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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​
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