Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Light spreads out equally in all directions from a point source. You stand two metres away, and your friend stands eight metres away, four times farther. So, is the electric field there four times weaker, or sixteen times weaker? Keep your answer in mind; we settle it in example one. Today is Part A of the unit problem bank on electromagnetic waves, at JEE Main level. So, let's begin.

0:32Tool 1: the displacement current

So, tool one: the displacement current. The electric flux, capital phi with a small E below, rises uniformly from zero to four point five two times ten to the power three volt metre, the V m, in two microseconds, the Greek mu followed by s. Find the displacement current, the i with a small d below. Which formula turns a changing flux into a current? It is Maxwell's displacement current, very good! i d equals epsilon naught, the Greek epsilon with a small zero, times d phi E by d t. It is not a flow of charge, and it exists only where the electric flux is changing, like between capacitor plates while they charge. Okay? The rise is uniform, so d phi E by d t is the change over the time: four point five two times ten to the power three, over two times ten to the power minus six. That gives two point two six times ten to the power nine volt metre per second, where s to the power minus one means per second. Right? Multiply by epsilon naught, eight point eight five times ten to the power minus twelve, and we get zero point zero two zero ampere, the capital A, or twenty milliamperes, the m A on screen. Forget the micro, and your answer is a million times too small. Clear?

1:51Tool 2: the wave in vacuum

Tool two: the wave itself. This wave in vacuum has E naught, the E with a small zero, of three point three kilovolts per metre, the k V m to the power minus one, and we need B naught. In the figure, the red arrows are the electric field, E with an arrow on top, and the dark blue ones the magnetic field, B with an arrow. With me? The wave travels along x, with E along y and B along z; lambda, the Greek letter below, 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 E cross B, with the times sign, points along the direction of travel. Okay? Now, one statement in Part A says that E and B differ in phase by ninety degrees. They are ninety degrees apart in direction, true. So are they also ninety degrees apart in phase? No, not at all; they oscillate in phase! In the figure, E and B reach their peaks together, and their zeros together. So that statement is false. Got it? In vacuum, all electromagnetic waves travel at c, one over the square root of mu naught times epsilon naught, whatever the frequency. Mu naught, the Greek mu with a small zero, is the permeability of free space, and c is three times ten to the power eight metres per second. Right? And the amplitudes are linked: E naught equals c times B naught, and in a medium the wave's speed v replaces c. So B naught is E naught over c, not c times E naught. Clear? So B naught is three point three times ten to the power three, over three times ten to the power eight. It gives one point one times ten to the power minus five tesla, the capital T on screen. Okay?

3:39Tool 3: intensity

Tool three: intensity, capital I, is the average energy crossing unit area, held perpendicular to the direction of travel, per unit time. Right? I equals the average energy density, the u inside angle brackets, times c. That is half epsilon naught E naught squared, times c, with E naught the peak value of the field. With me? So let me ask you this: if you double E naught, what happens to the intensity? It becomes four times, yes, exactly! I is proportional to E naught squared; the sign like a stretched alpha means proportional to. Clear? Now a point source of power, capital P, radiating uniformly. At a distance r, the power spreads over a sphere of area four pi r squared, so I equals P over four pi r squared, not over pi r squared, okay?

4:29Example 1: E field ratio at 2 m and 8 m from a point source

Now, example one, our opening puzzle. Find the ratio of the electric field amplitudes at two metres and at eight metres from a point source; the small m is the metre. Pause the video and try it first. Done? Okay, let's check. Call the nearer distance r one, which is two metres, and the farther one r two, which is eight metres; the small one and two below are just labels. Intensity goes as one over r squared, so I one over I two is r two over r one, whole squared, that is four squared, which is sixteen. Clear? So the field amplitudes are also in the ratio sixteen to one, right? No, no, careful, that is exactly the trap! Intensity goes as E naught squared, so E naught goes as the square root of the intensity, and that is one over r. So the field falls only as one over r. Got it? Then E naught one over E naught two is r two over r one, that is, eight over two, which gives four; so the answer is four to one. Puzzle solved: four times farther, the intensity is sixteen times less, but the field is only four times weaker. With me?

5:40Example 2: radiation force on a black solar panel

Now a harder one, example two. Here a black solar panel of area two square metres, the m squared, faces sunlight of intensity one kilowatt per square metre, the k W m to the power minus two. Find the force of the radiation on it. Pause the video and try it first. Done? Okay, let's check. Light carries momentum, and passes it to any surface that absorbs or reflects it. The figure shows an absorbing surface, a perfect reflector, and a reflector at an angle theta, not needed today. So, is our black panel an absorber or a reflector? An absorber, yes, exactly, since black absorbs all the light! And it faces the Sun, so the light falls straight on, at normal incidence. Under each picture, p is the radiation pressure: I over c for an absorber, two I over c for a perfect reflector. So for our panel, p rad, with a small rad below, is I over c. Okay? Pressure is force per unit area, so force is pressure times area: F equals I times capital A, over c, with capital A the area of the panel. Right? Now the units: one kilowatt per square metre is one thousand watts per square metre. Clear? So F equals one thousand times two, over three times ten to the power eight, which gives six point seven times ten to the power minus six newton, with capital N for newton. Use two I over c here, and you get double, wrong for a black panel. Got it?

7:13Recap: displacement current, E0 = cB0, field 1/r, force IA/c

So, quick revision. The displacement current is epsilon naught times d phi E by d t, only where the electric flux changes. In vacuum, E naught equals c times B naught, with E and B perpendicular and in phase. From a point source, the intensity falls as one over r squared, but the field only as one over r; and a black absorber feels a force of I times A over c. Now try all twelve problems of Part A on this page, then read the solutions. All the best!