Physics - EM Waves Question with Solution | TestHub

$E=E_0\cos \left(\omega t-\overrightarrow{K }·\overrightarrow{ r }\right)$

$\therefore$At any time$t$,

$\overrightarrow{K}=6\hat{\mathrm{i}}+8\hat{\mathrm{k}}$

$B_0=\frac{E_0}{c}=\frac{10}{c}$

$\mathrm{\lambda }=\frac{2\mathrm{\pi }}{k}=\frac{2\mathrm{\pi }}{10}=\frac{\mathrm{\pi }}{5}$

Also,$\overrightarrow{E}\times \overrightarrow{B}=\overrightarrow{c}$

$\therefore \mathrm{\omega }=2\mathrm{\pi }f=\frac{2\mathrm{\pi }c}{\mathrm{\lambda }}$

$\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 0& 1& 0\\ B_x& B_y& B_z\end{array}\right|=\frac{6\hat{\mathrm{i}}+8\hat{\mathrm{k}}}{10}$

$\mathrm{\omega }=\frac{2\mathrm{\pi }\times c}{\left(\frac{\mathrm{\pi }}{5}\right)}=10c$

$\therefore \hat{\mathrm{i}}\left[B_z-0\right]+\hat{\mathrm{j}}\left(0\right)+\hat{\mathrm{k}}\left(-B_x\right)=0.6\hat{\mathrm{i}}-0.8\hat{\mathrm{k}}$

Comparing the coefficients on both sides,

$\therefore B_z=0.6; B_x=-0.8$

$\left|\overrightarrow{B_0}\right|=\frac{10}{c}\left(-0.8\hat{\mathrm{i}}+0.6\hat{\mathrm{k}}\right)=\frac{-8\hat{\mathrm{i}}+6\hat{\mathrm{k}}}{c}$

So the magnetic field,$\overrightarrow{B}=\frac{1}{c}\left(-8\hat{\mathrm{i}}+6\hat{\mathrm{k}}\right)\cos \left(6x+8z-10ct\right) \mathrm{tesla}$