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
- A| E |^2 ₀ k
- B- | E |^2 ₀ k
- C| E |^2 ₀ E
- 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