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Additional: Properties of Electromagnetic Waves

Energy Carried by EM Waves

Electromagnetic waves carry energy as they propagate through space. This is evident from phenomena like the warming effect of sunlight (infrared and visible light), the heating of food in a microwave oven (microwaves), or the energy deposited in tissues during X-ray exposure. This energy is stored in the oscillating electric and magnetic fields of the wave.

Energy Density of Electric Field ($ u_E $)

We know from electrostatics that an electric field stores energy in space. The energy density (energy per unit volume) of an electric field $\vec{E}$ in vacuum is given by:

$ u_E = \frac{1}{2} \epsilon_0 E^2 $

Where $\epsilon_0$ is the permittivity of free space and $E$ is the magnitude of the electric field.

Energy Density of Magnetic Field ($ u_B $)

Similarly, a magnetic field also stores energy in space. The energy density of a magnetic field $\vec{B}$ in vacuum is given by:

$ u_B = \frac{B^2}{2\mu_0} $

Where $\mu_0$ is the permeability of free space and $B$ is the magnitude of the magnetic field.

Total Energy Density of an EM Wave

An electromagnetic wave consists of both electric and magnetic fields. The total instantaneous energy density ($u$) of an electromagnetic wave at any point in space is the sum of the energy densities of the electric and magnetic fields at that point:

$ u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{B^2}{2\mu_0} $

For a plane electromagnetic wave in vacuum, the magnitudes of the electric and magnetic fields are related by $E = cB$, where $c = 1/\sqrt{\mu_0 \epsilon_0}$. This means $c^2 = 1/(\mu_0 \epsilon_0)$.

Substitute $B = E/c$ into the expression for $u_B$:

$ u_B = \frac{(E/c)^2}{2\mu_0} = \frac{E^2}{2\mu_0 c^2} $

Substitute $c^2 = 1/(\mu_0 \epsilon_0)$ or $\mu_0 c^2 = 1/\epsilon_0$:

$ u_B = \frac{E^2}{2\mu_0 (1/(\mu_0 \epsilon_0))} = \frac{E^2}{2/\epsilon_0} = \frac{1}{2} \epsilon_0 E^2 $

Thus, in an electromagnetic wave in vacuum, the energy density of the magnetic field is equal to the energy density of the electric field at every instant and at every point in space.

$ u_B = u_E = \frac{1}{2} \epsilon_0 E^2 $

The total instantaneous energy density is then:

$ u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \epsilon_0 E^2 = \epsilon_0 E^2 $

We can also express this in terms of $B$: since $E=cB$, $E^2 = c^2 B^2 = (1/(\mu_0\epsilon_0)) B^2$. So $\epsilon_0 E^2 = \epsilon_0 \frac{B^2}{\mu_0 \epsilon_0} = \frac{B^2}{\mu_0}$. This doesn't match $u = B^2/\mu_0$. Let's recheck the relation $u_B = u_E$.

$u_B = \frac{B^2}{2\mu_0}$. Using $B=E/c$, $u_B = \frac{(E/c)^2}{2\mu_0} = \frac{E^2}{2\mu_0 c^2}$. Using $c^2 = 1/(\mu_0 \epsilon_0)$: $u_B = \frac{E^2}{2\mu_0 (1/(\mu_0 \epsilon_0))} = \frac{E^2 \mu_0 \epsilon_0}{2\mu_0} = \frac{1}{2} \epsilon_0 E^2$. This is correct.

So $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \epsilon_0 E^2 = \epsilon_0 E^2$.

Also $u = u_E + u_B = \frac{B^2}{2\mu_0} + \frac{B^2}{2\mu_0} = \frac{B^2}{\mu_0}$.

So, the total instantaneous energy density is $u = \epsilon_0 E^2 = B^2/\mu_0$. Wait, this doesn't look right. The initial formulas for $u_E$ and $u_B$ are correct. Their sum should be correct. $\epsilon_0 E^2 = \epsilon_0 (cB)^2 = \epsilon_0 c^2 B^2 = \epsilon_0 \frac{1}{\mu_0 \epsilon_0} B^2 = \frac{B^2}{\mu_0}$. Ah, yes, $\epsilon_0 E^2 = B^2/\mu_0$. So the total energy density can be written in terms of E alone or B alone.

$ u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{B^2}{2\mu_0} = \frac{1}{2} \epsilon_0 E^2 + \frac{\epsilon_0 E^2}{2} = \epsilon_0 E^2 $ (This seems right)

$ u = u_E + u_B = \frac{B^2}{2\mu_0} + \frac{B^2}{2\mu_0} = \frac{B^2}{\mu_0}$ (This also seems right)

Let's recheck the $u_B = \frac{1}{2}\epsilon_0 E^2$ derivation. $u_B = \frac{B^2}{2\mu_0}$. Using $B=E/c$: $u_B = \frac{(E/c)^2}{2\mu_0} = \frac{E^2}{2\mu_0 c^2}$. Using $c^2 = 1/(\mu_0 \epsilon_0)$, so $\mu_0 c^2 = 1/\epsilon_0$: $u_B = \frac{E^2}{2(1/\epsilon_0)} = \frac{1}{2} \epsilon_0 E^2$. This is correct.

So, instantaneous energy density $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \epsilon_0 E^2 = \epsilon_0 E^2$. Also $u = u_E + u_B = \frac{B^2}{2\mu_0} + \frac{B^2}{2\mu_0} = \frac{B^2}{\mu_0}$. There must be a factor of 2 difference.

Ah, $u = u_E + u_B = \frac{1}{2} \epsilon_0 E^2 + \frac{B^2}{2\mu_0}$. Using $B=E/c$ and $c=1/\sqrt{\mu_0\epsilon_0}$: $u = \frac{1}{2} \epsilon_0 E^2 + \frac{(E/c)^2}{2\mu_0} = \frac{1}{2} \epsilon_0 E^2 + \frac{E^2}{2\mu_0 c^2} = \frac{1}{2} \epsilon_0 E^2 + \frac{E^2}{2\mu_0 (1/\mu_0\epsilon_0)} = \frac{1}{2} \epsilon_0 E^2 + \frac{E^2 \mu_0 \epsilon_0}{2\mu_0} = \frac{1}{2} \epsilon_0 E^2 + \frac{1}{2} \epsilon_0 E^2 = \epsilon_0 E^2$. This confirms $u=\epsilon_0 E^2$.

Now using $E=cB$ and $c=1/\sqrt{\mu_0\epsilon_0}$: $u = \frac{1}{2} \epsilon_0 (cB)^2 + \frac{B^2}{2\mu_0} = \frac{1}{2} \epsilon_0 c^2 B^2 + \frac{B^2}{2\mu_0} = \frac{1}{2} \epsilon_0 \left(\frac{1}{\mu_0 \epsilon_0}\right) B^2 + \frac{B^2}{2\mu_0} = \frac{B^2}{2\mu_0} + \frac{B^2}{2\mu_0} = \frac{B^2}{\mu_0}$.

Okay, $u = \epsilon_0 E^2$ and $u = B^2/\mu_0$ are BOTH correct ways to write the total instantaneous energy density, because $\epsilon_0 E^2 = B^2/\mu_0$ due to the relation $E=cB$ and $c=1/\sqrt{\mu_0\epsilon_0}$.

Average Energy Density

Since the electric and magnetic fields in an electromagnetic wave vary sinusoidally, $E = E_0 \sin(kx - \omega t)$ and $B = B_0 \sin(kx - \omega t)$, the energy densities also vary with time and position.

$ u_E = \frac{1}{2} \epsilon_0 E_0^2 \sin^2(kx - \omega t) $

$ u_B = \frac{B_0^2}{2\mu_0} \sin^2(kx - \omega t) $

The average value of $\sin^2(kx - \omega t)$ over a cycle of time or space is $1/2$.

The average energy density is:

$ \langle u \rangle = \langle u_E \rangle + \langle u_B \rangle = \frac{1}{2} \epsilon_0 \langle E^2 \rangle + \frac{\langle B^2 \rangle}{2\mu_0} $

$ \langle E^2 \rangle = \langle E_0^2 \sin^2(kx - \omega t) \rangle = E_0^2 \langle \sin^2(kx - \omega t) \rangle = \frac{1}{2} E_0^2 $

$ \langle B^2 \rangle = \langle B_0^2 \sin^2(kx - \omega t) \rangle = B_0^2 \langle \sin^2(kx - \omega t) \rangle = \frac{1}{2} B_0^2 $

$ \langle u_E \rangle = \frac{1}{2} \epsilon_0 \left(\frac{1}{2} E_0^2\right) = \frac{1}{4} \epsilon_0 E_0^2 $

$ \langle u_B \rangle = \frac{(1/2) B_0^2}{2\mu_0} = \frac{B_0^2}{4\mu_0} $

Since $\frac{1}{2}\epsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$ (from $E_0=cB_0$), we have $\langle u_E \rangle = \langle u_B \rangle$.

The total average energy density is:

$ \langle u \rangle = \langle u_E \rangle + \langle u_B \rangle = \frac{1}{4} \epsilon_0 E_0^2 + \frac{1}{4} \epsilon_0 E_0^2 = \frac{1}{2} \epsilon_0 E_0^2 $

Also, $ \langle u \rangle = \langle u_E \rangle + \langle u_B \rangle = \frac{B_0^2}{4\mu_0} + \frac{B_0^2}{4\mu_0} = \frac{B_0^2}{2\mu_0} $.

So, the average energy density is $\langle u \rangle = \frac{1}{2} \epsilon_0 E_0^2 = \frac{B_0^2}{2\mu_0}$. This is also equal to $\frac{1}{2} \epsilon_0 E_{rms}^2 + \frac{B_{rms}^2}{2\mu_0}$, where $E_{rms} = E_0/\sqrt{2}$ and $B_{rms} = B_0/\sqrt{2}$.

Energy Flow and Intensity

Electromagnetic waves carry energy as they propagate. The rate of energy flow per unit area perpendicular to the direction of propagation is called the intensity ($I$) or irradiance of the wave.

Consider a volume of space traversed by the wave. If the wave travels at speed $c$, in a time $dt$, the wave covers a distance $dx = c \, dt$. The energy contained in a volume $dV = A \, dx = A c \, dt$ will pass through the area $A$ in time $dt$.

The energy in this volume is $dU = u \, dV = u A c \, dt$, where $u$ is the instantaneous energy density.

The rate of energy flow through area $A$ is $dP = dU/dt = uAc$.

The intensity $I$ is the power per unit area:

$ I = \frac{dP}{A} = \frac{uAc}{A} = uc $

Using the average energy density $\langle u \rangle = \frac{1}{2} \epsilon_0 E_0^2$, the average intensity is:

$ I_{avg} = \langle u \rangle c = \left(\frac{1}{2} \epsilon_0 E_0^2\right) c = \frac{1}{2} \epsilon_0 c E_0^2 $

Using $E_0 = cB_0$ and $c=1/\sqrt{\mu_0\epsilon_0}$:

$ I_{avg} = \frac{1}{2} \epsilon_0 \left(\frac{1}{\sqrt{\mu_0\epsilon_0}}\right) E_0^2 = \frac{1}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_0^2 $

Also, $ I_{avg} = \langle u \rangle c = \left(\frac{B_0^2}{2\mu_0}\right) c = \frac{c B_0^2}{2\mu_0} $.

Using $E_0 = cB_0$, $B_0 = E_0/c$. $ I_{avg} = \frac{c (E_0/c)^2}{2\mu_0} = \frac{c E_0^2}{2\mu_0 c^2} = \frac{E_0^2}{2\mu_0 c} $. Using $c=1/\sqrt{\mu_0\epsilon_0}$: $I_{avg} = \frac{E_0^2}{2\mu_0 (1/\sqrt{\mu_0\epsilon_0})} = \frac{E_0^2 \sqrt{\mu_0\epsilon_0}}{2\mu_0} = \frac{E_0^2 \sqrt{\epsilon_0}}{2\sqrt{\mu_0}}$. Hmm, should be $\sqrt{\epsilon_0/\mu_0}$.

Let's restart the average intensity expressions using $E_{rms}$ and $B_{rms}$. $\langle u \rangle = \frac{1}{2} \epsilon_0 E_0^2 = \epsilon_0 E_{rms}^2$. $\langle u \rangle = \frac{B_0^2}{2\mu_0} = \frac{B_{rms}^2}{\mu_0}$.

$ I_{avg} = \langle u \rangle c = \epsilon_0 E_{rms}^2 c $

$ I_{avg} = \langle u \rangle c = \frac{B_{rms}^2}{\mu_0} c $

Using $E_{rms} = cB_{rms}$:

$ I_{avg} = E_{rms} B_{rms} \frac{c}{c} = E_{rms} B_{rms} $ (This is wrong, should be $E_{rms} B_{rms} c$)

Let's use $E_0=cB_0$. $I_{avg} = \frac{V_0 I_0}{2} \cos\phi$. Wait, this was for AC circuits. For EM waves, there's no component like R, L, C in vacuum. The energy flow is related to the fields directly.

$ I_{avg} = \langle u \rangle c $. Using $\langle u \rangle = \frac{1}{2}\epsilon_0 E_0^2$: $I_{avg} = \frac{1}{2}\epsilon_0 E_0^2 c$. This is correct. Using $\langle u \rangle = \frac{B_0^2}{2\mu_0}$: $I_{avg} = \frac{B_0^2 c}{2\mu_0}$. This is correct. Using $E_0=cB_0$: $I_{avg} = \frac{E_0 (E_0/c) c}{2\mu_0} = \frac{E_0^2}{2\mu_0}$. No, this is wrong.

Let's retry $I_{avg}$ in terms of $E_0$ and $B_0$: $I_{avg} = \frac{V_0 I_0}{2}$ for a resistive circuit. This is not for EM waves in vacuum.

The power per unit area is $S = uc$. Average power per unit area $I_{avg} = \langle u \rangle c$. $\langle u \rangle = \frac{1}{2}\epsilon_0 E_{rms}^2 + \frac{B_{rms}^2}{2\mu_0}$. $E_{rms} = c B_{rms}$. So $\frac{1}{2}\epsilon_0 E_{rms}^2 = \frac{1}{2}\epsilon_0 (cB_{rms})^2 = \frac{1}{2}\epsilon_0 c^2 B_{rms}^2 = \frac{1}{2}\epsilon_0 \frac{1}{\mu_0\epsilon_0} B_{rms}^2 = \frac{B_{rms}^2}{2\mu_0}$.

$\langle u \rangle = \frac{1}{2}\epsilon_0 E_{rms}^2 + \frac{1}{2}\epsilon_0 E_{rms}^2 = \epsilon_0 E_{rms}^2$. $\langle u \rangle = \frac{B_{rms}^2}{2\mu_0} + \frac{B_{rms}^2}{2\mu_0} = \frac{B_{rms}^2}{\mu_0}$.

$I_{avg} = \langle u \rangle c = \epsilon_0 E_{rms}^2 c$. Using $E_{rms} = c B_{rms}$: $I_{avg} = \epsilon_0 (cB_{rms}) E_{rms} = \epsilon_0 c E_{rms} B_{rms}$. Using $B_{rms} = E_{rms}/c$: $I_{avg} = \epsilon_0 E_{rms}^2 c$. Using $E_{rms} = c B_{rms}$: $I_{avg} = \epsilon_0 (cB_{rms})^2 c = \epsilon_0 c^3 B_{rms}^2$. This seems off.

Let's use $I_{avg} = E_{rms} H_{rms} \cos\phi$ where $H_{rms} = B_{rms}/\mu_0$. For EM waves in vacuum, $E$ and $B$ are in phase, related by $E=cB$. $H=B/\mu_0 = E/(c\mu_0)$. $I_{avg} = E_{rms} H_{rms} = E_{rms} (B_{rms}/\mu_0)$. Using $B_{rms} = E_{rms}/c$. $I_{avg} = E_{rms} \frac{E_{rms}/c}{\mu_0} = \frac{E_{rms}^2}{\mu_0 c}$. Using $c=1/\sqrt{\mu_0\epsilon_0}$: $I_{avg} = \frac{E_{rms}^2}{\mu_0 (1/\sqrt{\mu_0\epsilon_0})} = \frac{E_{rms}^2 \sqrt{\mu_0\epsilon_0}}{\mu_0} = E_{rms}^2 \sqrt{\frac{\epsilon_0}{\mu_0}}$. This is correct.

Using $E_{rms} = c B_{rms}$: $I_{avg} = \frac{(cB_{rms})^2}{\mu_0 c} = \frac{c^2 B_{rms}^2}{\mu_0 c} = \frac{c B_{rms}^2}{\mu_0}$. This is also correct.

Using $E_{rms} = c B_{rms}$: $I_{avg} = E_{rms} B_{rms} c$. No, $I_{avg} = E_{rms} B_{rms}$ is not the formula. $I_{avg} = E_{rms} H_{rms} = E_{rms} (B_{rms}/\mu_0)$. This is correct.

Let's summarize the average intensity:

$ I_{avg} = \langle u \rangle c = \frac{1}{2} \epsilon_0 E_0^2 c = \epsilon_0 E_{rms}^2 c $

$ I_{avg} = \frac{B_0^2 c}{2\mu_0} = \frac{B_{rms}^2 c}{\mu_0} $

$ I_{avg} = \frac{E_0 B_0}{2\mu_0} $ (since $c = E_0/B_0$)

$ I_{avg} = \frac{E_{rms} B_{rms}}{\mu_0} $ (since $c = E_{rms}/B_{rms}$)

These formulas give the average intensity of the electromagnetic wave in vacuum.

Example 1. A monochromatic plane electromagnetic wave has a peak electric field amplitude of $100 \, V/m$. Calculate (a) the peak magnetic field amplitude, (b) the average energy density, and (c) the average intensity of the wave in vacuum. ($\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$, $c = 3 \times 10^8 \, m/s$, $\epsilon_0 = 8.854 \times 10^{-12} \, C^2/(N \cdot m^2)$).

Answer:

Given:

Peak electric field amplitude, $E_0 = 100 \, V/m$

Speed of light in vacuum, $c = 3 \times 10^8 \, m/s$

Permeability of free space, $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$

Permittivity of free space, $\epsilon_0 = 8.854 \times 10^{-12} \, C^2/(N \cdot m^2)$

(a) The peak magnetic field amplitude ($B_0$) is related to the peak electric field amplitude by $E_0 = cB_0$.

$ B_0 = \frac{E_0}{c} = \frac{100 \, V/m}{3 \times 10^8 \, m/s} $

$ B_0 = \frac{1}{3} \times 10^{-6} \, T \approx 0.333 \times 10^{-6} \, T = 3.33 \times 10^{-7} \, T $

The peak magnetic field amplitude is approximately $3.33 \times 10^{-7} \, T$.

(b) The average energy density $\langle u \rangle = \frac{1}{2} \epsilon_0 E_0^2$.

$ \langle u \rangle = \frac{1}{2} (8.854 \times 10^{-12} \, C^2/(N \cdot m^2)) (100 \, V/m)^2 $

$ \langle u \rangle = \frac{1}{2} \times 8.854 \times 10^{-12} \times 10000 \, J/m^3 $ (Unit check: $C^2/N \cdot m^2 \cdot V^2/m^2 = C^2 (J/C)^2 / (N m^4) = C^2 J^2 / (C^2 N m^4) = J^2 / (J/m \cdot m^4) = J^2 / (J m^3) = J/m^3$)

$ \langle u \rangle = 4.427 \times 10^{-12} \times 10^4 \, J/m^3 = 4.427 \times 10^{-8} \, J/m^3 $

The average energy density is $4.427 \times 10^{-8} \, J/m^3$.

(c) The average intensity $I_{avg} = \langle u \rangle c$.

$ I_{avg} = (4.427 \times 10^{-8} \, J/m^3) \times (3 \times 10^8 \, m/s) $

$ I_{avg} = (4.427 \times 3) \times 10^{0} \, W/m^2 $ (Unit check: $J/m^3 \cdot m/s = J/(m^2 s) = W/m^2$)

$ I_{avg} = 13.281 \, W/m^2 $

The average intensity of the wave is 13.281 Watts per square meter.


Momentum Carried by EM Waves (Radiation Pressure)

Besides carrying energy, electromagnetic waves also carry momentum. This might seem counter-intuitive since photons (quanta of EM waves) are massless particles, but they do possess momentum. When an electromagnetic wave interacts with a surface, it can transfer some or all of its momentum to the surface, exerting a force. This force per unit area is called radiation pressure.

Momentum Transfer

The momentum ($p$) carried by an electromagnetic wave is related to its energy ($U$) by:

$ p = \frac{U}{c} $

(This relation comes from the relativistic energy-momentum relation for massless particles, $E=pc$, where $E=U$ here).

Consider an electromagnetic wave incident on a surface.

Radiation Pressure

Radiation pressure ($P_{rad}$) is the force exerted per unit area by an electromagnetic wave on a surface.

The average intensity of the incident wave is $I_{avg} = P/A_{surface}$, where $P$ is the power incident on the surface area $A_{surface}$.

For a surface with reflectivity $r$ and absorptivity $a=1-r$, part of the energy is absorbed and part is reflected. The radiation pressure is $P_{rad} = (1+r) \frac{I_{avg}}{c} = (2-a) \frac{I_{avg}}{c}$.

Significance of Radiation Pressure

Radiation pressure is typically very small and is not noticeable in everyday life due to the high value of $c$. However, it is significant in certain contexts:

Example 1. Sunlight incident on the Earth's surface has an approximate average intensity of $1300 \, W/m^2$ (this is the solar constant, measured outside the atmosphere). Calculate the radiation pressure exerted by sunlight on a perfectly absorbing surface facing the Sun. ($c = 3 \times 10^8 \, m/s$).

Answer:

Given:

Average intensity of sunlight, $I_{avg} = 1300 \, W/m^2$

Speed of light, $c = 3 \times 10^8 \, m/s$

Surface is perfectly absorbing.

The radiation pressure for a perfectly absorbing surface is given by $P_{rad} = I_{avg}/c$.

$ P_{rad} = \frac{1300 \, W/m^2}{3 \times 10^8 \, m/s} $

$ P_{rad} = \frac{1300}{3} \times 10^{-8} \, N/m^2 $ (Unit check: $W/m^2 / (m/s) = (J/s)/m^3 = J/(s m^3)$. Wait, $W/m^2 / (m/s) = (N \cdot m/s)/m^3 = N/(m^2 s)$. Unit should be $N/m^2$. $W = J/s$, $J = N \cdot m$. $W/m^2 / (m/s) = (N \cdot m/s)/m^2 / (m/s) = N/m^2$. Yes, unit is correct).

$ P_{rad} \approx 433.33 \times 10^{-8} \, N/m^2 = 4.333 \times 10^{-6} \, N/m^2 $

The radiation pressure exerted by sunlight on a perfectly absorbing surface is approximately $4.333 \times 10^{-6}$ Pascals. This is a very small pressure compared to atmospheric pressure (about $10^5$ Pa).


Poynting Vector ($ \vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B} $)

Electromagnetic waves transport energy. The rate and direction of this energy flow are described by the Poynting vector. It is named after John Henry Poynting, who introduced it in 1884.

Definition of the Poynting Vector

The Poynting vector ($\vec{S}$) represents the instantaneous power per unit area carried by an electromagnetic field. Its magnitude gives the intensity of the wave (energy flow per unit area per unit time), and its direction gives the direction of energy flow, which is also the direction of wave propagation.

In vacuum, the Poynting vector is defined as:

$ \vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) $

Where $\vec{E}$ is the instantaneous electric field vector, $\vec{B}$ is the instantaneous magnetic field vector, and $\mu_0$ is the permeability of free space.

The SI unit of the Poynting vector is Watts per square meter ($W/m^2$), which is the unit of power per unit area (intensity).

Direction of Energy Flow

For a plane electromagnetic wave, the electric field $\vec{E}$ and the magnetic field $\vec{B}$ are perpendicular to each other and in phase (or $180^\circ$ out of phase depending on how you define the wave). The direction of propagation is perpendicular to both $\vec{E}$ and $\vec{B}$.

The direction of the cross product $\vec{E} \times \vec{B}$ is perpendicular to both $\vec{E}$ and $\vec{B}$. By the Right-Hand Rule, if $\vec{E}$ is along the y-axis and $\vec{B}$ is along the z-axis, then $\vec{E} \times \vec{B}$ is along the x-axis. This is the direction of propagation for a wave described by $E = E_0 \sin(kx - \omega t)\hat{j}$ and $B = B_0 \sin(kx - \omega t)\hat{k}$.

Thus, the direction of the Poynting vector $\vec{S}$ is the direction of propagation of the electromagnetic wave, consistent with the flow of energy in that direction.

Magnitude of the Poynting Vector (Instantaneous Intensity)

The magnitude of the Poynting vector $|S|$ is the instantaneous power per unit area (instantaneous intensity).

$ |S| = \left|\frac{1}{\mu_0} (\vec{E} \times \vec{B})\right| = \frac{1}{\mu_0} |\vec{E}| |\vec{B}| \sin\alpha $

where $\alpha$ is the angle between $\vec{E}$ and $\vec{B}$. For an electromagnetic wave, $\vec{E}$ and $\vec{B}$ are perpendicular, so $\alpha = 90^\circ$ and $\sin\alpha = 1$.

$ S = \frac{EB}{\mu_0} $

Using the relation $E = cB$ or $B = E/c$ for a wave in vacuum, and $c = 1/\sqrt{\mu_0\epsilon_0}$:

$ S = \frac{E (E/c)}{\mu_0} = \frac{E^2}{\mu_0 c} = \frac{E^2}{\mu_0 (1/\sqrt{\mu_0\epsilon_0})} = E^2 \sqrt{\frac{\epsilon_0}{\mu_0}} $

Also, $ S = \frac{(cB)B}{\mu_0} = \frac{c B^2}{\mu_0} = \frac{(1/\sqrt{\mu_0\epsilon_0}) B^2}{\mu_0} = \frac{B^2}{\sqrt{\mu_0\epsilon_0} \mu_0} = \frac{B^2}{\sqrt{\mu_0^3\epsilon_0}} $ (This doesn't seem like a useful form).

Let's stick to $S = EB/\mu_0$. Using $E=cB$: $S = (cB)B/\mu_0 = cB^2/\mu_0$. Using $B=E/c$: $S = E(E/c)/\mu_0 = E^2/(\mu_0 c)$. Using $c = 1/\sqrt{\mu_0\epsilon_0}$: $S = E^2/(\mu_0 \cdot 1/\sqrt{\mu_0\epsilon_0}) = E^2 \sqrt{\mu_0 \epsilon_0}/\mu_0 = E^2 \sqrt{\epsilon_0}/\sqrt{\mu_0} = E^2 \sqrt{\epsilon_0/\mu_0}$. This is consistent with $I_{avg} = E_{rms}^2 \sqrt{\epsilon_0/\mu_0}$.

The instantaneous Poynting vector is $S(x,t) = \frac{E_0 B_0}{\mu_0} \sin^2(kx - \omega t)$ (assuming fields are in phase, which they are in vacuum).

Average Poynting Vector (Average Intensity)

The average magnitude of the Poynting vector over a cycle is the average intensity $I_{avg}$:

$ I_{avg} = \langle S \rangle = \left\langle \frac{EB}{\mu_0} \right\rangle $

Since $E$ and $B$ are in phase and $E=cB$, $S = EB/\mu_0 = (cB)B/\mu_0 = cB^2/\mu_0$.

$ I_{avg} = \left\langle \frac{c B^2}{\mu_0} \right\rangle = \frac{c}{\mu_0} \langle B^2 \rangle = \frac{c}{\mu_0} \left(\frac{B_0^2}{2}\right) = \frac{c B_0^2}{2\mu_0} $

Using $B_0 = E_0/c$:

$ I_{avg} = \frac{c (E_0/c)^2}{2\mu_0} = \frac{c E_0^2}{2\mu_0 c^2} = \frac{E_0^2}{2\mu_0 c} $

Using $c = 1/\sqrt{\mu_0\epsilon_0}$:

$ I_{avg} = \frac{E_0^2}{2\mu_0 (1/\sqrt{\mu_0\epsilon_0})} = \frac{E_0^2 \sqrt{\mu_0\epsilon_0}}{2\mu_0} = \frac{E_0^2 \sqrt{\epsilon_0}}{2\sqrt{\mu_0}} = \frac{1}{2} E_0^2 \sqrt{\frac{\epsilon_0}{\mu_0}} $

Also, $ I_{avg} = \frac{E_{rms} B_{rms}}{\mu_0} $ where $E_{rms} = E_0/\sqrt{2}$ and $B_{rms} = B_0/\sqrt{2}$. $ \frac{(E_0/\sqrt{2})(B_0/\sqrt{2})}{\mu_0} = \frac{E_0 B_0}{2\mu_0} $, which is consistent with $\frac{c B_0^2}{2\mu_0} = \frac{(E_0/B_0) B_0^2}{2\mu_0} = \frac{E_0 B_0}{2\mu_0}$.

The average Poynting vector is sometimes denoted by $\langle \vec{S} \rangle$, and its magnitude is the average intensity $I_{avg}$.

The Poynting vector is a fundamental concept in electromagnetism, describing the energy flow associated with electric and magnetic fields. It is applicable not only to propagating electromagnetic waves but also to static fields and circuits where energy is being transferred.

Example 1. A plane electromagnetic wave in vacuum has an electric field given by $\vec{E}(x,t) = 200 \sin(kx - \omega t) \, \hat{j} \, V/m$. Calculate the magnitude and direction of the instantaneous Poynting vector at a point $(x,t)$. ($\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$, $c = 3 \times 10^8 \, m/s$).

Answer:

Given:

Electric field, $\vec{E}(x,t) = 200 \sin(kx - \omega t) \, \hat{j} \, V/m$. Peak electric field $E_0 = 200 \, V/m$. $\vec{E}$ is in the +y direction.

Wave is propagating in vacuum along the +x direction. In an EM wave, $\vec{B}$ is perpendicular to $\vec{E}$ and the direction of propagation. The direction of propagation is $\vec{E} \times \vec{B}$. Since propagation is $\hat{i}$ and $\vec{E}$ is $\hat{j}$, $\hat{j} \times (\text{direction of } \vec{B}) = \hat{i}$. This implies direction of $\vec{B}$ is $\hat{k}$ (+z direction).

The instantaneous magnetic field magnitude is $B(x,t) = E(x,t)/c$. The fields are in phase in vacuum.

$ B(x,t) = \frac{200 \, V/m}{3 \times 10^8 \, m/s} \sin(kx - \omega t) \, T = \frac{2}{3} \times 10^{-6} \sin(kx - \omega t) \, T $

$ \vec{B}(x,t) = \frac{2}{3} \times 10^{-6} \sin(kx - \omega t) \, \hat{k} \, T $

The instantaneous Poynting vector is $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B})$.

$ \vec{S} = \frac{1}{\mu_0} [ (E(x,t) \, \hat{j}) \times (B(x,t) \, \hat{k}) ] $

$ \vec{S} = \frac{1}{\mu_0} E(x,t) B(x,t) (\hat{j} \times \hat{k}) $

Since $\hat{j} \times \hat{k} = \hat{i}$ and $E(x,t)$ and $B(x,t)$ are both $\ge 0$ or $\le 0$ at the same time:

$ \vec{S} = \frac{E(x,t) B(x,t)}{\mu_0} \, \hat{i} $

The direction of the instantaneous Poynting vector is along the +x direction, which is the direction of wave propagation.

Now, calculate the magnitude of the instantaneous Poynting vector $S(x,t) = \frac{E(x,t) B(x,t)}{\mu_0}$.

$ S(x,t) = \frac{[200 \sin(kx - \omega t)] \times [\frac{2}{3} \times 10^{-6} \sin(kx - \omega t)]}{4\pi \times 10^{-7}} \, W/m^2 $

$ S(x,t) = \frac{200 \times (2/3) \times 10^{-6}}{4\pi \times 10^{-7}} \sin^2(kx - \omega t) \, W/m^2 $

$ S(x,t) = \frac{400/3 \times 10^{-6}}{4\pi \times 10^{-7}} \sin^2(kx - \omega t) \, W/m^2 $

$ S(x,t) = \frac{100/3}{\pi} \times \frac{10^{-6}}{10^{-7}} \sin^2(kx - \omega t) \, W/m^2 $

$ S(x,t) = \frac{100}{3\pi} \times 10 \sin^2(kx - \omega t) \, W/m^2 = \frac{1000}{3\pi} \sin^2(kx - \omega t) \, W/m^2 $

$ S(x,t) \approx \frac{1000}{3 \times 3.14159} \sin^2(kx - \omega t) \, W/m^2 \approx \frac{1000}{9.42477} \sin^2(kx - \omega t) \, W/m^2 $

$ S(x,t) \approx 106.1 \sin^2(kx - \omega t) \, W/m^2 $

The magnitude of the instantaneous Poynting vector is approximately $106.1 \sin^2(kx - \omega t) \, W/m^2$, and its direction is along the +x axis.