Question

Suppose that the electric field amplitude of an electromagnetic wave propagating along x-direction is ${{E}_{o}}=120N{{C}^{-1}}$ and that its frequency is $\upsilon =50.0MHz$,
A. The expression of electric field is $\overrightarrow{E}=120\sin \left( \dfrac{\pi }{3}x-100\pi \times {{10}^{6}}t \right)\widehat{j}$
B. The expression of electric field is $\overrightarrow{E}=60\sin \left( \dfrac{\pi }{3}x-100\pi \times {{10}^{6}}t \right)\widehat{j}$
C. The expression of electric field is $\overrightarrow{B}=40\times {{10}^{-8}}\sin \left( \dfrac{\pi }{3}x-\pi \times {{10}^{6}}t \right)\widehat{k}$
D. Both A and C

Answer 1

Hint: Electromagnetic wave is considered as a combination of electric field and magnetic field vectors. We will use the expression of EM waves to calculate different variables of an EM wave such as Amplitude, frequency and velocity of wave. By using the calculated values, we can draw the expression for the Electric and Magnetic field of the EM wave.

Formulae used:
$\dfrac{{{E}_{o}}}{{{B}_{o}}}=c$
${{\overrightarrow{E}}_{y}}={{E}_{o}}\sin (kx-\omega t)\widehat{j}$
\[{{\overrightarrow{B}}_{z}}={{B}_{o}}\sin \left( kx-\omega t \right)\widehat{k}\]

Complete step by step answer:
Electromagnetic waves are the waves that are created as a result of vibrations between an electric field and a magnetic field. We can say that electromagnetic waves are composed of oscillating electric and magnetic fields. Electromagnetic waves travel with a constant value of velocity in vacuum. That speed is equal to the speed of light, that is, $3\times {{10}^{8}}m{{s}^{-1}}$. They are not deflected either by electric field or magnetic field. EM waves are also capable of showing interference and diffraction patterns.
Electromagnetic waves do not require any medium for their propagation. They can travel through anything like solids, liquids, air or even vacuum. Electromagnetic waves are transverse waves, meaning that they are measured by their amplitude and wavelength.
Electromagnetic waves consist of both electric and magnetic fields waves. These waves oscillate in perpendicular planes with respect to each other. The creation of an electromagnetic wave begins with an oscillating charged particle, such as an electron, which creates oscillating and magnetic fields.
Representation of an Electromagnetic wave:

We are given that the EM wave is propagating along the x-axis, and the amplitude of the electric field vector is ${{E}_{o}}=120N{{C}^{-1}}$.
Using the relation between electric and magnetic field, we have,
$\dfrac{{{E}_{o}}}{{{B}_{o}}}=c$
$\begin{align}
  & {{B}_{o}}=\dfrac{{{E}_{o}}}{c}=\dfrac{120}{3\times {{10}^{8}}}T \\
 & =40\times {{10}^{-8}}T \\
 & =400\times {{10}^{-9}}T
\end{align}$
${{B}_{o}}=400nT$
The required amplitude of the magnetic field is $400nT$.
Angular frequency is given by,
$\begin{align}
  & \omega =2\pi \upsilon \\
 & =2\pi \times 50\times {{10}^{6}} \\
 & =3.14\times {{10}^{8}}\text{rad }{{\text{s}}^{-1}}
\end{align}$
Wave vector is given by,
$k=\dfrac{2\pi }{\lambda }$
Also,
$k=\dfrac{2\pi \upsilon }{\upsilon \lambda }=\dfrac{2\pi \upsilon }{c}$
As $2\pi \upsilon =\omega $
$k=\dfrac{\omega }{c}=\dfrac{\pi \times {{10}^{8}}}{3\times {{10}^{8}}}\text{rad }{{\text{m}}^{-1}}$
$k=\dfrac{\pi }{3}\text{rad }{{\text{m}}^{-1}}$
Wavelength of the electromagnetic wave is given by,
$c=\upsilon \lambda $
\[\lambda =\dfrac{c}{\upsilon }=\dfrac{3\times {{10}^{8}}}{50\times {{10}^{6}}}\text{m}\]
$\lambda =6\text{m}$
Let the electromagnetic wave travel along +x-axis, and $\overrightarrow{E}$and $\overrightarrow{B}$are along y-axis and z-axis respectively.
Then,
${{\overrightarrow{E}}_{y}}={{E}_{o}}\sin (kx-\omega t)\widehat{j}$
Putting values, we get,
\[{{\overrightarrow{E}}_{y}}=120\sin \left( 1.05x-3.14\times {{10}^{8}}t \right)\widehat{j}N{{C}^{-1}}\]
$\overrightarrow{E}=120\sin \left( \dfrac{\pi }{3}x-100\pi \times {{10}^{6}}t \right)\widehat{j}$
Also,
\[{{\overrightarrow{B}}_{z}}={{B}_{o}}\sin \left( kx-\omega t \right)\widehat{k}\]
Putting values, we get,
${{\overrightarrow{B}}_{z}}=400\sin \left( 1.05x-3.14\times {{10}^{8}}t \right)\widehat{k}nT$
$\overrightarrow{B}=40\times {{10}^{-8}}\sin \left( \dfrac{\pi }{3}x-\pi \times {{10}^{6}}t \right)\widehat{k}$
Comparing obtained expressions with the given options,
Both A and C are correct.
Hence, the correct option is D.

Note: Electromagnetic waves are produced when an electric field couples with a magnetic field. Because the divergence of the electric and magnetic fields are zero, there are no fields in the direction of propagation of waves. The direction of electric field, magnetic field and the direction of propagation; all three are mutually perpendicular to each other.