Energy density and intensity

Sunlight on your hand

Hold your hand in the sunlight. It warms up. The light has given energy to your skin, and the energy came across empty space in the fields of a wave. In this lesson we find how much energy a wave holds, and how fast it flows.

Energy in the two fields

You know two energy densities from earlier chapters. An electric field $E$ stores $u_E = \tfrac{1}{2}\varepsilon_0 E^2$ in each cubic metre. A magnetic field $B$ stores $u_B = \dfrac{B^2}{2\mu_0}$ in each cubic metre.

In a wave in vacuum, $E = cB$ and $c^2 = 1/\mu_0\varepsilon_0$. Put $B = E/c$ into $u_B$:

The two parts are equal at every point and every instant. The total energy density is

The field $E$ is a sine wave, and the average of $\sin^2$ over a cycle is $\tfrac{1}{2}$. So the average energy density is

Intensity

The intensity $I$ is the average energy that crosses one square metre, at right angles to the wave, in one second. In a time $t$ the wave moves a distance $ct$, so the energy in a column of area $1$ m² and length $ct$ crosses the square metre. Hence $I = \bar{u}\,c$:

Some books use the root mean square (rms) field, $E_{rms} = E_0/\sqrt{2}$. Then $I = \varepsilon_0 c E_{rms}^2$.

The Poynting vector, in words

The energy flows in a definite direction. The Poynting vector $\vec{S}$ tells us the direction and the rate. Its direction is the direction of $\vec{E}\times\vec{B}$, which is the direction of travel. Its size is the energy that crosses a square metre each second:

$S$ changes with time as $\sin^2$ and is never negative. Its average over a cycle is the intensity $I$.

From a power to the fields

A source of power $P$ that sends its waves out equally in all directions spreads the power over a sphere of area $4\pi r^2$. At a distance $r$,

The intensity falls like $1/r^2$, so $E_0$ and $B_0$ fall like $1/r$.

Double the distance and read the peak electric field. Does it halve or fall to a quarter? It halves, since $E_0 \propto 1/r$.

A worked example. Sunlight at the Earth has an intensity of $1361$ W/m². Find $E_0$, $B_0$ and the average energy density. Use $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)$ and $c = 3.00 \times 10^{8}$ m/s.

1. The electric field. We want $E_0$, so $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}}$. In SI units, $\varepsilon_0 c = 8.85 \times 10^{-12} \times 3.00 \times 10^{8} = 2.655 \times 10^{-3}$. So $E_0 = \sqrt{\dfrac{2 \times 1361}{2.655 \times 10^{-3}}}\ \text{V/m} = 1013\ \text{V/m}$.
2. The magnetic field. $B_0 = \dfrac{E_0}{c} = \dfrac{1013\ \text{V/m}}{3.00 \times 10^{8}\ \text{m/s}} = 3.38\ \mu\text{T}$.
3. The energy density. $\bar{u} = \dfrac{I}{c} = \dfrac{1361\ \text{W/m}^2}{3.00 \times 10^{8}\ \text{m/s}} = 4.54\ \mu\text{J/m}^3$.
4. Check. The electric part is $\tfrac{1}{4}\varepsilon_0 E_0^2 = 2.27\ \mu\text{J/m}^3$, the magnetic part is $\dfrac{B_0^2}{4\mu_0} = 2.27\ \mu\text{J/m}^3$, and the two are equal.

The electric field of sunlight is large, about a kilovolt for each metre, but it changes direction $10^{14}$ times each second. The magnetic field is only a few microtesla.

A second example. A radio transmitter of power $50$ kW sends its waves out equally in all directions. At a distance of $10$ km:

1. The intensity. $I = \dfrac{P}{4\pi r^2} = \dfrac{50000\ \text{W}}{4\pi \times (10000\ \text{m})^2} = 39.8\ \mu\text{W/m}^2$.
2. The electric field. $E_0 = \sqrt{\dfrac{2I}{\varepsilon_0 c}} = 0.173$ V/m. Its rms value is $\dfrac{E_0}{\sqrt{2}} = 0.122$ V/m.
3. The magnetic field. $B_0 = \dfrac{E_0}{c} = 0.577$ nT.

Other books. Many books write the intensity as $S$ or $\langle S\rangle$ and call it the magnitude of the average Poynting vector. NCERT writes $I$. Where a book says "the amplitude of the electric field", it means $E_0$.

Practise with new numbers

Try it in Python

The program finds the intensity, the peak electric field and the peak magnetic field of a $50$ kW transmitter at four distances. It then does the same sums for sunlight.

Common mistakes

• Forgetting the factor $\tfrac{1}{2}$. $I = \tfrac{1}{2}\varepsilon_0 c E_0^2$ uses the peak field. Without the $\tfrac{1}{2}$ you would have the largest value of $S$, not the average.
• Putting the whole energy in the electric field. In a wave the energy is shared equally: half in $\vec{E}$ and half in $\vec{B}$.
• Using $E_{rms}$ and $E_0$ in the same formula. $I = \tfrac{1}{2}\varepsilon_0 cE_0^2 = \varepsilon_0 cE_{rms}^2$. Never mix them.
• Thinking $E_0$ falls as $1/r^2$. It is the intensity that falls as $1/r^2$. The field strength falls as $1/r$.

Remember these

• $u_E = u_B$ in a wave. $\bar{u} = \tfrac{1}{2}\varepsilon_0 E_0^2 = B_0^2/2\mu_0$.
• $I = \bar{u}c = \tfrac{1}{2}\varepsilon_0 cE_0^2 = \varepsilon_0 cE_{rms}^2$.
• Poynting vector: $\vec{S} = \vec{E}\times\vec{B}/\mu_0$, along the direction of travel. Its average is $I$.
• A point source: $I = P/4\pi r^2$, so $E_0 \propto 1/r$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A beam of light has an intensity of $0.80$ W/m². What is the peak electric field?

2. A plane wave has $E_0 = 150$ V/m. What is its intensity?

3. The $50$ kW transmitter of the example is $40$ km away. What is the peak electric field now?

4. In a wave, what fraction of the energy is in the magnetic field?

References

• OpenStax, University Physics Volume 2, Chapter 16 (Electromagnetic Waves): energy carried by electromagnetic waves.
• NCERT Physics Class 12, Chapter 8 (Electromagnetic Waves): energy and momentum of electromagnetic waves.