If Electric field intensity of a uniform plane electromagnetic wave is given as E = -301.6sin (kz - ω t)ax + 452.4sin (kz - ω t)ay 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 μ₀ = 4π× 10⁻⁷ NA ⁻² ] (a) +0.8sin (kz - ω t)ay + 0.8sin (kz - ω t)ax (b) +1.0× 10⁻⁶sin (kz - ω t)ay + 1.5× 10⁻⁶sin (kz - ω t)ax (c) -0.8sin (kz - ω t)ay - 1.2sin (kz - ω t)ax (d) -1.0× 10⁻⁶sin (kz - ω t)ay - 1.5× 10⁻⁶sin (kz - ω t)ax

Answer: (c). H = (B)/(μ₀) = (E)/(μ₀ c), with μ₀ c = 4π×10⁻⁷×3×10⁸ = 377 Ω. (301.6)/(377) = 0.8, (452.4)/(377) = 1.2 A/m. Wave moves along +z, H = ( k× E)/(377): k×(-0.8 ax) = -0.8 ay, k×(1.2 ay) = -1.2 ax. So the answer is (c) -0.8sin (kz-ω t) ay - 1.2sin (kz-ω t) ax.

If Electric field intensity of a uniform plane electromagnetic wave… | JEE Main 2022 (26 Jun, Shift 1)

Q36JEE Main 2022 · 26 JUN · SHIFT 1Medium

Nature of EM Waves & Wave Equation

If Electric field intensity of a uniform plane electromagnetic wave is given as

$E = -301.6\sin(kz - \omega t)\hat{a}_x + 452.4\sin(kz - \omega t)\hat{a}_y\,\frac{\text{V}}{\text{m}}$ . 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 ${}^{-2}$ ]

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Answer: (c)

$H = \dfrac{B}{\mu_0} = \dfrac{E}{\mu_0 c}$ , with $\mu_0 c = 4\pi\times10^{-7}\times3\times10^{8} = 377\ \Omega$ .
$\dfrac{301.6}{377} = 0.8$ , $\dfrac{452.4}{377} = 1.2$ A/m.
Wave moves along $+z$ , $\vec H = \dfrac{\hat k\times\vec E}{377}$ : $\hat k\times(-0.8\hat a_x) = -0.8\hat a_y$ , $\hat k\times(1.2\hat a_y) = -1.2\hat a_x$ .
So the answer is (c) $-0.8\sin(kz-\omega t)\hat a_y - 1.2\sin(kz-\omega t)\hat a_x$ .

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