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JEE MainPhysicsElectromagnetic Waves

An electromagnetic wave propagating in vacuum has an instantaneous electric field E , propagation vector k , and angular frequency . If ₀ is the permeability of free space, the instantaneous Poynting vector S associated with this wave is given by :

Options

  1. A| E |^2 ₀ k
  2. B- | E |^2 ₀ k
  3. C| E |^2 ₀ E
  4. D| E |^2 ₀ k

Correct answer

A. | E |^2 ₀ k

Step-by-step solution

The instantaneous Poynting vector is defined as S = 1 ₀ ( E B ) . For an electromagnetic wave, the magnetic field vector is related to the electric field and propagation vector by B = k E . Substituting this into the expression for the Poynting vector gives: S = 1 ₀ [ E ( k E ) ] = 1 ₀ [ E ( k E ) ] Using the vector triple product identity A ( B C ) = B ( A C ) - C ( A B ) , we can expand the cross product: E ( k E ) = k ( E E ) - E ( E k ) Since electromagnetic waves are transverse, the electric field is perpendic

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