Electromagnetic Waves

JEE Main level, single correct, medium

75 of 165

Question 75 of 165

The electric field in a plane electromagnetic wave is given by, E=50sin⁡(500x−10×1010t)Vm−1E=50\sin \left(500x-10\times {10}^{10}t\right)V{m}^{-1}. The velocity of an electromagnetic wave in this medium is: (Given c=c= the speed of light in vacuum).

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Answer: (D) 23c\dfrac{2}{3}c

The equation of electric field is given in the question, E=50sin⁡(500x−10×1010t)Vm−1E=50\sin \left(500x-10\times {10}^{10}t\right)V{m}^{-1} now compare it the equation of electric field in an electromagnetic wave, E=E0sin⁡(kx−ωt)Vm−1E={E}_{0}\sin \left(kx-\omega t\right)V{m}^{-1}
⇒ω=1010rads−1\Rightarrow \omega ={10}^{10}rad{s}^{-1} and, k=50m−1k=50{m}^{-1}, so the velocity of electromagnetic waves is, v=ωk=101050=2×108=23cv=\dfrac{\omega }{k}=\dfrac{{10}^{10}}{50}=2\times {10}^{8}=\dfrac{2}{3}c.

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