EM Fields
Fields
Accelerating charges emit electromagnetic waves. A changing electric field generates a magnetic field and a changing magnetic field generates an electric field. This is the principle behind the propagation of electromagnetic waves, because electromagnetic waves, unlike sound waves, do not need a medium to travel through.
EM waves propagate when an electric field oscillating in one plane produces a magnetic field oscillating in a plane at right angles to it, which produces an oscillating electric field, and so on. The propagation of electromagnetic waves can be described as mutual induction. The changing electric field is responsible for inducing the magnetic field and vice versa.
We use E to denote electric fields and B to denote magnetic fields.
These concepts will be covered in more detail in a later lesson.
These mutually regenerating fields, most commonly described as self-propagating, travel through empty space at a constant speed of approximately \(\text{3} \times \text{10}^{\text{8}}\) \(\text{m·s$^{-1}$}\), represented by c. We will use \(\text{3} \times \text{10}^{\text{8}}\) \(\text{m·s$^{-1}$}\) for all of our calculations in this book. Although an electromagnetic wave can travel through empty space, it can also travel through a medium (such as water and air). When an electromagnetic wave travels through a medium, it travels slower than it would through empty space (vacuum).
A diagram showing the mutually regenerating electric field (red (solid) line) and magnetic field (blue (dashed) line).
Since an EM radiation is a wave, the following equation still applies:
\[v = f \cdot \lambda\]
Except that we can replace \(v\) with \(c\):
\[c = f \cdot \lambda\]
Example: Em Radiation I
Question
Calculate the frequency of an electromagnetic wave with a wavelength of \(\text{4.2} \times \text{10}^{-\text{7}}\) \(\text{m}\).
Step 1: Wave equation
We use the formula: \(c = f \cdot \lambda\) to calculate frequency. The speed of light is a constant \(\text{3} \times \text{10}^{\text{8}}\) \(\text{m·s$^{-1}$}\).
Step 2: Calculate
\begin{align*} c & = f\lambda \\ \text{3} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$} & = f \times \text{4.2} \times \text{10}^{-\text{7}}\text{ m} \\ f & = \text{7.14} \times \text{10}^{\text{14}}\text{ Hz} \end{align*}Step 3: Quote the final answer
The frequency is \(\text{7} \times \text{10}^{\text{14}}\) \(\text{Hz}\).
Example: Em Radiation II
Question
An electromagnetic wave has a wavelength of \(\text{200}\) \(\text{nm}\). What is the frequency of the radiation?
Step 1: What do we know?
Recall that all radiation travels at the speed of light (\(c\)) in vacuum. Since the question does not specify through what type of material the wave is travelling, one can assume that it is travelling through a vacuum. We can identify two properties of the radiation - wavelength (\(\text{200}\) \(\text{nm}\)) speed (\(c\)).
Step 2: Apply the wave equation
\begin{align*} c & = f\lambda \\ \text{3} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$} & = f \times \text{200} \times \text{10}^{-\text{9}}\text{ m} \\ f & = \text{1.5} \times \text{10}^{\text{15}}\text{ Hz} \end{align*}Step 3: Quote the final answer
The frequency is \(\text{1.5} \times \text{10}^{\text{15}}\) \(\text{Hz}\).
This lesson is part of:
Introducing Electromagnetic Waves