Suppose that the electric field part of an electromagnetic wave in vacuum

Solution:

(a) 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 $y$ direction i.e., $-\hat{j}$.

(b) It is given that,

$\vec{E}=3.1 \mathrm{~N} / \mathrm{C} \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^{8} \mathrm{rad} / \mathrm{s}\right) t\right] \hat{i}$    ...(1)

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

$\vec{E}=E_{0} \sin (k x-\omega t) \hat{i}$     ...(2)

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

Electric field amplitude, E0 = 3.1 N/C

Angular frequency, ω = 5.4 × 108 rad/s

Wave number, k = 1.8 rad/m

Wavelength, $\lambda=\frac{2 \pi}{1.8}=3.490 \mathrm{~m}$

(c) Frequency of wave is given as:

$v=\frac{\omega}{2 \pi}$

$=\frac{5.4 \times 10^{8}}{2 \pi}=8.6 \times 10^{7} \mathrm{~Hz}$

(d) Magnetic field strength is given as:

$B_{0}=\frac{E_{0}}{c}$

Where,

c = Speed of light = 3 × 108 m/s

$\therefore B_{0}=\frac{3.1}{3 \times 10^{8}}=1.03 \times 10^{-7} \mathrm{~T}$

(e) 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:

$\vec{B}=B_{0} \cos (k y+\omega t) \hat{k}$

$=\left\{\left(1.03 \times 10^{-7} \mathrm{~T}\right) \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^{6} \mathrm{rad} / \mathrm{s}\right) t\right]\right\} \hat{k}$