# 16.3 Energy carried by electromagnetic waves  (Page 2/5)

 Page 2 / 5

The energy passing through area A in time $\text{Δ}t$ is

$u\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{volume}=uAc\text{Δ}t.$

The energy per unit area per unit time passing through a plane perpendicular to the wave, called the energy flux and denoted by S , can be calculated by dividing the energy by the area A and the time interval $\text{Δ}t$ .

$S=\frac{\text{Energy passing area}\phantom{\rule{0.2em}{0ex}}A\phantom{\rule{0.2em}{0ex}}\text{in time}\phantom{\rule{0.2em}{0ex}}\text{Δ}t}{A\text{Δ}t}=uc={\epsilon }_{0}c{E}^{2}=\frac{1}{{\mu }_{0}}EB.$

More generally, the flux of energy through any surface also depends on the orientation of the surface. To take the direction into account, we introduce a vector $\stackrel{\to }{S}$ , called the Poynting vector    , with the following definition:

$\stackrel{\to }{S}=\frac{1}{{\mu }_{0}}\stackrel{\to }{E}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}.$

The cross-product of $\stackrel{\to }{E}$ and $\stackrel{\to }{B}$ points in the direction perpendicular to both vectors. To confirm that the direction of $\stackrel{\to }{S}$ is that of wave propagation, and not its negative, return to [link] . Note that Lenz’s and Faraday’s laws imply that when the magnetic field shown is increasing in time, the electric field is greater at x than at $x+\text{Δ}x$ . The electric field is decreasing with increasing x at the given time and location. The proportionality between electric and magnetic fields requires the electric field to increase in time along with the magnetic field. This is possible only if the wave is propagating to the right in the diagram, in which case, the relative orientations show that $\stackrel{\to }{S}=\frac{1}{{\mu }_{0}}\stackrel{\to }{E}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{B}$ is specifically in the direction of propagation of the electromagnetic wave.

The energy flux at any place also varies in time, as can be seen by substituting u from [link] into [link] .

$S\left(x,t\right)=c{\epsilon }_{0}{E}_{0}^{2}{\text{cos}}^{2}\left(kx-\omega t\right)$

Because the frequency of visible light is very high, of the order of ${10}^{14}\phantom{\rule{0.2em}{0ex}}\text{Hz,}$ the energy flux for visible light through any area is an extremely rapidly varying quantity. Most measuring devices, including our eyes, detect only an average over many cycles. The time average of the energy flux is the intensity I of the electromagnetic wave and is the power per unit area. It can be expressed by averaging the cosine function in [link] over one complete cycle, which is the same as time-averaging over many cycles (here, T is one period):

$I={S}_{\text{avg}}=c{\epsilon }_{0}{E}_{0}^{2}\frac{1}{T}\underset{0}{\overset{T}{\int }}{\text{cos}}^{2}\left(2\pi \frac{t}{T}\right)dt.$

We can either evaluate the integral, or else note that because the sine and cosine differ merely in phase, the average over a complete cycle for ${\text{cos}}^{2}\left(\xi \right)$ is the same as for ${\text{sin}}^{2}\left(\xi \right)$ , to obtain

$⟨{\text{cos}}^{2}\xi ⟩=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\left[⟨{\text{cos}}^{2}\xi ⟩+⟨{\text{sin}}^{2}\xi ⟩\right]=\frac{1}{2}⟨1⟩=\frac{1}{2}.$

where the angle brackets $⟨\text{⋯}⟩$ stand for the time-averaging operation. The intensity of light moving at speed c in vacuum is then found to be

$I={S}_{\text{avg}}=\frac{1}{2}c{\epsilon }_{0}{E}_{0}^{2}$

in terms of the maximum electric field strength ${E}_{0},$ which is also the electric field amplitude. Algebraic manipulation produces the relationship

$I=\frac{c{B}_{0}^{2}}{2{\mu }_{0}}$

where ${B}_{0}$ is the magnetic field amplitude, which is the same as the maximum magnetic field strength. One more expression for ${I}_{\text{avg}}$ in terms of both electric and magnetic field strengths is useful. Substituting the fact that $c{B}_{0}={E}_{0},$ the previous expression becomes

$I=\frac{{E}_{0}{B}_{0}}{2{\mu }_{0}}.$

We can use whichever of the three preceding equations is most convenient, because the three equations are really just different versions of the same result: The energy in a wave is related to amplitude squared. Furthermore, because these equations are based on the assumption that the electromagnetic waves are sinusoidal, the peak intensity is twice the average intensity; that is, ${I}_{0}=2I.$

Maxwell's stress tensor is
if 6.0×10^13 electrons are placed on a metal sphere of charge 9.0micro Coulombs, what is the net charge on the sphere
18.51micro Coulombs
ASHOK
Is it possible to find the magnetic field of a circular loop at the centre by using ampere's law?
Is it possible to find the magnetic field of a circular loop at it's centre?
yes
Brother
The density of a gas of relative molecular mass 28 at a certain temperature is 0.90 K kgmcube.The root mean square speed of the gas molecules at that temperature is 602ms.Assuming that the rate of diffusion of a gas in inversely proportional to the square root of its density,calculate the density of
A hot liquid at 80degree Celsius is added to 600g of the same liquid originally at 10 degree Celsius. when the mixture reaches 30 degree Celsius, what will be the total mass of the liquid?
what is electrostatics
Study of charges which are at rest
himanshu
Explain Kinematics
Two equal positive charges are repelling each other. The force on the charge on the left is 3.0 Newtons. Using your notes on Coulomb's law, and the forces acting on each of the charges, what is the force on the charge on the right?
Using the same two positive charges, the left positive charge is increased so that its charge is 4 times LARGER than the charge on the right. Using your notes on Coulomb's law and changes to the charge, once the charge is increased, what is the new force of repulsion between the two positive charges?
Nya
A mass 'm' is attached to a spring oscillates every 5 second. If the mass is increased by a 5 kg, the period increases by 3 second. Find its initial mass 'm'
a hot water tank containing 50,000g of water is heated by an electric immersion heater rated at 3kilowatt,240volt, calculate the current
what is charge
product of current and time
Jaffar
Why always amber gain electrons and fur loose electrons? Why the opposite doesn't happen?
A closely wound search coil has an area of 4cm^2,1000 turns and a resistance of 40ohm. It is connected to a ballistic galvanometer whose resistance is 24 ohm. When coil is rotated from a position parallel to uniform magnetic field to one perpendicular to field,the galvanometer indicates a charge
Using Kirchhoff's rules, when choosing your loops, can you choose a loop that doesn't have a voltage?
how was the check your understand 12.7 solved?