If Electric field intensity of a uniform plane electromagnetic wave is given as $$E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}{V \over m}$$. Then magnetic intensity 'H' of this wave in Am^$-$1 will be :

[Given : Speed of light in vacuum $c = 3 \times {10^8}$ ms^$-$1, Permeability of vacuum ${\mu _0} = 4\pi \times {10^{ - 7}}$ NA^$-$2]

<p>$$\overrightarrow E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}$$</p> <p>${E_{0x}} = 301.6$</p> <p>${E_{0y}} = + 452.4$</p> <p>${E_0} = \sqrt {E_{0x}^2 + E_{0y}^2}$</p> <p>Now, $${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C} = {{\sqrt {E_{0x}^2 + E_{0y}^2} } \over C}$$</p> <p>Also, $$\widehat B = \widehat C \times \widehat E \Rightarrow \widehat k \times {{({E_{0x}}\widehat i + {E_{0y}}\widehat j)} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}$$</p> <p>$$\widehat B = {{ - {E_{0x}}\widehat j - {E_{0y}}\widehat i} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}$$</p> <p>$$\overrightarrow B = - {{{E_{0x}}} \over C}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over C}\sin (kz - \omega t){\widehat a_x}$$</p> <p>$\overrightarrow H = {{\overrightarrow B } \over {{\mu _0}}}$</p> <p>$$\overrightarrow H = - {{{E_{0x}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_x}$$</p>