Energy flow (current)

...and intensity of E-M waves.

Read sections 8.1.1 and 8.1.2

The rate of work done by E-M fields on charges in a volume ${\cal V}$ can be written as $$\frac{d W}{dt}=-\frac{d}{dt}\int_{\cal V}\frac{1}{2}\left(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\right) d\tau-\frac{1}{\mu_0}\oint_{\cal S}\left(\myv E \times \myv B\right)\cdot d\myv a.$$ Poynting's Theorem:

  • The rate of work done on charges in {\cal V} by the electromagnetic force is equal to:
  • the rate of decrease of the energy stored in fields in volume ${\cal V}$,
  • minus the power that flowed out through the surface bounding ${\cal V}$.

So apparently the quantity in the surface integral is the "power flux" per unit time leaving volume ${\cal V}$: $$\myv S\equiv\frac{1}{\mu_0}(\myv E\times\myv B).$$ Since power is energy/time, we could also think of $S$ as energy/time/unit area.

Energy loss in a wire

Example 8.1

The power dissipated in a resistor is the product $P=VI$. We *should* get the same thing if we integrate the Poynting vector over the surface of the cylindrical wire shown...right?

  • $\myv E$ points in the direction of the current, and has magnitude $$ E=\frac{V}{L}$$
  • $\myv B$ points in the $\uv \phi$ direction, tangent to the surface, and has magnitude $$ B=\frac{\mu_0I}{2\pi a}$$
  • Find $|\myv S$, and
  • Integrate over the cylinder surface $\int \myv S\cdot d\myv a$ to get the power based on our calculation of $\myv E$ and $\myv B$...

Energy flow of a plane wave

What direction does the Poynting vector point? $$\myv{S}=\frac{1}{\mu_0}\myv E\times\myv B? $$

  1. Pick a direction for $\myv k$: let's say it points in the $\uv x$ direction.
  2. We showed above the $\myv E\cdot \myv k=0$. So, we have some freedom as far as the direction of $\myv E$ goes, but it has to be at right angles to $\myv k$. Since $$\myv E(\myv r,t) = \myv{\cal E}\cos(\myv k\cdot \myv r - \omega t),$$ this means that $\myv{\cal E} \perp \myv k$. So let's say that ${\cal E}$ points in the $\uv y$ direction.
  3. As time passes the instantaneous $\myv{E}$ at some point in space will oscillate sinusoidally, sometimes pointing in the $+\uv y$ and sometimes in the $-\uv y$ direction.
  4. Once we've set a particular direction for $\myv{\cal E}$ (a "polarization" state), we do not have similar freedom to pick any ol' $\myv B$ field orientation, because $\myv E$ and $\myv B$ are linked through $$\omega \myv B = \myv k \times \myv E.$$ Since both fields for a particular plane wave solution are oscillating with the same cosine dependence, this implies that $$\omega \myv{\mathcal B} = \myv k \times \myv{\mathcal E}.$$

The questions for you to figure out are...

  1. What direction is $\myv{\cal B}$ pointing if $\myv{\cal E}={\cal E}\uv y$ and $\myv k = k\uv x$?
  2. What direction is $\myv S$ pointing when $\myv{E}$ is temporarily pointing in the $+\uv y$ direction?
  3. But $\myv E$ and $\myv B$ are both oscillating between positive and negative values in their respective directions. Does this also mean that $\myv S$ oscillates between + and - directions or not?

Power of a plane wave

We previously found that, for a plane wave solution of Maxwell's equations, $\myv E \perp \myv B \perp \myv k$ , and $\myv E$ and $\myv B$ are in-phase, and the magnitude of the magnetic field is $${\cal B}=(k/\omega){\cal E}=(1/v){\cal E}.$$

In free space, the Poynting vector, which is the power per unit area (area perpendicular to the direction of propagation) is $$S=\frac{1}{\mu_0}(\myv E\times \myv B)$$

If the wave is propagating in the $\uv z$ directions, $$\myv S=\frac{1}{\mu_0}{\cal E}{\cal B}\cos^2(kz-\omega t+\delta)\uv z$$ The constants can be written: $$\frac{1}{\mu_0}{\cal E}{\cal B}=\frac{\sqrt{\mu_0\epsilon_0}}{\mu_0}{\cal E}^2=c\epsilon_0{\cal E}^2$$

The average value of the $\cos^2(...)$ function over the course of on cycle is 1/2. So, the average value of the Poynting vector is $$\langle \myv S \rangle = \frac 12 c\epsilon_0{\cal E}^2\uv z.$$ The average power of an electromagnetic wave per unit area is called its intensity, I: $$I\equiv \langle S \rangle = \frac 12 c\epsilon_0{\cal E}^2.$$