Let’s talk about the relationship between electric and mangetic fields in a electromagnetic wave. Consider a plain harmonic wave:
The magnitudes of the magnetic and electric field of an electromagnetic wave are related by:
\[ E = cB \]
There exist electromagnetic waves that span 20 orders or more of magnitude – the spectrum varies in wavelength (and hence frequency). Different types of waves include:
The Doppler Shift is the fact that an observer will observe a higher frequency of electromagnetic waves as the distance between the observer and the source of the waves decreases.
See Formulas for some insight on the key formula related to this section.
Electromagnetic waves have energy. How do we know? Well they produce electric and magnetic fields, and we know the energy densities in those fields (see Formulas). As a result, we can find out the energy of an electromagnetic wave at a given point and a given time.
Using some math, we can find out that the energy density of the magnetic field is equal to energy density of the electric field. Thus, the total energy is:
\[ \mu = 2 * \mu_{E} = 2 * \frac{1}{2} \varepsilon_{0} E^{2} = \varepsilon_{0} E^{2} \]
This equation and more are provided in the Formulas section
Sunlight Intensity on Earth
\[ I = 100 \frac{mW}{cm^{2}} = 1 \frac{kW}{m^{2}} \]
Sunlight Electric Field Strength
\[ E^{2}_{rms} = \mu_{0} c I \]
\[ E_{rms} = 614 \frac{V}{m} \]
Impedance of Free Space
\[ Z_{0} = \mu_{0} c = 377 \Omega \]
How is energy transported in E-M waves? Photons. These are the quanta of electromagnetic radiation; photons are quantum objects that exhibit properties of waves and particles.
Photons exhibit an energy proportional to their frequency:
\[ E = hf \]
Photons also carry momentum:
\[ p = \frac{E}{c} = \frac{h}{\lambda} \]
Where:
A wave’s magnetic field vector formula is given below.
\[ \vec{B} = (5 \mu T) sin ( ky - \omega t) \vec{z} \]
What is the vector expression for the corresponding Electric Field?
Determine magnitude of electric field
\[ E_{0} = c B_{0} = ( 3 \times 10^{8} ) ( 5 \times 10^{-6} ) \]
\[ E_{0} = 1500 N/C \]
Determine direction of the propagation of the wave: from the magnetic field oscillation formula we see it is in the positive y direction.
We can find the electric field due to the fact that \(\vec{E} \times \vec{B}\) points in the direction of propagation. Thus, the direction of the electric field points in the negative x direction.
The vector expression for the corresponding Electric Field is then:
\[ \vec{E} = ( 1500 N/C ) sin ( ky - \omega t ) ( - \hat{x} ) \]
This problem was just a reflection on how the electric and magnetic fields are related for an electromagnetic wave. This relationship is due to Faraday’s and Ampere’s law.
\[ \frac{E_{0}}{B_{0}} = c \]
It is also worth noting that we knew the phase of the electric field and magnetic field were in phase with each other since this was an electromagnetic wave.
We also knew that \(\vec{E} \times \vec{B}\) gave us the direction of propagation for the wave, which was needed to determine the direction of the final vector.
Velocity
\[ c = \frac{\omega}{k} = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} = \frac{E_{0}}{B_{0}} = 3 \times 10^{8} m/s \]
Doppler Shift
\[ f' = f \sqrt{\frac{1 \pm \beta}{1 \mp \beta}} \]
Where:
Notes:
\[ f' \approx f (1 \pm \beta ) \]
Energy of an electric field:
\[ \mu_{E} = \frac{1}{2} \varepsilon_{0} E^{2} \]
Energy of a magnetic field:
\[ \mu_{B} = \frac{1}{2} \frac{B^{2}}{\mu_{0}} \]
Energy of electromagnetic wave:
\[ \mu = \varepsilon_{0} E^{2} \]
Average Energy Density
\[ \langle u \rangle = \frac{1}{2} \varepsilon_{} E_{0}^{2} \]
Intensity (Average power per area)
\[ I = \frac{1}{2} c \varepsilon_{0} E_{0}^{2} \]
Poynting Vector (Instantaneous power per wave per unit area)
Vector \[ \vec{S} \equiv \frac{\vec{E} \times \vec{B}}{\mu_{0}} \]
Magnitude \[ S = c \varepsilon_{0} E^{2} \]
Average Magnitude \[ \langle S \rangle = I \]