Physics - EM Waves Question with Solution | TestHub

Given, $\mathrm{B}=3\times 10^{-8}\cos (1.6\times 10^{3}x$ $+48\times 10^{10}t)$ On comparing equations (1) & (2), we get; $\begin{array}{rl}{\mathrm{B}}_0& =3\times 10^{-8}\\ c& =\frac{\omega }{k}=\frac{48\times 10^{10}}{1.6\times 10^3}\\ & =3\times 10^8\end{array}$ And also, $c=\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}$ Hence, $E_0=B_0\times c$ $\begin{array}{rl}& =3\times 10^{-8}\times 3\times 10^8\\ & =9\end{array}$ So, the required equation is: $\mathrm{E}=9\cos \left(1.6\times 10^{3}x+48\times 10^{10}t\right)$