Physics - EM Waves Question with Solution | TestHub

Electric field $\vec{E}=5\hat{j}\cos (3x+4z)$ phase angle is independent of time phase angle at $\mathrm{t}=0,3\mathrm{x}+4\mathrm{z}$. So phase angle for $\vec{B}$ will also be $3x+4z$ because $\vec{E}\&\vec{B}$ oscillate in same phase.

Direction of wave propagation $=\frac{3\hat{i}+4\hat{k}}{\sqrt{3^2+4^2}}=\frac{3\hat{i}+4\hat{k}}{5}$

Let magnetic field vector $\overrightarrow{\mathrm{B}}=a\hat{\mathrm{i}}+b\hat{\mathrm{j}}+d\hat{\mathrm{k}}$

direction of wave propagation $\frac{\vec{E}\times \vec{B}}{|\vec{E}||\vec{B}|}=\frac{5\hat{j}\times (a\hat{i}+b\hat{j}+d\hat{k})}{5\times {\left(a^2+b^2+d^2\right)}^{\frac{1}{2}}}=\frac{-5a\hat{k}+5d\hat{i}}{5{\left(a^2+b^2+d^2\right)}^{\frac{1}{2}}}$

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{c}}=\frac{5}{\mathrm{c}}$

$\mathrm{B}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{d}}^2}=\frac{5}{\mathrm{c}}$

$\frac{3\hat{i}+4\hat{k}}{5}=\frac{c\times 5}{5\times 5}(d\hat{i}-a\hat{k})$

$3\hat{i}+4\hat{k}=c(d\hat{i}-a\hat{k})$

$\mathrm{d}=\frac{3}{\mathrm{c}},\mathrm{a}=-\frac{4}{\mathrm{c}}$

$\vec{B}=\frac{3}{c}\hat{k}-\frac{4}{c}\hat{i}=\frac{1}{c}(3\hat{k}-4\hat{i})$

so general equation of $\vec{B}$

$\vec{B}=\frac{1}{c}(3\hat{k}-4\hat{i})\cos (3x+4z-5ct)$