Intensity and the Poynting Vector

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Sunlight reaching the Earth carries about one point four kilowatts of power through every square metre held facing the Sun. Light is an electromagnetic wave, so it carries an electric field. So how strong is that field? Just a few volts per metre, or about a thousand? By the end of this video, you will calculate it. Questions on intensity come up often in JEE, so let's begin.

0:29Intensity: energy per area per time

So, the intensity, capital I, is the average energy crossing unit area per unit time. Its unit is watt per square metre, the W m to the power minus two on screen. Okay? And the area is held perpendicular to the direction of travel, so the wave crosses it straight on. Right? Now let me ask you this. The wave moves at c, the speed of light. In a time t, how far does it travel? Yes, exactly, c times t! So in time t, all the energy in a column of length c times t crosses the area. We call the area capital A, as on screen. With me? Now, u is the energy per unit volume, the energy density, and the angle brackets mean its average. So the column holds the average u, times A times c times t. Divide by A and by t, and the intensity is the average u, times c. Simple, isn't it?

1:29The Poynting vector

For the energy flow at each instant, we use the Poynting vector, S with an arrow on top, which means a vector. S vector equals E vector cross B vector, over mu naught. The cross sign is the cross product, and mu naught, the Greek letter mu with a small zero below, is the permeability of free space. Okay? Remember, E and B are perpendicular to each other and to the direction of travel. So which way does E cross B point? Yes, exactly, along the direction of travel! So S gives the instantaneous energy flow per unit area, and it points the way the wave moves. Got it? But the fields oscillate, so S keeps changing from instant to instant. The average of its magnitude is exactly the intensity. Clear?

2:19Intensity in terms of the fields

E naught, capital E with a small zero below, is the amplitude of the electric field. From the last lesson, what is the average energy density in terms of E naught? Yes, half epsilon naught E naught squared! Epsilon naught, the curly e with a small zero, is the permittivity of free space, and the small two on top means squared. Multiply by c, and the intensity is half epsilon naught E naught squared, times c. Okay? B naught is the magnetic amplitude, and in vacuum, E naught equals c times B naught. So the intensity is also c times B naught squared over two mu naught, or E naught times B naught over two mu naught. Right? Now here is a trap. At its peak, the Poynting vector has the value S max, the S with a small max below it, equal to E naught times B naught over mu naught. So is this peak value the intensity? No, no, careful, the peak is not the intensity! The intensity is the average, which is half the peak. So S max, the maximum, is two times the intensity. With me? With the r m s value, the root mean square value shown by the small r m s below the E, there is no half. The intensity is epsilon naught, times c, times E r m s squared. Okay? Now a topper's trick: the square root of mu naught over epsilon naught is three hundred and seventy seven ohms; the capital omega on screen is the ohm. So, as the double arrow shows, in S I units the intensity is E naught squared over two times three hundred and seventy seven, right?

4:04Point source: the inverse square law

Now, a point source gives out power, capital P, uniformly in all directions. At a distance r, over what area is this power spread? Yes, over a sphere of radius r, with area four pi r squared; pi is the Greek letter on screen. So the intensity is P over four pi r squared, the same all over a sphere centred on the source. Clear? Now tell me this, if you go twice as far from the source, what happens to the intensity? And what happens to E naught? The intensity falls to a quarter, because it is proportional to one over r squared; the stretched alpha sign means proportional to. But be careful, E naught does not fall to a quarter. Intensity goes as E naught squared, so E naught and B naught go as one over r, and they only halve. That is a common trap, okay? And in a medium, epsilon naught becomes epsilon, the permittivity of the medium, and c becomes v, the speed in that medium. So the intensity is half epsilon E naught squared, times v. Got it?

5:18Example 1: visible light intensity 2 m from a 60 W bulb

Now our first example. Here, a sixty watt bulb converts five percent of its power into visible light. Find the intensity of visible light two metres away. The capital W is watts, and the small m is metres. Pause the video and try it first. Done? Okay, let's check. The bulb is sixty watts, so P equals sixty watts, right? No, no, wait, that is the trap! We want visible light only, which is just five percent of the power. Five percent of sixty watts is three watts, so P is three watts. Okay? Treat the bulb as a point source. The intensity is P over four pi r squared, so three over four pi times two squared, which is three over sixteen pi. Right? That comes to about zero point zero six zero watt per square metre; the wavy equals sign means approximately. Easy, isn't it?

6:20Example 2: E0 and B0 of sunlight from I = 1.4 kW/m2

Now a harder one, and it is our puzzle from the start. The intensity of sunlight at the Earth is one point four kilowatts per square metre; the k W on screen is kilowatts. Find the amplitudes of its electric and magnetic fields, treating it as a single wave. Pause the video and try it first. Done? Okay, let's check. Kilo means a thousand, so the intensity is one thousand four hundred watts per square metre. With me? Now, how do we get E naught from the intensity? Yes, just turn the formula around! E naught equals the square root of two I over epsilon naught times c. Put in two times one thousand four hundred, over eight point eight five times ten to the power minus twelve, times three times ten to the power eight. Okay? Work it out, and E naught is about one point zero three times ten to the power three volts per metre, the V m to the power minus one on screen. That is about a thousand volts per metre, not a few! So that answers our puzzle. Surprised? Now, B naught equals E naught over c, so one point zero three times ten to the power three, over three times ten to the power eight. That is about three point four times ten to the power minus six tesla, the capital T on screen. If you got both, very good!

7:49Quick revision: I from E0, inverse square law, E0 from I

So, quick revision. Intensity is the average u times c, which is half epsilon naught E naught squared, times c, which is the average magnitude of the Poynting vector. For a point source, it is P over four pi r squared, and E naught falls as one over r. And from intensity back to amplitude, E naught is the square root of two I over epsilon naught c, and B naught is E naught over c. Now go and try the practice problems on this page. All the best!