Question

A plane electromagnetic wave having a frequency $\nu = 23.9GHz$ propagates along the positive z-direction in free space. The peak value of the electric field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave?
$
  {\text{A}}{\text{. }}\overrightarrow B = 2 \times {10^7}\sin \left( {0.5 \times {{10}^3}z + 1.5 \times {{10}^{11}}t} \right)\widehat i \\
  {\text{B}}{\text{. }}\overrightarrow B = 2 \times {10^{ - 7}}\sin \left( {1.5 \times {{10}^2}z + 0.5 \times {{10}^{11}}t} \right)\widehat j \\
  {\text{C}}{\text{. }}\overrightarrow B = 2 \times {10^{ - 7}}\sin \left( {0.5 \times {{10}^3}z - 1.5 \times {{10}^{11}}t} \right)\widehat i \\
  {\text{D}}{\text{. }}\overrightarrow B = 60\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k \\
 $

Answer 1

Hint: The peak value of the magnetic field is equal to the peak value of the electric field divided by the velocity of light. We can find out the values of wave vector angular frequency from the given value of frequency. Matching these values with the given options, we can find the correct answer.

Complete answer:
We are given a plane electromagnetic wave whose frequency is given as
$\nu = 23.9GHz = 23.9 \times {10^9}Hz$
It is propagating along the z-direction in free space. An electromagnetic wave has electric field and magnetic field oscillating perpendicular to each other along the direction of propagation of the electromagnetic wave. We are given the peak value of the electric field which is
${E_0} = 60V/m$
We need to find out the expression for the magnetic field by using the given information. First of all, we need to know the general expression for the magnetic field of an electromagnetic wave which is given as
$B = {B_0}\sin \left( {kz - \omega t} \right)$
It is propagating along the z direction. Here ${B_0}$ is the peak value of the magnetic field, k is known as the wave vector and is defined as follows:
$k = \dfrac{{2\pi }}{\lambda } = \dfrac{{2\pi \nu }}{c} = \dfrac{\omega }{c}$
The angular frequency is given as $\omega = 2\pi \nu $.
The peak value of the magnetic field can be calculated from the peak value of the electric field in the following way.
${B_0} = \dfrac{{{E_0}}}{c} = \dfrac{{60}}{{3 \times {{10}^8}}} = 2 \times {10^{ - 7}}T$
The angular frequency can be calculated from the frequency of the wave as follows:
$\omega = 2\pi \nu = 2\pi \times 23.9 \times {10^9} = 150.1 \times {10^9} = 1.5 \times {10^{11}}rad/s$
The value of the wave vector can be calculated from it in the following way.
$k = \dfrac{\omega }{c} = \dfrac{{1.5 \times {{10}^{11}}}}{{3 \times {{10}^8}}} = 0.5 \times {10^3}{m^{ - 1}}$
Out of the given options, the option C has the matching values of peak value, wave vector and the angular frequency.

Hence, option C should be the correct answer.

Note:
It should be noted that we can calculate the direction of the magnetic field from the direction of the electric field but we are not given the direction of the electric field. Due to this, we will have relied on the values of the peak value, wave vector and the angular frequency to find the correct answer.