Question

The Poynting vector of an electromagnetic wave in vacuum is

\(S = \left\{ {\left( {120W/{m^2}} \right){{\sin }^2}\left[ {8.0rad/m} \right)z + \left( {2.4 \times {{10}^9}rad/s} \right)t]} \right\}k\)

What is the frequency of the wave?
1. 382 MHz
2. 172 MHz
3. 241 MHz
4. 54.6 MHz

Answer 1

Correct Answer - Option 1 : 382 MHz

Concept:

Light may be described as a wave. The electric field vector propagating in the x-direction in a vacuum may be written as:

\(E\left( {x,t} \right) = {E_{max}}\cos \left( {kx - \omega t + \phi } \right)\)

There is also a magnetic field associated with the electric field when light propagates. The magnetic field is perpendicular to the direction of propagation as well as to the electric field E.

 \(B\left( {x,t} \right) = {B_{max}}\cos \left( {kx - \omega t + \phi } \right)\)

Such a combination of mutually perpendicular electric and magnetic fields in referred to as an electromagnetic wave in a vacuum.

The wavenumber is k = 2π/λ, where λ is the wavelength of the wave. 

The frequency f of the wave is ν = ω/2π, ω is the angular frequency. 

The speed of any periodic wave is the product of its wavelength and frequency. V = λν.

Poynting Vector:

When an electromagnetic wave advances, the electromagnetic energy flows in the direction of E × H. The total energy flowing perpendicularly per second per unit area into the surface in free space is called the Poynting vector, where

\(\vec S = {c^2}\epsilon_0\left( {\vec E × \vec B} \right) = \frac{1}{{{\mu _o}}}\left( {\vec E × \vec B} \right)\;W/{m^2}\)

Calculation:

\(S = \left\{ {\left( {120W/{m^2}} \right){{\sin }^2}\left[ {8.0rad/m} \right)z + \left( {2.4 × {{10}^9}rad/s} \right)t]} \right\}k\)

Here ω = 2.4 × 10⁹ rad/s

Frequency ν = ω/2π = 382 × 10⁶ Hz = 382 MHz