Electromagnetic Waves • Topic 2 of 3

Nature & Properties of EM Waves

An electromagnetic wave consists of an electric field and a magnetic field that oscillate together, at right angles to each other and to the direction of travel. Maxwell's equations, when applied to empty space, predict exactly this kind of wave — and they fix its speed to be the speed of light. That was the moment physicists realised light is electromagnetic.

Transverse nature. In an EM wave the fields do not point along the direction of motion; they point sideways. The electric field $\vec{E}$ and the magnetic field $\vec{B}$ are mutually perpendicular, and both are perpendicular to the direction of propagation. In short, $\vec{E}\perp\vec{B}\perp$ (direction of travel). The direction of propagation is along $\vec{E}\times\vec{B}$. Because the fields are perpendicular to the travel direction, EM waves are transverse and can therefore be polarised — a property longitudinal waves like sound do not have.

For a wave travelling along the $x$-axis, the fields can be written as $E_y=E_0\sin(kx-\omega t)$ and $B_z=B_0\sin(kx-\omega t)$. The two fields oscillate in phase — they reach their maxima and zeros at the same instant.

Speed in vacuum. Maxwell's equations give the speed of an EM wave in free space as:

  • $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}$, where $\mu_0=4\pi\times10^{-7}\ \text{T·m/A}$ and $\epsilon_0=8.85\times10^{-12}\ \text{F/m}$.
  • Putting in the numbers gives $c\approx3\times10^{8}\ \text{m/s}$ — exactly the measured speed of light, which clinched the electromagnetic theory of light.
  • In a medium of permittivity $\epsilon$ and permeability $\mu$, the speed is $v=\frac{1}{\sqrt{\mu\epsilon}}$, always less than $c$.

Relation between field amplitudes. The electric and magnetic field magnitudes in an EM wave are locked together. At every instant:

  • $\frac{E_0}{B_0}=c$ (also $\frac{E}{B}=c$ for the instantaneous values).
  • Because $c$ is large, $E_0$ is numerically much bigger than $B_0$; this is why the electric field is responsible for most everyday effects of light, such as the force on electrons in a detector.

Energy carried by the wave. An EM wave transports energy stored in both fields. The energy density (energy per unit volume) of the electric field is $u_E=\frac{1}{2}\epsilon_0 E^2$ and of the magnetic field is $u_B=\frac{B^2}{2\mu_0}$. A key result is that the two contributions are equal: $u_E=u_B$ at every point, because $E=cB$. The total energy density is $u=u_E+u_B=\epsilon_0 E^2=\frac{B^2}{\mu_0}$.

Intensity, momentum and radiation pressure. The intensity is the average energy crossing unit area per unit time: $I=\frac{1}{2}\epsilon_0 E_0^2 c$ (using the average of $\sin^2$, which is $\tfrac12$). Crucially, EM waves carry not only energy $U$ but also momentum $p=\frac{U}{c}$. When the wave hits a surface and is absorbed, it delivers this momentum and exerts a radiation pressure $P=\frac{I}{c}$ (for a perfectly absorbing surface; it is $\frac{2I}{c}$ for a perfect reflector). This tiny pressure is what pushes the tails of comets away from the Sun and is the principle behind solar sails.

Perpendicular E and B fields of an EM wave propagating along xx (travel)EB (into / out of page)E perpendicular B perpendicular travelE0 / B0 = c
1
Worked Example
The electric field amplitude of an EM wave in vacuum is $E_0=120\ \text{N/C}$. Find the amplitude of its magnetic field.
Solution
  1. Step 1: Use $\frac{E_0}{B_0}=c$, so $B_0=\frac{E_0}{c}$.
  2. Step 2: Substitute: $B_0=\frac{120}{3\times10^{8}}$.
  3. Step 3: Compute: $B_0=4\times10^{-7}\ \text{T}$.

Answer: $B_0=4\times10^{-7}\ \text{T}=0.4\ \mu\text{T}$.

2
Worked Example
Verify that $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}$ gives about $3\times10^{8}\ \text{m/s}$. ($\mu_0=4\pi\times10^{-7}$, $\epsilon_0=8.85\times10^{-12}$.)
Solution
  1. Step 1: $\mu_0\epsilon_0=(4\pi\times10^{-7})(8.85\times10^{-12})\approx1.11\times10^{-17}$.
  2. Step 2: $\sqrt{\mu_0\epsilon_0}\approx3.33\times10^{-9}$.
  3. Step 3: $c=\frac{1}{3.33\times10^{-9}}\approx3.0\times10^{8}\ \text{m/s}$.

Answer: $c\approx3\times10^{8}\ \text{m/s}$, the speed of light.

3
Worked Example
An EM wave in vacuum has electric field amplitude $E_0=48\ \text{V/m}$. Find the average intensity of the wave. ($\epsilon_0=8.85\times10^{-12}$, $c=3\times10^{8}$.)
Solution
  1. Step 1: Use $I=\frac{1}{2}\epsilon_0 E_0^2 c$.
  2. Step 2: $I=\frac{1}{2}\times8.85\times10^{-12}\times(48)^2\times3\times10^{8}$.
  3. Step 3: $(48)^2=2304$; $I=\frac{1}{2}\times8.85\times10^{-12}\times2304\times3\times10^{8}\approx3.06\ \text{W/m}^2$.

Answer: $I\approx3.06\ \text{W/m}^2$.

4
Worked Example
Sunlight reaches the Earth with an intensity of about $1360\ \text{W/m}^2$. Find the radiation pressure on a perfectly absorbing surface. ($c=3\times10^{8}\ \text{m/s}$.)
Solution
  1. Step 1: For a perfect absorber, $P=\frac{I}{c}$.
  2. Step 2: Substitute: $P=\frac{1360}{3\times10^{8}}$.
  3. Step 3: Compute: $P\approx4.53\times10^{-6}\ \text{Pa}$.

Answer: $P\approx4.5\times10^{-6}\ \text{Pa}$ (very small).

5
Worked Example
The magnetic field amplitude of an EM wave is $B_0=2\times10^{-8}\ \text{T}$. Find the electric field amplitude.
Solution
  1. Step 1: Use $\frac{E_0}{B_0}=c$, so $E_0=cB_0$.
  2. Step 2: $E_0=3\times10^{8}\times2\times10^{-8}$.
  3. Step 3: Compute: $E_0=6\ \text{V/m}$.

Answer: $E_0=6\ \text{V/m}$.

6
Worked Example
Find the momentum delivered when $30\ \text{J}$ of light is fully absorbed by a surface. ($c=3\times10^{8}\ \text{m/s}$.)
Solution
  1. Step 1: An EM wave carries momentum related to its energy by $p=\frac{U}{c}$ (a result from Maxwell's theory).
  2. Step 2: Substitute $U=30\ \text{J}$ and $c=3\times10^{8}\ \text{m/s}$: $p=\frac{30}{3\times10^{8}}$.
  3. Step 3: Compute: $p=1\times10^{-7}\ \text{kg·m/s}$.

Answer: $p=1\times10^{-7}\ \text{kg·m/s}$.

Key Points

  • EM waves are transverse: $\vec{E}$, $\vec{B}$ and the direction of travel are mutually perpendicular ($\vec{E}\perp\vec{B}\perp$ propagation), with travel along $\vec{E}\times\vec{B}$.
  • Speed in vacuum is $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}\approx3\times10^{8}\ \text{m/s}$; in a medium $v=\frac{1}{\sqrt{\mu\epsilon}}
  • The field amplitudes are locked: $\frac{E_0}{B_0}=c$; E and B oscillate in phase.
  • The electric and magnetic energy densities are equal: $u_E=\frac{1}{2}\epsilon_0 E^2=u_B=\frac{B^2}{2\mu_0}$; total $u=\epsilon_0 E^2$.
  • EM waves carry momentum $p=\frac{U}{c}$ and exert radiation pressure $P=\frac{I}{c}$ (absorber) or $\frac{2I}{c}$ (reflector); intensity $I=\frac{1}{2}\epsilon_0 E_0^2 c$.
Tap an option to check your answer0 / 4
Q1.In an electromagnetic wave, the electric and magnetic fields are:
Explanation: $\vec{E}$, $\vec{B}$ and the direction of propagation are mutually perpendicular; the wave is transverse.
Q2.The speed of an EM wave in vacuum is given by:
Explanation: Maxwell's equations give $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}\approx3\times10^{8}\ \text{m/s}$.
Q3.The ratio of the electric to magnetic field amplitudes in an EM wave equals:
Explanation: At every instant $\frac{E_0}{B_0}=c$, so $E_0$ is numerically much larger than $B_0$.
Q4.The radiation pressure on a perfectly reflecting surface for an incident intensity $I$ is:
Explanation: A reflector reverses the wave's momentum, so it receives twice the momentum of an absorber: $P=\frac{2I}{c}$.