If Electric field intensity of a uniform plane electromagnetic wave is given as E=-301.6 \sin (k z-\omega t) \stackrel{\^}{a}_x+452.4 \sin (k z-\omega t) \stackrel{\^}{a}_y \frac{V}{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}] [26-Jun-2022-Shift-1]

Correct Answer is : −0.8sin(kz−ωt)∧ay−1.2sin(kz−ωt)∧ax

+0.8 \sin (k z-\omega t) \hat{a}_y+0.8 \sin (k z-\omega t) \hat{a}_x

+1.0 \times 10^{-6} \sin (k z-\omega t) \hat{a}_y+1.5 \times 10^{-6}(k z-\omega t) \hat{a}_x

-0.8 \sin (k z-\omega t) \hat{a}_y-1.2 \sin (k z-\omega t) \hat{a}_x

-1.0 \times 10^{-6} \sin (k z-\omega t) \hat{a}_y-1.5 \times 10^{-6} \sin (k z-\omega t) \hat{a}_x

Electromagnetic Waves Part 1

Section: Physics

Question : 23 of 100

Marks: +1, -0

If Electric field intensity of a uniform plane electromagnetic wave is given as E=−301.6‌sin(kz−ωt)‌

x+452.4‌sin(kz−ωt)‌

y‌

. Then magnetic intensity ' H′ of this wave in Am−1 will be:
[Given : Speed of light in vacuum c=3×108ms−1, Permeability of vacuum .µ0=4π×10−7NA−2]

[26-Jun-2022-Shift-1]

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