A plane electromagnetic wave travelling along the x – direction has a…

Here, $E_0=66 Vm^{-1}, \lambda=3 mm=3 \times 10^{-3} m$

∴ $B_0 =\frac{E_0}{c}=\frac{66}{3 \times 10^8}=2.2 \times 10^{-7} T$

$\omega =2 \pi v=\frac{2 \pi c}{\lambda}=\frac{2 \pi \times 3 \times 10^8}{3 \times 10^{-3}}=2 \pi \times 10^{11}$

$\vec{E}$ is along y - direction and the wave propagates along x - direction. Therefore, $\vec{B}$ should be in a direction perpendicular to both x and y - axes. Using vector algebra $\vec{E} \times \vec{B}$ should be along x - direction.

Since $(\hat{i})=(+\hat{j}) \times(+\hat{k}), \vec{B}$ is along the $z-$ direction.

∴ The equation for the electric field along $y$ - direction in the electromagnetic wave is given by

$E_y=E_0 \cos \omega\left(t-\frac{x}{c}\right)=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$

and the equation for the magnetic field along z - direction in the electromagnetic wave is given by

$B_{z}=B_0 \cos \omega\left(t-\frac{x}{c}\right)=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$