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Electromagnetic Waves — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Waves5 Marks
Question
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} Hz$ and amplitude $48 V m^{-1}.$
  1. What is the wavelength of the wave?
  2. What is the amplitude of the oscillating magnetic field?
  3. Show that the average energy density of the $E$ field equals the average energy density of the $B$ field. $[c = 3 \times 10^8 m s^{-1}.]$
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✓

Answer

Frequency of the electromagnetic wave $, ν = 2.0 \times 10^{10} Hz$ Electric field amplitude $, E_0 = 48 V m^{-1}$ Speed of light $, c = 3 \times 10^8 m/ s$
  1. Wavelength of a wave is given as:
$\lambda=\frac{\text{c}}{\text{v}}$
$=\frac{3\times10^8}{2\times10^{10}}=0.015 \ \text{m}$
  1. Magnetic field strength is given as:
$\text{B}_0=\frac{\text{E}_0}{\text{c}}$
$=\frac{48}{3\times10^{8}}=1.6\times10^{-7}\ \text{T}$
  1. Energy density of the electric field is given as:
$\text{U}_E=\frac{1}{2}\in_0\text{E}^2$
And, energy density of the magnetic field is given as:
$\text{U}_\text{B}=\frac{1}{2\mu_0}\text{B}^2$
Where, $\in_0$ = Permittivity of free space
$\mu_0 =$ Permeability of free space
We have the relation connecting $E$ and $B$ as:
$E = cB … (1)$
Where,
$\text{c}=\frac{1}{\sqrt{\in_0 \ \mu_0}}\dots(2)$
Putting equation $(2)$ in equation $(1),$ we get
$\text{E}=\frac{1}{\sqrt{\in_0 \ \mu_0}}\text{B}$
Squaring both sides, we get
$\text{E}^2=\frac{1}{\in_0\ \mu_0}\text{B}^2$
$\in_0\text{E}^2=\frac{\text{B}^2}{\mu_0}$
$\frac{1}{2}\in_0\text{E}^2=\frac{1}{2}\frac{\text{B}^2}{\mu_0}$
$\Rightarrow \ \text{U}_\text{E}=\text{U}_\text{B}$

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