Answer to Question #102518 in Optics for VV

Question #102518
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.
1
Expert's answer
2020-02-11T10:11:27-0500

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


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


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


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


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


×B=μ0ϵ0dEdt\nabla \times \overrightarrow{B}=\mu_0 \epsilon_0\frac{d\overrightarrow{E}}{dt}


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


dEdx=dBdt\frac{dE}{dx}=-\frac{dB}{dt}


d2Bdx2=ddx(μ0ϵ0dEdt)=μ0ϵ0ddtdEdx=μ0ϵ0ddtdBdt=μ0ϵ0d2Bdt2\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


d2Bdx2=μ0ϵ0d2Bdt2\frac{d^2B}{dx^2}=\mu_0\epsilon_0\frac{d^2B}{dt^2}


The general wave equation for a wave ψ(x,t)\psi(x,t) traveling in the xx direction with speed vv


d2ψdx2=1v2d2ψdt2\frac{d^2\psi}{dx^2}=\frac{1}{v^2}\frac{d^2\psi}{dt^2}


So,


μ0ϵ0=1v2c=v=1μ0ϵ0=2.997108m/s\mu_0\epsilon_0=\frac{1}{v^2}\to c=v=\frac{1}{\sqrt{\mu_0\epsilon_0}}=2.997\cdot10^8 m/s









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