Suppose that the electric field part of an electromagnetic wave in vacuum is $E = 3.1 \ \text{N/C} \cos[(1.8…

1. From the given electric field vector, it can be inferred that the electric field is directed along the negative x direction. Hence, the direction of motion is along the negative y

diraction i.e., \(-\hat{\text{j}}\).

2. It is given that,

\(\vec{\text{E}}=3.1 \ \text{N}/\text{C}\cos\Big[(1.8 \ \text{rad}/ \text{m})\text{y}+(5.4\times10^8 \ \text{rad}/\text{s})\text{t}\Big]\hat{\text{i}}\dots(1)\)

The general equation for the electric field vector in the positive x direction can be written as:

\(\vec{\text{E}}=\text{E}_0\sin(\text{kx}-\omega\text{t})\dots(2)\)

On comparing equations (1) and (2), we get

Electric field amplitude, E₀ = 3.1 N/C

Angular frequency, \(\omega\) = 5.4 x 10⁸ rad/s

Wave number, k = 1.8 rad/m

 wavelength, \(\lambda=\frac{2\pi}{1.8}=3.490 \ \text{m}\)

3. Frequency of wave is given as:

\(\text{v}=\frac{\omega}{2\pi}\)

\(=\frac{5.4\times10^8}{2\pi}=8.6\times10^7 \ \text{Hz}\)

4. Magnetic field strength is given as:

\(\text{B}_0=\frac{\text{E}_0}{\text{c}}\)

Where,

c = Speed of light = 3 × 10⁸ m/s

\(\therefore\ \text{B}_0=\frac{3.1}{3\times10^8}=1.03\times10^{-7} \ \text{T}\)

5. On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:

\(\vec{\text{B}}=\text{B}_0\cos(\text{ky}+\omega\text{t})\hat{\text{k}}\)

\(=\Big\{(1.03\times10^{-7}\text{T})\cos\Big[(1.8 \ \text{rad}/\text{m})\text{y}+(5.4\times10^6\text{rad}/\text{s})\text{t}\Big]\Big\}\hat{\text{k}}\)