Electromagnetic Waves

JEE Main level, single correct, medium

66 of 165

Question 66 of 165

If Electric field intensity of a uniform plane electro magnetic wave is given as E=−301.6sin⁡(kz−ωt)a^x+452.4sin⁡(kz−ωt)a^yVm−1.E=-301.6\sin \left(kz-\omega t\right){\hat{a}}_{x}+452.4\sin \left(kz-\omega t\right){\hat{a}}_{y}V{m}^{-1}. Then, magnetic intensity HH of this wave in Am−1A{m}^{-1} will be [Given : Speed of light in vacuum c=3×108ms−1c=3\times {10}^{8}{ms}^{-1}, Permeability of vacuum μ0=4π×10−7NA−2{\mu }_{0}=4\pi \times {10}^{-7}N{A}^{-2}]

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Answer: (C) −0.8sin⁡(kz−ωt)a^y−1.2sin⁡(kz−ωt)a^x-0.8\sin \left(kz-\omega t\right){\hat{a}}_{y}-1.2\sin \left(kz-\omega t\right){\hat{a}}_{x}

The relation between magnetic field and electric field is given by,
c=E0B0⇒B0=E0cc=\dfrac{{E}_{0}}{{B}_{0}}\Rightarrow {B}_{0}=\dfrac{{E}_{0}}{c}
And magnetic intensity is given by,
H0=B0μ0⇒H0=E0cμ0{H}_{0}=\dfrac{{B}_{0}}{{\mu }_{0}} \Rightarrow {H}_{0}=\dfrac{{E}_{0}}{c{\mu }_{0}}
Also, E⃗\vec{E} and B⃗\vec{B} are perpendicular to each other.
Therefore,
∣Hy0∣=301.63×108×4π×10−7=0.8Am−1∣Hx0∣=452.43×108×4π×10−7=1.2Am−1\left|{H}_{y0}\right|=\dfrac{301.6}{3\times {10}^{8}\times 4\pi \times {10}^{-7}}=0.8A{m}^{-1} \left|{H}_{x0}\right|=\dfrac{452.4}{3\times {10}^{8}\times 4\pi \times {10}^{-7}}=1.2A{m}^{-1}
Hence, C is the answer.

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