Suppose that the electric field part of an electromagnetic wave in vacuum is $E = \{(3.1 \; N/C) \cos [(1.8 \; rad/m) y + (5.4 \times 10^{6} \; rad/s) t] \} \hat{i}$.
$(a)$ What is the direction of propagation?
$(b)$ What is the wavelength $\lambda$?
$(c)$ What is the frequency $\nu$?
$(d)$ What is the amplitude of the magnetic field part of the wave?
$(e)$ Write an expression for the magnetic field part of the wave.

(A) The electric field is oscillating along the $x$-axis and varies with $y$ and $t$. Since the argument of the cosine function is $(ky + \omega t)$,the wave propagates in the negative $y$-direction,i.e.,$-\hat{j}$.
$(b)$ Comparing the given equation with the standard form $E = E_0 \cos(ky + \omega t)$,we have $k = 1.8 \; rad/m$. The wavelength $\lambda = \frac{2\pi}{k} = \frac{2 \times 3.14}{1.8} \approx 3.49 \; m$.
$(c)$ The angular frequency $\omega = 5.4 \times 10^{6} \; rad/s$. The frequency $\nu = \frac{\omega}{2\pi} = \frac{5.4 \times 10^{6}}{2 \times 3.14} \approx 8.6 \times 10^{5} \; Hz$.
$(d)$ The amplitude of the magnetic field $B_0 = \frac{E_0}{c}$,where $E_0 = 3.1 \; N/C$ and $c = 3 \times 10^{8} \; m/s$. Thus,$B_0 = \frac{3.1}{3 \times 10^{8}} \approx 1.03 \times 10^{-8} \; T$.
$(e)$ Since $\vec{E}$ is along $\hat{i}$ and propagation is along $-\hat{j}$,the magnetic field $\vec{B}$ must be along $\hat{k}$ (as $\vec{E} \times \vec{B}$ gives the direction of propagation). The expression is $\vec{B} = \{(1.03 \times 10^{-8} \; T) \cos [(1.8 \; rad/m) y + (5.4 \times 10^{6} \; rad/s) t] \} \hat{k}$.