Electromagnetic Waves

JEE Main level, single correct, medium

103 of 165

Question 103 of 165

The electric field of a plane electromagnetic wave is given by
E⃗=E0i^cos⁡⁡(kz)cos⁡(ωt)\vec{E}={E}_{0}\hat{i}\cos ⁡\left(kz\right)cos⁡(\omega t)
The corresponding magnetic field B⃗\vec{B} is then given by:

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Answer: (C) B⃗=E0Cj^sin⁡⁡(kz)sin(ωt)\vec{B}=\dfrac{{E}_{0}}{C}\hat{j}\sin ⁡\left(kz\right)sin(\omega t)

E⃗=E0i^cos(kz)cos(ωt)\vec{E}={E}_{0} \hat{i} cos \left(kz\right)cos\left(\omega t\right)
E⃗=E02i^cos(kz−ωt)+E02i^cos(kz+ωt)\vec{E}=\dfrac{{E}_{0}}{2} \hat{i} cos \left(kz-\omega t\right)+\dfrac{{E}_{0}}{2}\hat{i}cos\left(kz+\omega t\right)
B⃗=E02Cj^cos(kz−ωt)+E02Cj^cos(kz+ωt)\vec{B}=\dfrac{{E}_{0}}{2C} \hat{j} cos \left(kz-\omega t\right)+\dfrac{{E}_{0}}{2C}\hat{j}cos\left(kz+\omega t\right)
=E02Cj^[2sin(kz)sin(ωt)]=\dfrac{{E}_{0}}{2C}\hat{j}\left[2 sin \left(kz\right)sin\left(\omega t\right)\right]
=E0Cj^[sin(kz)sin(ωt)]=\dfrac{{E}_{0}}{C}\hat{j} \left[ sin \left(kz\right)sin\left(\omega t\right)\right]

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