Set 1

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Picture a laser beam of one point five kilowatts falling straight on a mirror, and the mirror sends all of it back. Light pushes the mirror. But how hard? One student says five times ten to the power minus six newton. Another says ten to the power minus five newton. So who is right? Hold that thought. Today is our first mixed problem set on electromagnetic waves: twelve problems in twenty minutes. Let us pick up the tools, and spot the traps. Problems like these come up often in JEE Main. So, let's begin.

0:39Tool 1: the displacement current

So, first tool: the displacement current. i d, the i with a small d below, equals epsilon naught times d phi E by d t. Epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space, and phi E, capital phi with a small E below, is the electric flux. For a capacitor, this is the same as C times d V by d t, where C is the capacitance and V the voltage. Okay? But remember the condition. It exists only where the electric flux changes, like the gap of a capacitor while it is charging. Once the capacitor is fully charged, nothing changes, so i d is zero. Right? Now problem one of the set. The voltage across an eight microfarad capacitor rises at two hundred and fifty volts per second. The Greek mu means micro, capital F is the farad, and V s to the power minus one is volt per second. Find the displacement current. But no charge crosses the gap, so the answer is zero, right? No, no, careful, that is the trap, because the voltage is still rising. So the flux between the plates is changing, and that is all we need. i d equals eight times ten to the power minus six, times two hundred and fifty, which gives two times ten to the power minus three ampere, capital A on screen. That is two milliamperes, the m A you see. Got it?

2:10Tool 2: the wave in vacuum

Second tool: the wave itself. In the figure, E with an arrow on top, in red, is the electric field, and B with an arrow, in blue, is the magnetic field. The wave travels along x, with E along y and B along z, and lambda, the Greek letter at the bottom, is the wavelength. The upside down T sign means perpendicular: E and B are perpendicular to each other and to the direction of travel, so the wave is transverse. And they are in phase: they peak together. Okay? Problem two gives B naught, the B with a small zero, which is the amplitude of the magnetic field in vacuum: one point two times ten to the power minus seven tesla, the capital T on screen. We need E naught, the electric amplitude. So, do we multiply by c, the speed of light, or divide by c? We multiply, yes, exactly! In vacuum, E naught equals c times B naught, where c, the speed of light, is three times ten to the power eight metres per second. Many students write B naught equals c times E naught, and that is a classic trap. Clear? So, for problem two, E naught is three times ten to the power eight, times one point two times ten to the power minus seven. Okay? That gives thirty six volts per metre, the V m to the power minus one on screen. See, in S I units the number for E naught comes out far bigger than the number for B naught, three times ten to the power eight times bigger, and that is a quick check on your answer. Right? Now the direction. E cross B, where the times sign means the cross product, points along the direction of travel, as the label in the figure says. Turn it round, and B equals one over c, times n hat cross E, where n hat, the n with a little cap on top, is the unit vector along the direction of travel. With me? Problem four: the wave travels along minus y, and its electric field points along plus x. So which way does B point? Let us check. Here n hat is minus j hat, and the field is along i hat, where i hat, j hat and k hat are the unit vectors along x, y and z. Now j hat cross i hat is minus k hat, and the minus sign in front turns it into plus k hat. So B is along plus z. If you got minus z, you dropped one minus sign. Got it?

4:48Example 1: intensity 4 m from an 80 W point source

Now, example one: problem seven of the set. An eighty watt point source radiates uniformly; capital W is the watt. Find the intensity four metres away. Pause the video and try it first. Done? Okay, let's check. The source radiates uniformly, so the eighty watts go out equally in all directions. So, at four metres, over what area are they spread? Over a sphere of radius four metres, yes, exactly! Its area is four pi r squared, with pi the Greek letter, where r is the distance from the source. Not pi r squared, which is the area of a circle. Okay? Intensity, capital I, is the average energy per second crossing each square metre held perpendicular to the direction of travel. So I equals P over four pi r squared, with capital P the power. That is eighty, over four pi times sixteen, because four squared is sixteen. Right? Four times sixteen is sixty four, so I equals eighty over sixty four pi. That is about zero point four zero watt per square metre; the wavy equals sign means approximately, and m to the power minus two means per square metre. Clear?

6:08Example 2: force of a 1.5 kW laser on a mirror

Now the harder one, and it is our opening puzzle: problem eight. This time, a one point five kilowatt laser beam, the k W on screen, is reflected completely by a mirror at normal incidence, which means straight on. Find the force on the mirror. Pause the video and try it first. Done? Okay, let's check. Look at the figure. On the left, light falls on an absorbing surface; in the middle, on a perfect mirror; on the right, light falls on a mirror at an angle theta, the Greek letter, to the normal, and that formula with cos squared theta we do not need today. So tell me, which gets the bigger push, the absorber or the mirror? The mirror, twice as much, that is correct! The light arrives with its momentum and leaves with that momentum reversed, so the change is double. Under each picture, small p is the radiation pressure: I over c for the absorber, and two I over c for the mirror. For a whole beam of power capital P falling straight on the surface, the force, capital F, is P over c if it is absorbed, and two P over c if it is reflected. Okay? Now the units: one point five kilowatts is one thousand five hundred watts. Always change to watts first. Right? So F equals two times one thousand five hundred, over three times ten to the power eight. The top is three thousand. With me? Three thousand over three times ten to the power eight gives ten to the power minus five newton, with capital N for newton. Puzzle solved, and the second student is right! The first one used P over c, which is the answer for an absorber, not for a mirror. Got it?

7:56Quick revision: id = C dV/dt, E0 = cB0, intensity, P/c, 2P/c

So, quick revision. The displacement current is C times d V by d t, and it exists only where the electric flux changes. In vacuum, E naught equals c times B naught, and E cross B points along the direction of travel. From a point source, I equals P over four pi r squared, and a beam falling straight on a surface pushes with P over c on an absorber, and two P over c on a mirror. Now try all twelve problems of this set on this page, in twenty minutes, then check the solutions. All the best!