Practicing Success
A plane electromagnetic wave travelling along the x – direction has a wavelength of 3 mm. The variation in the electric field occurs in the y – direction with an amplitude 66 V m-1. The equations for the electric and magnetic fields as a function of x and t are respectively: |
$E_{y}=33 \cos \pi × 10^{11}\left(t-\frac{x}{c}\right), B_{z}=1.1 × 10^{-7} \cos \pi × 10^{11}\left(t-\frac{x}{c}\right)$ $E_y=11 \cos 2 \pi × 10^{11}\left(t-\frac{x}{c}\right), B_y=11 × 10^{-7} \cos 2 \pi × 10^{11}\left(t-\frac{x}{c}\right)$ $E_{x}=33 \cos \pi × 10^{11}\left(t-\frac{x}{c}\right), B_{x}=11 × 10^{-7} \cos \pi × 10^{11}\left(t-\frac{x}{c}\right)$ $E_{y}=66 \cos 2 \pi × 10^{11}\left(t-\frac{x}{c}\right), B_{z}=2.2 × 10^{-7} \cos 2 \pi × 10^{11}\left(t-\frac{x}{c}\right)$ |
$E_{y}=66 \cos 2 \pi × 10^{11}\left(t-\frac{x}{c}\right), B_{z}=2.2 × 10^{-7} \cos 2 \pi × 10^{11}\left(t-\frac{x}{c}\right)$ |
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 $B_{z}=B_0 \cos \omega\left(t-\frac{x}{c}\right)=2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$ |