AP Physics 2- 14.4 Electromagnetic Waves- Study Notes- New Syllabus
AP Physics 2- 14.4 Electromagnetic Waves – Study Notes
AP Physics 2- 14.4 Electromagnetic Waves – Study Notes – per latest Syllabus.
Key Concepts:
- Properties of an Electromagnetic Wave
Properties of an Electromagnetic Wave
Electromagnetic (EM) waves are transverse waves consisting of oscillating electric and magnetic fields that are mutually perpendicular and also perpendicular to the direction of wave propagation.
Key Properties:
Transverse Nature:
The electric field (\(\mathrm{\vec{E}}\)) and magnetic field (\(\mathrm{\vec{B}}\)) oscillate perpendicular to each other and to the direction of propagation (\(\mathrm{\vec{k}}\)).
Speed in Vacuum:
$ \mathrm{c = \dfrac{1}{\sqrt{\mu_0 \epsilon_0}} \approx 3.0 \times 10^8 \, m/s} $ w
here \(\mathrm{\mu_0}\) is the permeability and \(\mathrm{\epsilon_0}\) the permittivity of free space.
Relationship Between Fields:
$ \mathrm{E = cB} $
Energy Transport:
EM waves carry energy and momentum. The rate of energy transfer per unit area is given by the Poynting vector:
$ \mathrm{\vec{S} = \dfrac{1}{\mu_0} \vec{E} \times \vec{B}} $
Average intensity of a sinusoidal wave: $ \mathrm{I = \dfrac{1}{2} \, c \epsilon_0 E_0^2} $
- Polarization: EM waves can be polarized since they are transverse.
- Do Not Require a Medium: Unlike mechanical waves, EM waves can travel through a vacuum.
Electromagnetic Spectrum:
All EM waves travel at speed \(\mathrm{c}\) in vacuum, but they differ in wavelength (\(\mathrm{\lambda}\)), frequency (\(\mathrm{f}\)), and photon energy (\(\mathrm{E = hf}\)).
| Region | Wavelength Range | Frequency Range | Relative Photon Energy | Examples/Uses |
|---|---|---|---|---|
| Radio | \(\mathrm{> 10^{-1} \, m}\) | \(\mathrm{< 10^9 \, Hz}\) | Lowest | Broadcasting, communication |
| Microwave | \(\mathrm{10^{-3} – 10^{-1} \, m}\) | \(\mathrm{10^9 – 10^{11} \, Hz}\) | Low | Radar, cooking, satellites |
| Infrared | \(\mathrm{10^{-6} – 10^{-3} \, m}\) | \(\mathrm{10^{11} – 10^{14} \, Hz}\) | Medium | Heat radiation, remote controls |
| Visible Light | \(\mathrm{4 \times 10^{-7} – 7 \times 10^{-7} \, m}\) | \(\mathrm{4 \times 10^{14} – 7.5 \times 10^{14} \, Hz}\) | Medium | Human vision (ROYGBIV spectrum) |
| Ultraviolet | \(\mathrm{10^{-8} – 4 \times 10^{-7} \, m}\) | \(\mathrm{10^{15} – 10^{16} \, Hz}\) | High | Sunburn, sterilization |
| X-rays | \(\mathrm{10^{-11} – 10^{-8} \, m}\) | \(\mathrm{10^{16} – 10^{19} \, Hz}\) | Very High | Medical imaging |
| Gamma Rays | \(\mathrm{< 10^{-11} \, m}\) | \(\mathrm{> 10^{19} \, Hz}\) | Highest | Nuclear reactions, cancer therapy |
Trend:
- As \(\mathrm{\lambda}\) decreases → \(\mathrm{f}\) and photon energy (\(\mathrm{E = hf}\)) increase.
- All regions travel at the same speed \(\mathrm{c}\) in vacuum.
Example :
An electromagnetic wave in vacuum has an electric field amplitude of \(\mathrm{120 \, V/m}\). Find the corresponding magnetic field amplitude.
▶️ Answer/Explanation
Step 1: Relation between fields: \(\mathrm{E = cB}\)
Step 2: \(\mathrm{B = \dfrac{E}{c} = \dfrac{120}{3.0 \times 10^8}}\)
Step 3: \(\mathrm{B = 4.0 \times 10^{-7} \, T}\)
Final Answer: Magnetic field amplitude = \(\mathrm{4.0 \times 10^{-7} \, T}\).
Example :
Light of average intensity \(\mathrm{600 \, W/m^2}\) is incident on a surface. Calculate the amplitude of the electric field of the wave.
▶️ Answer/Explanation
Step 1: Use intensity relation: \(\mathrm{I = \dfrac{1}{2} c \epsilon_0 E_0^2}\)
Step 2: Rearrange: \(\mathrm{E_0 = \sqrt{\dfrac{2I}{c \epsilon_0}}}\)
Step 3: Substitute values: \(\mathrm{E_0 = \sqrt{\dfrac{2(600)}{(3.0 \times 10^8)(8.85 \times 10^{-12})}}}\)
\(\mathrm{E_0 \approx 674 \, V/m}\)
Final Answer: Electric field amplitude = \(\mathrm{674 \, V/m}\).