If Electric field intensity of a uniform plane electro magnetic wave is given as $E = - 301.6 s i n (k z - ω t) {\hat{a}}_{x} + 452.4 s i n (k z - ω t) \hat{a} t \frac{V}{m} .$

Then, magnetic intensity ‘H’ of this wave in Am^-¹ will be:

Given : Speed of light in vacuum c = 3 × 10⁸ ms^-¹, Permeability of vacuum $μ_{0} = 4 π \times 1 0^{- 7} N A^{- 2}$ ]

Option 1 -

$+ 0.8 s i n (k z - ω t) {\hat{a}}_{y} + 0.8 s i n (k z - ω t) {\hat{a}}_{x} .$

Option 2 -

$+ 1.0 * 1 0^{- 6} s i n (k z - ω t) {\hat{a}}_{y} + 1.5 * 1 0^{- 6} (k z - ω t) {\hat{a}}_{x}$

Option 3 -

$- 0.8 s i n (k z - ω t) {\hat{a}}_{y} - 1.2 s i n (k z - ω t) {\hat{a}}_{x}$

Option 4 -

$- 1.0 * 1 0^{- 6} s i n (k z - ω t) {\hat{a}}_{y} - 1.5 * 1 0^{- 6} s i n (k z - ω t) {\hat{a}}_{x}$

2 Views | Posted 2 weeks ago

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 2

Detailed Solution:

Since

$\vec{E} . \vec{B} = 0 (\vec{E} ⊥ \vec{B})$

$& \frac{E_{0}}{B_{0}} = C = 3 \times 1 0^{8} m / s e c$

for option D

$\frac{E_{0}}{B_{0}} = \frac{\sqrt[]{301 . 6^{2} + 452 . 4^{2}}}{1 0^{- 6} \sqrt[]{1 . 5^{2} + 1^{2}}} \begin{matrix} = 301.6 \times 1 0^{6} & = 3.01 \times 1 0^{8} m / s e c \end{matrix}$

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