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Energy density, flux and power.

Figure 7.8:

We now turn our attention to the energy carried by the electromagnetic wave. For simplicity we shall initially restrict ourselves to points located along the $x$ axis for the situation shown in Figure 7.7.

The energy density in the electric and magnetic fields is given by

U=\frac{1}{2} \epsilon_0 E^2 + \frac{1}{2 \mu_0} B^2
\end{displaymath} (7.18)

For an electromagnetic wave the electric and magnetic fields are related. The energy density can then be written in terms of only the electric field as
U=\frac{1}{2} \epsilon_0 E^2 + \frac{1}{2 c^2 \mu_0} E^2
\end{displaymath} (7.19)

The speed of light $c$ is related to $\epsilon_0$ and $\mu_0$ as $c^2=1/\epsilon_0 \mu_0$. Using this we find that the energy density has the form
U= \epsilon_0 E^2
\end{displaymath} (7.20)

The instantaneous energy density of the electromagnetic wave oscillates with time. The time average of the energy density is often a more useful quantity. We have already discussed how to calculate the time average of an oscillating quantity. This is particularly simple in the complex notation where the electric field
\tilde{E}_y(x,t)= \tilde{E} e^{i (\omega t - k x)}
\end{displaymath} (7.21)

has a mean squared value $\langle E^2 \rangle= \tilde{E} \tilde{E}^*/2$. Using this we find that the average energy density is
\langle U \rangle = \frac{1}{2} \epsilon_0 \tilde{E} \tilde{E}^*=
\frac{1}{2} \epsilon_0 E^2
\end{displaymath} (7.22)

where $E$ is the amplitude of the electric field.

We next consider the energy flux of the electromagnetic radiation. The radiation propagates along the $x$ axis at the point where we want to calculate the flux. Consider a surface which is perpendicular to direction in which the wave is propagating as shown in Figure 7.8. The energy flux $S$ refers to the energy which crosses an unit area of this surface in unit time. It has units $Watt \, m^{-2}$. The flux $S$ is the power that would be received by collecting the radiation in an area $1 \, m^2$ placed perpendicular to the direction in which the wave is propagating as shown in Figure 7.8.

The average flux can be calculated by noting that the wave propagates along the $x$ axis with speed $c$. The average energy $\langle U \rangle$ contained in an unit volume would take a time to cross the surface. The flux is the energy which would cross in one second which is

\langle S \rangle = \langle U \rangle \, c = \frac{1}{2} c \epsilon_0 E^2
\end{displaymath} (7.23)

The energy flux is actually a vector quantity representing both the direction and the rate at which the wave carries energy. Referring back to equation (7.8) and (7.15) we see that at any point the average flux $\langle \vec{S} \rangle$ is pointed radially outwards and has a value
\langle \vec{S} \rangle = \left( \frac{q^2 y_0^2 \omega^4}{...
...3 \epsilon_0} \right) \frac{\sin^2 \theta}{r^2} \hat{r} \,.
\end{displaymath} (7.24)

Note that the flux falls as $1/r^2$ as we move away from the source. This is a property which may already be familiar to some of us from considerations of the conservation of energy. Note that the total energy crossing a surface enclosing the source will be constant irrespective of the shape and size of the surface.

Figure 7.9:

Let us know shift our point of reference to the location of the dipole and ask how much power is radiated in any given direction. This is quantified using the power emitted per solid angle. Consider a solid angle $d \Omega$ along a direction $\hat{r}$ at an angle $\theta$ to the dipole as shown in Figure 7.9. The power $d P$ radiated into this solid angle can be calculated by multiplying the flux with the area corresponding to this solid angle

d P = \langle \vec{S} \rangle \cdot \hat{r} \, r^2 \, d \Omega
\end{displaymath} (7.25)

which gives us the power radiated per unit solid angle to be
\frac{d }{d \Omega} \langle \vec{P} \rangle(\theta) = \left...^4}{32 \pi^2
c^3 \epsilon_0} \right) \sin^2 \theta \,.
\end{displaymath} (7.26)

This tells us the radiation pattern of the dipole radiation, ie.. the directional dependence of the radiation is proportional to $\sin^2 \theta$. The radiation is maximum in the direction perpendicular to the dipole while there is no radiation emitted along the direction of the dipole. The radiation pattern is shown in Figure 7.10. Another important point to note is that the radiation depends on $\omega^4$ which tells us that the same dipole will radiate significantly more power if it is made to oscillate at a higher frequency, doubling the frequency will increase the power sixteen times.
Figure 7.10:

The total power radiated can be calculated by integrating over all solid angles. Using $d \Omega = sin \theta \, d \theta \, d \phi$ and

\int_0^{2 \pi} d \phi \, \int_0^{\pi} sin^3 \theta \, d \theta =
\frac{8 \pi}{3}
\end{displaymath} (7.27)

gives the total power $P$ to be

It is often convenient to express this in terms of the amplitude of the current in the wires of the oscillator as
\langle \vec{P} \rangle =\frac{I^2 l^2 \omega^2}{12 \pi c^3 \epsilon_0} \,.
\end{displaymath} (7.29)

The power radiated by the electric dipole is proportional to the square of the current. This behaviour is exactly the same as that of a resistance except that the oscillator emits the power as radiation while the resistance converts it to heat. We can express the radiated power in terms of an equivalent resistance with
\langle \vec{P} \rangle = \frac{1}{2} R I^2
\end{displaymath} (7.30)

R=\left( \frac{l}{\lambda} \right)^2 \, 790 \, \Omega
\end{displaymath} (7.31)

$l$ being the length of the dipole (Figure 7.5) and $\lambda$ the wavelength of the radiation.
  1. An oscillating current of amplitude $2 \, Amps$ and $\omega= 3 \,
{\rm GHz}$ is fed into a dipole antenna of length $1 \, m$ oriented along the $y$ axis and located at $(0,0,0)$. All coordinates are in $km$ and $\epsilon_0=8.85 \times 10^{-12} \, C^2
\, N^{-1} \, m^{-2}$.
    What is the amplitude of the electric field at the point $(1,2,0)$? ( $1.2 \times 10^{-1} {\rm V}{\rm m}^{-1}$)
    In which direction does the electric oscillate at the point $(1,2,0)$? [Give the angles with respect to the $x$,$y$ and $z$ axis.] ($26.6^{\circ}$, $63.4^{\circ}$, $0^{\circ}$
    What is the total power radiated? ( $4.00 \times 10^3 {\rm W}$)
    How does the total power change if the frequency is doubled? ($4$ times)
  2. A charge $q$ moves in a circular orbit with period $T$ in the $y-z$ plane with center at the origin $(0,0,0)$. The charge is at at $t=0$, and the motion is counter-clockwise as seen from the point P $(r,0,0)$ where $r \gg a$. At a later time $t$
    What is the acceleration of the charge?
    For the point P, what is the retarded acceleration of the charge?
    What is the electric field at P?
    What is the time averaged power radiated by the charge?
  3. In which direction does an oscillating electric dipole radiate maximum power? What is the FWHM of the radiation pattern?
  4. For the same peak current and frequency, how does the total power change if the length of the dipole is halved.
  5. Consider a conducting wire of length $1 {\rm m}$ and $1 \, {\rm mm}$ diameter. At what frequency does the radiative resistance become comparable to the usual Ohmic resistance. Use $\sigma=10^{-7}
(\Omega m)^{-1}$ which is the typical value of conductivity for conducting metals.
  6. An electric dipole oscillator radiates $1 {\rm KW}$ power. What is the flux $1 {\rm km}$ away in the direction perpendicular to the dipole and at $30{\circ}$ to the dipole.
  7. Particles of charge $q$ and mass $m$ with kinetic energy are injected perpendicular to a magnetic field . The charges experiences an acceleration as they go around in circles in the magnetic field. Calculate the rate at which the energy is radiated. Show that the energy of the particles falls as $E(t)=E_0 e^{-t/\tau}$, where $\tau$ the decay time is related to $q,B$ and $m$.

    Cyclotrons typically have magnetic fields of $1 {\rm Tesla}$ or higher. Using this value, calculate the frequency at which the radiation will be emitted for electrons $(q,m)=(1.6\time 10^{-19} C, 9.1 \time
10^{-31} \, kg)$ and protons . Calculate $\tau$ for protons and electrons, and use this to determine how long it will take for them to radiate away half the initial kinetic energy.

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Next: The vector nature of Up: Electromagnetic Waves. Previous: Sinusoidal Oscillations.   Contents
Physics 1st Year 2009-01-06