Question

The amplitude of electric field vector of a plane electromagnetic wave is \(E_0 = 150\) N/C and frequency \(\nu = 50\) MHz. Find out
(i) Amplitude of magnetic field (\(B_0\))
(ii) Angular frequency (\(\omega\))
(iii) Wavelength (\(\lambda\))

Hint:

A very useful and easy way to remember the relationship between E and B in an EM wave is \(E = cB\). Always ensure that all quantities are in their base SI units (Hz for frequency, m/s for speed, etc.) before starting calculations to avoid errors.

Answer 1

Step 1: Understanding the Concept:
In a plane electromagnetic (EM) wave propagating in a vacuum, the electric field (\(E\)) and magnetic field (\(B\)) are mutually perpendicular and also perpendicular to the direction of wave propagation. Their amplitudes (\(E_0\) and \(B_0\)) are related by the speed of light (\(c\)). The frequency (\(\nu\)), angular frequency (\(\omega\)), and wavelength (\(\lambda\)) are related through the wave speed equation.

Step 2: Key Formulas:
The formulas needed to solve this problem are: \begin{enumerate} \item Relation between electric and magnetic field amplitudes: \(c = \frac{E_0}{B_0}\) \item Relation between angular frequency and linear frequency: \(\omega = 2\pi\nu\) \item Relation between wave speed, frequency, and wavelength: \(c = \nu\lambda\) \end{enumerate} We will use the constant value for the speed of light in vacuum, \(c = 3 \times 10^8\) m/s.

Step 3: Detailed Calculations:
First, convert the given frequency to SI units: \[ \nu = 50 \, \text{MHz} = 50 \times 10^6 \, \text{Hz} \] Given \(E_0 = 150\) N/C. (i) Amplitude of magnetic field (\(B_0\)):
From the formula \(c = E_0 / B_0\), we can rearrange to find \(B_0\): \[ B_0 = \frac{E_0}{c} \] Substitute the given values: \[ B_0 = \frac{150 \, \text{N/C}}{3 \times 10^8 \, \text{m/s}} = 50 \times 10^{-8} \, \text{T} \] \[ B_0 = 5 \times 10^{-7} \, \text{T} \] (ii) Angular frequency (\(\omega\)):
Using the formula \(\omega = 2\pi\nu\): \[ \omega = 2\pi \times (50 \times 10^6 \, \text{Hz}) = 100\pi \times 10^6 \, \text{rad/s} \] \[ \omega = \pi \times 10^8 \, \text{rad/s} \] Substituting \(\pi \approx 3.14\): \[ \omega \approx 3.14 \times 10^8 \, \text{rad/s} \] (iii) Wavelength (\(\lambda\)):
From the formula \(c = \nu\lambda\), we can rearrange to find \(\lambda\): \[ \lambda = \frac{c}{\nu} \] Substitute the values: \[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{50 \times 10^6 \, \text{Hz}} = \frac{300 \times 10^6 \, \text{m/s}}{50 \times 10^6 \, \text{Hz}} \] \[ \lambda = \frac{300}{50} \, \text{m} = 6 \, \text{m} \]

Step 4: Final Answer:
\begin{itemize} \item (i) Amplitude of magnetic field, \(B_0 = 5 \times 10^{-7}\) T. \item (ii) Angular frequency, \(\omega = \pi \times 10^8\) rad/s. \item (iii) Wavelength, \(\lambda = 6\) m. \end{itemize}