Electromagnetic Waves

JEE Main level, single correct, medium

121 of 165

Question 121 of 165

Magnetic field in a plane electromagnetic wave is given by,

B⃗=B0sin⁡(kx+ωt)j^T\vec{B}={B}_{0}\sin \left(kx+\omega t\right)\hat{j}T. Expression for corresponding electric field will be:

(Where cc is speed of light)

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Answer: (C) E⃗=B0csin⁡(kx+ωt)k^Vm−1\vec{E}={B}_{0}c\sin \left(kx+\omega t\right)\hat{k}V{m}^{-1}

For electromagnetic wave speed of light is given by

c=E0B0c=\dfrac{{E}_{0}}{{B}_{0}}
Electric field is expressed as E0=cB0,E⃗is⊥to directionxand directionofB⃗{E}_{0}=c{B}_{0},\vec{E}is\perp \text{to direction}x\text{and direction}of\vec{B}

The expression of electric field is E⃗=cB0sin⁡(kx+wt)k^Vm−1\vec{E}=c{B}_{0}sin⁡(kx+wt)\hat{k}V{m}^{-1}

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