QJEEM22Phaf1e: JEE Main 2022 Physics - Class 12 JEEM22Phaf1e Solution: If Electric field intensity of a uniform plane electromagnetic…

$\overrightarrow E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}$

${E_{0x}} = 301.6$

${E_{0y}} = + 452.4$

${E_0} = \sqrt {E_{0x}^2 + E_{0y}^2}$

Now, ${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C} = {{\sqrt {E_{0x}^2 + E_{0y}^2} } \over C}$

Also, $\widehat B = \widehat C \times \widehat E \Rightarrow \widehat k \times {{({E_{0x}}\widehat i + {E_{0y}}\widehat j)} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}$

$\widehat B = {{ - {E_{0x}}\widehat j - {E_{0y}}\widehat i} \over {\sqrt {E_{0x}^2 + E_{0y}^2} }}$

$\overrightarrow B = - {{{E_{0x}}} \over C}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over C}\sin (kz - \omega t){\widehat a_x}$

$\overrightarrow H = {{\overrightarrow B } \over {{\mu _0}}}$

$\overrightarrow H = - {{{E_{0x}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_y} - {{{E_{0y}}} \over {{\mu _0}C}}\sin (kz - \omega t){\widehat a_x}$