Question

Light waves are propagating in vacuum. Derive the wave equation for the associated magnetic field vector. On the basis of this equation, calculate the speed of light.

Answer 1

$\overrightarrow{E}(x,t)=E(x,t)\overrightarrow{j}$

$\overrightarrow{B}(x,t)=B(x,t)\overrightarrow{k}$

$\nabla \cdot \overrightarrow{E}=0$

$\nabla \cdot \overrightarrow{B}=0$

$\nabla \times \overrightarrow{E}=-\frac{d\overrightarrow{B}}{dt}$

$\nabla \times \overrightarrow{B}=\mu_0 \epsilon_0\frac{d\overrightarrow{E}}{dt}$

$\nabla \times E(x,t)\overrightarrow{j}=\frac{dE}{dx}\overrightarrow{k}$

$\frac{dE}{dx}=-\frac{dB}{dt}$

$\frac{d^2B}{dx^2}=-\frac{d}{dx}(\mu_0\epsilon_0\frac{dE}{dt})=-\mu_0\epsilon_0\frac{d}{dt}\frac{dE}{dx}=\mu_0\epsilon_0\frac{d}{dt}\frac{dB}{dt}=\mu_0\epsilon_0\frac{d^2B}{dt^2}$

So, we have

$\frac{d^2B}{dx^2}=\mu_0\epsilon_0\frac{d^2B}{dt^2}$

The general wave equation for a wave $\psi(x,t)$ traveling in the $x$ direction with speed $v$

$\frac{d^2\psi}{dx^2}=\frac{1}{v^2}\frac{d^2\psi}{dt^2}$

So,

$\mu_0\epsilon_0=\frac{1}{v^2}\to c=v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=2.997\cdot10^8 m/s$