Question 95: EM waves

Question

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 $$c = 3 	imes {10^8}$$ ms$$-$$1, Permeability of vacuum $${\mu _0} = 4\pi  	imes {10^{ - 7}}$$ NA$$-$$2]

Accepted Answer

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

Answer

$$ + 0.8\sin (kz - \omega t){\widehat a_y} + 0.8\sin (kz - \omega t){\widehat a_x}$$

Answer

$$ + 1.0 	imes {10^{ - 6}}\sin (kz - \omega t){\widehat a_y} + 1.5 	imes {10^{ - 6}}(kz - \omega t){\widehat a_x}$$

Answer

$$ - 0.8\sin (kz - \omega t){\widehat a_y} - 1.2\sin (kz - \omega t){\widehat a_x}$$

Answer

$$ - 1.0 	imes {10^{ - 6}}\sin (kz - \omega t){\widehat a_y} - 1.5 	imes {10^{ - 6}}\sin (kz - \omega t){\widehat a_x}$$

EM waves

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