The electric field of a plane polarized electromagnetic wave in...

E = {E}_{0}\cos \left(\omegat - \overset{\to}{K} \cdot \overset{\to}{r}\right)$<br /><br />$∴$At any time$t$,<br /><br />$\overset{\to}{K} = 6\overset{^}{i} + 8\overset{^}{k}$<br /><br />${B}_{0} = \dfrac{{E}_{0}}{c} = \dfrac{10}{c}$<br /><br />$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{10} = \dfrac{\pi}{5}$<br /><br />Also,$\overset{\to}{E} \times \overset{\to}{B} = \overset{\to}{c}$<br /><br />$∴ \omega = 2\pif = \dfrac{2\pic}{\lambda}<br /><br />\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 0 & 1 & 0 \\ {B}_{x} & {B}_{y} & {B}_{z}\end{cases}\right| = \dfrac{6\overset{^}{i} + 8\overset{^}{k}}{10}

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

∴ \overset{^}{i}\left[{B}_{z} - 0\right] + \overset{^}{j}\left(0\right) + \overset{^}{k}\left(- {B}_{x}\right) = 0.6\overset{^}{i} - 0.8\overset{^}{k}

Comparing the coefficients on both sides,

$∴ {B}_{z} = 0.6; {B}_{x} = - 0.8$

\left|\overset{\to}{{B}_{0}}\right| = \dfrac{10}{c}\left(- 0.8\overset{^}{i} + 0.6\overset{^}{k}\right) = \dfrac{- 8\overset{^}{i} + 6\overset{^}{k}}{c}$<br /><br />So the magnetic field,$\overset{\to}{B} = \dfrac{1}{c}\left(- 8\overset{^}{i} + 6\overset{^}{k}\right)\cos \left(6x + 8z - 10ct\right)tesla