first. Done? Okay, let's check. Omega is two times ten to the power seven, and k is one over fifteen. So omega over k is two times ten to the power seven, times fifteen, which gives three times ten to the power eight metres per second: c, as it should be in vacuum. Okay? Now the direction: inside the bracket, omega t and x over fifteen are added. So which way does this wave travel? Along minus x, yes, well done! When omega t and k x carry the same sign, the wave travels along minus x. So n hat, the unit vector of travel, with a little cap, is minus i hat, where i hat points along x. Right? E naught equals c times B naught: three times ten to the power eight, times five times ten to the power minus eight, which gives fifteen volts per metre. Clear? Now, which way does E point? For a wave along plus x, E along y goes with B along z. Here B is along z, so E is along plus y too, right? No, no, wait, this wave goes along minus x! E is parallel to B cross n hat; the two upright lines mean parallel. k hat, along z, cross minus i hat, gives minus j hat, because k hat cross i hat is j hat, and j hat points along y. So E points along minus y. With me? So E y equals minus fifteen, times sine of the same bracket, in volts per metre. The same bracket keeps E and B in phase. Got it?

8:19Quick revision: 100 s a problem, E0 as 1/r, E along B x n

So, quick revision. Twenty minutes a set, about one hundred seconds a problem. From a point source, intensity falls as one over r squared, but the field amplitude only as one over r. And for directions, read the signs, then use E parallel to B cross n hat. Now open Set one, start your timer, and try it. All the best!