Chapter 8: Electromagnetic Waves

In an electromagnetic wave travelling in free space, the magnitudes of the electric and magnetic fields are related through the speed of light: $$c = \dfrac{E}{B}$$.

Step 1 - Magnitude of B.

$$B = \dfrac{E}{c} = \dfrac{6.3 \, \mathrm{V/m}}{3 \times 10^{8} \, \mathrm{m/s}} = 2.1 \times 10^{-8} \, \mathrm{T}$$

Step 2 - Direction of B.

The wave propagates along $$+x$$ (the $$\hat{\mathbf{i}}$$ direction) and the electric field points along $$+y$$ (the $$\hat{\mathbf{j}}$$ direction). In an electromagnetic wave the vectors $$\mathbf{E}$$, $$\mathbf{B}$$ and the direction of propagation form a right-handed set, with $$\mathbf{E} \times \mathbf{B}$$ pointing along the direction of propagation.

We therefore need $$\mathbf{E} \times \mathbf{B}$$ to point along $$+x$$. Since $$\hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}$$, the magnetic field must lie along $$+z$$ (the $$\hat{\mathbf{k}}$$ direction).

Hence $$\mathbf{B} = 2.1 \times 10^{-8} \, \hat{\mathbf{k}} \, \mathrm{T}$$.

The given magnetic field is $$B_y = (2 \times 10^{-7}) \, \mathrm{T} \, \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t)$$.

Compare this with the standard form of a plane wave $$B_y = B_0 \sin(kx + \omega t)$$. This gives the angular wave number and the angular frequency:

$$k = 0.5 \times 10^{3} \, \mathrm{rad/m} = 500 \, \mathrm{rad/m}, \qquad \omega = 1.5 \times 10^{11} \, \mathrm{rad/s}$$

(a) Wavelength and frequency.

$$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{500 \, \mathrm{rad/m}} = 1.26 \times 10^{-2} \, \mathrm{m}$$

$$\nu = \dfrac{\omega}{2\pi} = \dfrac{1.5 \times 10^{11}}{2\pi} \, \mathrm{Hz} = 2.39 \times 10^{10} \, \mathrm{Hz}$$

(As a check, $$\nu\lambda = (2.39 \times 10^{10})(1.26 \times 10^{-2}) \approx 3 \times 10^{8} \, \mathrm{m/s} = c$$.)

(b) Expression for the electric field.

The amplitude of the electric field is $$E_0 = cB_0 = (3 \times 10^{8})(2 \times 10^{-7}) = 60 \, \mathrm{V/m}$$.

The phase is $$kx + \omega t$$, so the wave moves along the $$-x$$ direction, and $$\mathbf{B}$$ oscillates along $$y$$. For $$\mathbf{E} \times \mathbf{B}$$ to point along $$-x$$, the electric field must oscillate along the $$z$$-axis, since $$\hat{\mathbf{k}} \times \hat{\mathbf{j}} = -\hat{\mathbf{i}}$$.

Therefore $$E_z = 60 \, \mathrm{V/m} \, \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t)$$, with $$E_x = E_y = 0$$.

All three — X-rays, visible red light and radio-waves — are electromagnetic (EM) waves.
For every electromagnetic wave propagating through vacuum (or through air to an excellent approximation) the speed is a universal constant
$$v = c = 3.0 \times 10^{8}\;\text{m\,s}^{-1}.$$

The speed v of any wave is linked to its wavelength $$\lambda$$ and frequency $$\nu$$ by $$v = \nu\,\lambda.$$

Because the wavelengths are different, their frequencies are necessarily different, viz.

• For the X-ray: $$\nu_X = \dfrac{c}{\lambda_X} = \dfrac{3.0\times10^{8}}{1.0\times10^{-10}} \text{ Hz} \approx 3.0\times10^{18}\;\text{Hz},$$
• For red light: $$\nu_R = \dfrac{c}{\lambda_R} = \dfrac{3.0\times10^{8}}{6.8\times10^{-7}} \text{ Hz} \approx 4.4\times10^{14}\;\text{Hz},$$
• For the radio-wave: $$\nu_{rad} = \dfrac{c}{\lambda_{rad}} = \dfrac{3.0\times10^{8}}{500} \text{ Hz} \approx 6.0\times10^{5}\;\text{Hz},$$

Although the frequencies differ widely, inserting any pair $$\nu$$ and $$\lambda$$ into $$v = \nu\lambda$$ always yields the same numerical value $$3.0\times10^{8}\;\text{m\,s}^{-1}$$.

Hence the only physical quantity that is identical for all three given waves is their speed in vacuum (or air).

Directions of $$\vec{E}$$ and $$\vec{B}$$. For a plane electromagnetic wave the wave-vector $$\vec{k}$$, the electric field $$\vec{E}$$ and the magnetic field $$\vec{B}$$ are mutually perpendicular and obey the right-hand rule $$\vec{k} \propto \vec{E} \times \vec{B}$$. Both $$\vec{E}$$ and $$\vec{B}$$ are therefore perpendicular to the direction of propagation.

The wave travels along the $$+z$$ axis, i.e. $$\vec{k} = k\,\hat{z}$$. Hence both $$\vec{E}$$ and $$\vec{B}$$ lie in the $$x$$-$$y$$ plane. A convenient choice that satisfies the right-hand rule is

• $$\vec{E}$$ along $$+x$$, i.e. $$\vec{E} = E_0\,\hat{x}$$
• $$\vec{B}$$ along $$+y$$, i.e. $$\vec{B} = B_0\,\hat{y}$$

(any other pair of mutually perpendicular transverse directions for which $$\vec{E} \times \vec{B}$$ points along $$+z$$ is equally valid).

Wavelength. For an electromagnetic wave in vacuum, $$c = \nu\,\lambda$$, so

$$\lambda = \dfrac{c}{\nu}$$

With $$c = 3.0 \times 10^{8}\,\mathrm{m\,s^{-1}}$$ and $$\nu = 30\,\mathrm{MHz} = 3.0 \times 10^{7}\,\mathrm{Hz}$$,

$$\lambda = \dfrac{3.0 \times 10^{8}\,\mathrm{m\,s^{-1}}}{3.0 \times 10^{7}\,\mathrm{Hz}} = 10\,\mathrm{m}$$

Given: Frequency range that the radio can receive is from $$f_{\min}=7.5\,\text{MHz}$$ to $$f_{\max}=12\,\text{MHz}$$.

Convert the limits to hertz so that the speed–frequency–wavelength relation can be used:

$$f_{\min}=7.5\,\text{MHz}=7.5\times10^{6}\,\text{Hz}$$
$$f_{\max}=12\,\text{MHz}=12\times10^{6}\,\text{Hz}$$

The speed of electromagnetic waves in free space is
$$c=3.0\times10^{8}\,\text{m\,s}^{-1}$$

The wavelength corresponding to any frequency is obtained from
$$\lambda=\dfrac{c}{f}$$

For the lower-frequency end (7.5 MHz):

$$\lambda_{\max}=\dfrac{3.0\times10^{8}}{7.5\times10^{6}}\,\text{m}$$

$$\lambda_{\max}=0.4\times10^{2}\,\text{m}=40\,\text{m}$$

For the higher-frequency end (12 MHz):

$$\lambda_{\min}=\dfrac{3.0\times10^{8}}{12\times10^{6}}\,\text{m}$$

$$\lambda_{\min}=0.25\times10^{2}\,\text{m}=25\,\text{m}$$

Therefore, the radio’s wavelength band is from 25 m to 40 m.

An accelerating charge radiates an electromagnetic wave. When the charge undergoes simple harmonic motion at frequency $$\nu$$, the time-dependent electric dipole moment oscillates at the same frequency, and the radiated electric and magnetic fields therefore vary in time with that same frequency.

Hence the frequency of the emitted electromagnetic wave equals the mechanical oscillation frequency of the source:

$$\nu_{\mathrm{em}} = \nu = 10^{9}\,\mathrm{Hz}$$

So the oscillator radiates electromagnetic waves of frequency $$1\,\mathrm{GHz}$$ (in the microwave region of the spectrum).

Given data

• Amplitude of the magnetic field: $$B_0 = 510\;\text{nT}$$

Step 1 · Convert $$B_0$$ to tesla

Because $$1\;\text{nT}=10^{-9}\;\text{T}$$,

$$B_0 = 510\,\text{nT} = 510\times10^{-9}\;\text{T} = 5.10\times10^{-7}\;\text{T}$$

Step 2 · Use the relation between field amplitudes in vacuum

For an electromagnetic (EM) wave moving in vacuum the electric and magnetic field amplitudes are related by

$$E_0 = c\,B_0$$

where $$c = 3.0\times10^{8}\;\text{m\,s}^{-1}$$ is the speed of light.

Step 3 · Calculate $$E_0$$

$$E_0 = (3.0\times10^{8}\;\text{m\,s}^{-1})(5.10\times10^{-7}\;\text{T})$$

Combine the numerical factors and powers of ten:

$$E_0 = (3.0\times5.10)\times10^{8-7}\;\text{V\,m}^{-1}$$

$$E_0 = 15.3\times10^{1}\;\text{V\,m}^{-1}$$

Simplifying,

$$E_0 = 153\;\text{V\,m}^{-1}$$

Thus the electric field amplitude of the EM wave is 153 V m⁻¹.

Given data

Electric–field amplitude: $$E_0 = 120\;\mathrm{N\,C^{-1}}$$
Frequency: $$\nu = 50.0\;\mathrm{MHz}=50.0\times10^{6}\;\mathrm{Hz}$$
Speed of light in vacuum: $$c = 3.00\times10^{8}\;\mathrm{m\,s^{-1}}$$

The angular frequency, wave number and wavelength of a uniform electromagnetic (EM) wave in free space are related by

$$\omega = 2\pi\nu,\qquad k = \dfrac{\omega}{c}=\dfrac{2\pi}{\lambda},\qquad \lambda = \dfrac{c}{\nu}.$$

(a) Evaluation of $$B_0,\;\omega,\;k,\;\lambda$$

1. Magnetic–field amplitude

   The amplitudes satisfy $$E_0 = cB_0\;\Rightarrow\;B_0 = \dfrac{E_0}{c}.$$
   $$B_0 = \dfrac{120}{3.00\times10^{8}}\;\mathrm{T} = 4.0\times10^{-7}\;\mathrm{T}.$$

2. Angular frequency

   $$\omega = 2\pi\nu = 2\pi(50.0\times10^{6})\;\mathrm{rad\,s^{-1}} = 3.14\times10^{8}\;\mathrm{rad\,s^{-1}}.$$

3. Wave number

   $$k = \dfrac{\omega}{c} = \dfrac{3.14\times10^{8}}{3.00\times10^{8}}\;\mathrm{rad\,m^{-1}} = 1.05\;\mathrm{rad\,m^{-1}}.$$

4. Wavelength

   $$\lambda = \dfrac{c}{\nu} = \dfrac{3.00\times10^{8}}{50.0\times10^{6}}\;\mathrm{m} = 6.0\;\mathrm{m}.$$

(b) Space–time expressions for $$\mathbf E$$ and $$\mathbf B$$

Choose the wave to travel in the +x–direction. Let the electric field oscillate along +y and the magnetic field along +z, so that $$\mathbf E,\;\mathbf B,\;\mathbf k$$ form a right–handed set.

With this choice, the phase of both fields is $$kx-\omega t$$. Hence

• Electric field:
  $$\mathbf E(x,t)=\hat{\jmath}\,E_0\sin(kx-\omega t).$$

• Magnetic field:
  $$\mathbf B(x,t)=\hat{\mathbf k}\,B_0\sin(kx-\omega t).$$

Substituting the numerical amplitudes,

$$\mathbf E(x,t)=\hat{\jmath}\,(120\;\mathrm{N\,C^{-1}})\sin\!(1.05x-3.14\times10^{8}t),$$
$$\mathbf B(x,t)=\hat{\mathbf k}\,(4.0\times10^{-7}\;\mathrm{T})\sin\!(1.05x-3.14\times10^{8}t),$$

where $$x$$ is in metres and $$t$$ in seconds.

Setting up the conversion. The energy of one photon of frequency $$\nu$$ is

$$E = h\,\nu, \qquad h = 6.626 \times 10^{-34}\,\mathrm{J\,s}$$

To express $$E$$ in electron-volts, divide by $$1\,\mathrm{eV} = 1.602 \times 10^{-19}\,\mathrm{J}$$:

$$\dfrac{h}{1\,\mathrm{eV}} = \dfrac{6.626 \times 10^{-34}}{1.602 \times 10^{-19}}\,\mathrm{eV\,s} = 4.136 \times 10^{-15}\,\mathrm{eV\,s}$$

Hence, for any frequency $$\nu$$ in hertz,

$$E\;(\mathrm{eV}) = 4.136 \times 10^{-15}\,\nu\;(\mathrm{Hz})$$

Photon energies across the electromagnetic spectrum.

Relation between energy scale and source. Each photon carries an energy that matches the characteristic energy change of the physical process producing it. The progression from radio to gamma rays parallels the energy scale of the source:

• Low-frequency electrical oscillations in antennas radiate nano-eV radio photons.
• Molecular rotational transitions ($$\sim 10^{-5}$$ to $$10^{-2}$$ eV) and vibrational transitions ($$\sim 10^{-2}$$ to $$10^{-1}$$ eV) emit microwaves and infra-red, respectively.
• Outer-shell electron transitions in atoms involve a few eV, so atoms emit and absorb visible and ultraviolet light.
• Sudden deceleration of fast (keV) electrons in X-ray tubes converts that kinetic energy into keV X-ray photons; inner-shell electronic transitions also emit X-rays.
• Transitions between nuclear states involve $$10^{5}$$ to $$10^{7}$$ eV, producing gamma rays.

Thus, as one moves from the radio end to the gamma end of the spectrum, both the photon energy and the characteristic energy scale of the radiating mechanism rise by many orders of magnitude.