Electromagnetic Waves

JEE Main level, single correct, medium

115 of 165

Question 115 of 165

A plane electromagnetic wave of wavelength λ\lambda has an intensity II. It is propagating along the positive Y−Y-direction. The allowed expressions for the electric and magnetic fields are given by:

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Answer: (A) E⃗=2Iϵ0ccos⁡[2πλ(y−ct)]k^\vec{E}=\sqrt{\dfrac{2I}{{\epsilon }_{0}c}}\cos \left[\dfrac{2\pi }{\lambda }\left(y-ct\right)\right]\hat{k},

B⃗=+1cEi^\vec{B}=+\dfrac{1}{c}E\hat{i}

If E0{E}_{0} is the amplitude of the electric field then the intensity of the electromagnetic wave is given as I=12ϵ0E02×cI=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}\times c. From the above relation, we get E0=2Icϵ0{E}_{0}=\sqrt{\dfrac{2I}{c{\epsilon }_{0}}}. The relation between electric and magnetic fields is B0=E0c{B}_{0}=\dfrac{{E}_{0}}{c}. The direction of E⃗×B⃗\vec{E}\times \vec{B} will be along +j^+\hat{j}.

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