Nature Of Electromagnetic Waves

Electromagnetic waves are a fascinating aspect of our daily lives, often without us realizing it. They encompass a broad spectrum of waves, including visible light that allows us to see, radio waves that enable us to communicate over long distances, and microwaves that heat our food. These waves travel through space, carrying energy and information, making modern technology like smartphones, Wi-Fi, and GPS possible. Understanding the nature of electromagnetic waves not only enhances our grasp of physics but also highlights their integral role in everyday conveniences and the advancement of various technologies.

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This Story also Contains

1. Nature of Electromagnetic Waves
2. Solved Examples Based on Nature of Electromagnetic Waves
3. Summary

Nature of Electromagnetic Waves

From Maxwell’s equations, we can observe that electric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation. Also from our discussion of the displacement current, in that capacitor, the electric field inside the plates of the capacitor is directed perpendicular to the plates. The figure given below shows a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate, at a given time t). The electric field E_x is along the x-axis, and varies sinusoidally with z, at a given time. The magnetic field B_y is along the y-axis and again varies sinusoidally with z. The electric and magnetic fields E_x and B_y are perpendicular to each other and to the direction z of propagation.

Now from the Lorentz equation -

$
\begin{gathered}
\vec{F}=q(\vec{E}+\vec{v} \times \vec{B}) \\
E_z=E z_0 \sin (\omega t-k y)
\end{gathered}
$
$B_x=B x_0 \sin (\omega t-k y)$, where $\frac{\omega}{k}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

since, $\omega=2 \pi f$, where f is the frequency and $k=\frac{2 \pi}{\lambda}$, where $\lambda$ is the wavelength.
Therefore, $\frac{\omega}{k}=\frac{2 \pi f}{2 \pi / \lambda}=f \lambda$

But $f \lambda$ gives the velocity of the wave. So $f \lambda=c=\omega k$, So we can write -
$
c=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}
$

It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as -

$B_0=\frac{E_o}{c}$

In a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes,

$v=\frac{1}{\sqrt{\mu \varepsilon}}$

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Solved Examples Based on Nature of Electromagnetic Waves

Example 1: If the magnetic field of a plane electromagnetic wave is given by (The speed of light $=3 \times 10^8 \mathrm{~m} / \mathrm{s}$ )
$B=100 \times 10^{-6} \sin \left[2 \pi \times 2 \times 10^{15}\left(t-\frac{x}{c}\right)\right]_{\text {then the maximum electric field }}$ associated with it is:

1) $6 \times 10^4 \mathrm{~N} / \mathrm{C}$
2) $3 \times 10^4 \mathrm{~N} / \mathrm{C}$
3) $4 \times 10^4 \mathrm{~N} / \mathrm{C}$
4) $4.5 \times 10^4 \mathrm{~N} / \mathrm{C}$

Solution:

Relation between Eo and Bo -
$
E_o=c \cdot B_o
$
$E_o=$ Electric field amplitude
$B_o=$ Magnetic field amplitude
$\mathrm{C}=$ Speed of light in vacuum

i.e $C=3 \times 10^8 \mathrm{~m} / \mathrm{s} B_0=100 \times 10^{-6} \mathrm{~T}$ so we have: $E_0=B_0 \times C=100 \times 10^{-6} \times 3 \times 10^8 E_0=3 \times 10^4 \mathrm{~N} / \mathrm{C}$

Hence, the answer is option (2).

Example 2: A plane electromagnetic wave propagating along a y-direction electric field $\vec{E}$ and magnetic field $\vec{B}$ components
1) $E_x, B_z$ or $E_z, B_x$
2) $E_x, B_y$ or $E_y, B_x$
3) $E_y, B_x$ or $E_x, B_x$
4) $E_y, B_y$ or $E_z, B_z$

Solution:

or,


Since the direction of propagation of the wave is in y direction the magnetic field and electric field are either in x or in the z-direction.

Example 3: The electric field of a plane electromagnetic wave propagating along the $x$ direction in a vacuum is $\vec{E}=E_0 \hat{j} \cos (\omega t-k x)$. The magnetic field $\vec{B}$, at the moment, is :

1) $\vec{B}=\frac{E_0}{\sqrt{\mu_0 \epsilon_0}} \cos (k x) \hat{k}$
2) $\vec{B}=E_0 \sqrt{\mu_0 \epsilon_0} \cos (k x) \hat{j}$
3) $\vec{B}=E_0 \sqrt{\mu_0 \epsilon_0} \cos (k x) \hat{k}$
4) $\vec{B}=\frac{E_0}{\sqrt{\mu_0 \epsilon_0}} \cos (k x) \hat{j}$

Solution:

$\begin{aligned} & E=E_0 \cos (\omega t-k x) \hat{j} \\ & E_0=B_0 C \\ & B_0=\frac{E}{C}=\frac{E_0}{\frac{1}{\mu_0 \epsilon_0}} \\ & B_0=E_0 \sqrt{\mu_0 \epsilon_0} \\ & \mathrm{~B}=\mathrm{E}_0 \sqrt{\mu_0 \epsilon_0} \cos (\omega \mathrm{t}-\mathrm{kx}) \hat{\mathrm{k}} \\ & \text { at } \mathrm{t}=0 \\ & \overrightarrow{\mathrm{B}}=\mathrm{E}_0 \sqrt{\mu_0 \epsilon_0} \cos (\mathrm{kx}) \hat{\mathrm{k}}\end{aligned}$

Hence, the answer is option (3).

Example 4: In a plane electromagnetic wave, the directions of the electric field and magnetic field are represented $\hat{k}$ and $2 \hat{i}-2 \hat{j}$ respectively. What is the unit vector along the direction of propagation of the wave?

$\begin{aligned}
& \text { 1) } \frac{1}{\sqrt{2}}(\hat{j}+\hat{k}) \\
& \text { 2) } \frac{1}{\sqrt{5}}(2 \hat{i}+\hat{j}) \\
& \text { 3) } \frac{1}{\sqrt{5}}(\hat{i}+2 \hat{j}) \\
& \text { 4) } \frac{1}{\sqrt{2}}(\hat{i}+\hat{j})
\end{aligned}$

Solution:

$\begin{aligned} & A s \quad \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}} \| \overrightarrow{\mathrm{C}} \\ & \hat{\mathrm{E}} \times \hat{\mathrm{B}}=\frac{1}{\sqrt{2}}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & 0 & 1 \\ 1 & -1 & 0\end{array}\right|=\frac{\hat{\mathrm{i}}+\hat{\mathrm{i}}}{\sqrt{2}} \\ & \Rightarrow \hat{\mathrm{C}}=\frac{\mathrm{i}+\mathrm{i}}{\sqrt{2}}\end{aligned}$

Hence, the answer is option (4).

Example 5: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency $2 \times 10^{10} \mathrm{~Hz}$ and amplitude $48 \frac{\mathrm{v}}{\mathrm{m}}$. The amplitude of the oscillating magnetic field will be:

1) $\frac{1}{16} \times 10^{-8} \frac{w b}{m^2}$
2) $16 \times 10^{-8} \frac{w b}{m^2}$
3) $12 \times 10^{-7} \frac{\mathrm{wb}}{\mathrm{m}^2}$
4) $\frac{1}{12} \times 10^{-7} \frac{w b}{m^2}$

Solution:

Relation between Eo and Bo -
$
E_o=c \cdot B_o
$
- wherein
$E_o=$ Electric field amplitude
$B_o=$ Magnetic field amplitude
$\mathrm{C}=$ Speed of light in vacuum
So the magnitude of Oscillating mag. field
$
B=\frac{E}{C}=\frac{48}{3 \times 10^8}=16 \times 10^{-8}
$

Hence, the answer is option (2).

Summary

Electromagnetic waves are 'waves' of electric and magnetic fields propagating through space. They travel at light speed; hence, they are truly non-medium dependent, meaning they can travel through the vacuum of space. Such waves result from charge acceleration and vary in wavelength and frequency. It ranges from radio waves and microwaves through infrared and visible light to ultraviolet, X-rays, and gamma rays. All those bear individual differences in their properties and applicability—technologies like wireless communications and medical imaging—up to astronomy itself.