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Electromagnetic Waves question

2022 · 26 Jun · Shift 1 · Q55
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  5. /2022 · 26 Jun · Shift 1 · Q55

Electromagnetic Waves question

2022 · 26 Jun · Shift 1 · Q55

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
If Electric field intensity of a uniform plane electromagnetic wave is given as E=−301.6sin⁡(kz−ωt)a^x+452.4sin⁡(kz−ωt)a^yVmE = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}{V \over m}E=−301.6sin(kz−ωt)ax​+452.4sin(kz−ωt)ay​mV​. Then magnetic intensity 'H' of this wave in Am −-− 1 will be : [Given : Speed of light in vacuum c=3×108c = 3 \times {10^8}c=3×108 ms −-− 1, Permeability of vacuum μ0=4π×10−7{\mu _0} = 4\pi \times {10^{ - 7}}μ0​=4π×10−7 NA −-− 2]
  1. A
    +0.8sin⁡(kz−ωt)a^y+0.8sin⁡(kz−ωt)a^x+ 0.8\sin (kz - \omega t){\widehat a_y} + 0.8\sin (kz - \omega t){\widehat a_x}+0.8sin(kz−ωt)ay​+0.8sin(kz−ωt)ax​
  2. B
    +1.0×10−6sin⁡(kz−ωt)a^y+1.5×10−6(kz−ωt)a^x+ 1.0 \times {10^{ - 6}}\sin (kz - \omega t){\widehat a_y} + 1.5 \times {10^{ - 6}}(kz - \omega t){\widehat a_x}+1.0×10−6sin(kz−ωt)ay​+1.5×10−6(kz−ωt)ax​
  3. C
    −0.8sin⁡(kz−ωt)a^y−1.2sin⁡(kz−ωt)a^x- 0.8\sin (kz - \omega t){\widehat a_y} - 1.2\sin (kz - \omega t){\widehat a_x}−0.8sin(kz−ωt)ay​−1.2sin(kz−ωt)ax​
  4. D
    −1.0×10−6sin⁡(kz−ωt)a^y−1.5×10−6sin⁡(kz−ωt)a^x- 1.0 \times {10^{ - 6}}\sin (kz - \omega t){\widehat a_y} - 1.5 \times {10^{ - 6}}\sin (kz - \omega t){\widehat a_x}−1.0×10−6sin(kz−ωt)ay​−1.5×10−6sin(kz−ωt)ax​
View written solutionFree

Correct answer: C

  1. Given electric field
E⃗=−301.6sin⁡(kz−ωt) a^x+452.4sin⁡(kz−ωt) a^yV/m\vec E = -301.6\sin(kz-\omega t)\,\hat a_x + 452.4\sin(kz-\omega t)\,\hat a_y \quad \text{V/m}E=−301.6sin(kz−ωt)a^x​+452.4sin(kz−ωt)a^y​V/m

So the wave is of the form sin⁡(kz−ωt)\sin(kz-\omega t)sin(kz−ωt), hence it is propagating in the +z+z+z-direction.

  1. Use the relation between E⃗\vec EE and H⃗\vec HH

For a uniform plane electromagnetic wave in free space,

H⃗=1η0(a^z×E⃗)\vec H = \frac{1}{\eta_0}(\hat a_z \times \vec E)H=η0​1​(a^z​×E)

where intrinsic impedance of free space is

η0=μ0ϵ0=μ0c\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = \mu_0 cη0​=ϵ0​μ0​​​=μ0​c

Given,

μ0=4π×10−7,c=3×108\mu_0 = 4\pi \times 10^{-7}, \qquad c=3\times 10^8μ0​=4π×10−7,c=3×108

Therefore,

η0=μ0c=4π×10−7×3×108=120π≈376.99 Ω\eta_0 = \mu_0 c = 4\pi \times 10^{-7} \times 3\times 10^8 = 120\pi \approx 376.99\,\Omegaη0​=μ0​c=4π×10−7×3×108=120π≈376.99Ω

So,

η0≈377 Ω\eta_0 \approx 377\,\Omegaη0​≈377Ω
  1. Find direction of H⃗\vec HH

Write

E⃗=Exa^x+Eya^y\vec E = E_x \hat a_x + E_y \hat a_yE=Ex​a^x​+Ey​a^y​

with

Ex=−301.6sin⁡(kz−ωt),Ey=452.4sin⁡(kz−ωt)E_x = -301.6\sin(kz-\omega t), \qquad E_y = 452.4\sin(kz-\omega t)Ex​=−301.6sin(kz−ωt),Ey​=452.4sin(kz−ωt)

Now,

a^z×a^x=a^y,a^z×a^y=−a^x\hat a_z \times \hat a_x = \hat a_y, \qquad \hat a_z \times \hat a_y = -\hat a_xa^z​×a^x​=a^y​,a^z​×a^y​=−a^x​

Hence,

a^z×E⃗=Exa^y−Eya^x\hat a_z \times \vec E = E_x\hat a_y - E_y\hat a_xa^z​×E=Ex​a^y​−Ey​a^x​

Substituting ExE_xEx​ and EyE_yEy​,

a^z×E⃗=−301.6sin⁡(kz−ωt)a^y−452.4sin⁡(kz−ωt)a^x\hat a_z \times \vec E = -301.6\sin(kz-\omega t)\hat a_y - 452.4\sin(kz-\omega t)\hat a_xa^z​×E=−301.6sin(kz−ωt)a^y​−452.4sin(kz−ωt)a^x​

Therefore,

H⃗=1377[−301.6sin⁡(kz−ωt)a^y−452.4sin⁡(kz−ωt)a^x]\vec H = \frac{1}{377}\left[-301.6\sin(kz-\omega t)\hat a_y - 452.4\sin(kz-\omega t)\hat a_x\right]H=3771​[−301.6sin(kz−ωt)a^y​−452.4sin(kz−ωt)a^x​]
  1. Compute components

For a^y\hat a_ya^y​ component:

301.6377=0.8\frac{301.6}{377} = 0.8377301.6​=0.8

For a^x\hat a_xa^x​ component:

452.4377=1.2\frac{452.4}{377} = 1.2377452.4​=1.2

Thus,

H⃗=−0.8sin⁡(kz−ωt)a^y−1.2sin⁡(kz−ωt)a^x A/m\boxed{\vec H = -0.8\sin(kz-\omega t)\hat a_y -1.2\sin(kz-\omega t)\hat a_x\ \text{A/m}}H=−0.8sin(kz−ωt)a^y​−1.2sin(kz−ωt)a^x​ A/m​
  1. Match with options

This corresponds to:

Option C\boxed{\text{Option C}}Option C​
  1. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

They match.

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