A plane EM wave travelling along z direction is described by
`E=E_0sin (kz-omegat) hati and B=B_0 sin (kz-omegat)hatj`.
show that

Options

A. The average energy density of the wave is given by `u_(av)=(1)/(4)epsilon_(0)E_(0)^(2)+(1)/(4)(B_(0)^(2))/(mu_(0))`
B. The time averaged intensity of the wave is given by `I_(av)=(1)/(2)c epsilon_(0)E_(0)^(2)`
C. Both (a) and (b)
D. None of these

Answer

C

Explanation

Energy density of electric field E is
`u_(E) = (1)/(2) epsilon_(0)E^(2)`
Energy density due to magnetic field B is
`u_(B) = (1)/(2) (B^(2))/(mu_(0))`
Total energy density of wave `= (1)/(2) epsilon_(0)E^(2) +(1)/(2) (B^(2))/(mu_(0))`
Now `(u_(E))_(av) = (1)/(T) int_(0)^(T) u_(epsilon) dt = (1)/(T) int_(0)^(T) (1)/(2) epsilon_(0)E^(2)`
`= (epsilon_(0))/(2T) int_(0)^(T) E_(0)^(2) sin^(2) (k x - omega t) dt`
`= (epsilon_(0))/(2T) (E_(0)^(2)xx (T)/(2)) =(1)/(4) epsilon_(0)E_(0)^(2) (int_(0)^(T) sin^(2) (kx - omega t)dt = T//2)`
Similarly, `(u_(B))_(av) = (1)/(4) (B_(0)^(2))/(mu_(0))`
`:. u_(av) = (1)/(2) epsilon_(0) E_(0)^(2) +(1)/(4) (B_(0)^(2))/(mu_(0))`
`:' E = cB_(0)` and `c = (1)/(sqrt(mu_(0)epsilon_(0)))`
`:. (1)/(4) (B_(0)^(2))/(mu_(0)) = (1)/(4) (E_(0)^(2)//c^(2))/(mu_(0)) = (E_(0)^(2))/(4mu_(0)) xx mu_(0) epsilon_(0) = (1)/(4) epsilon_(0) E_(0)^(2)`
`u_(av) = (1)/(4) epsilon_(0) E_(0)^(2) +(1)/(4) (B_(0)^(2))/(mu_(0)) = (1)/(4) epsilon_(0)E_(0)^(2) + (1)/(4) epsilon_(0)E_(0)^(2) = (1)/(2) epsilon_(0)E_(0)^(2) = (1)/(2) (B_(0)^(2))/(mu_(0))`
Time averaged intensity
`I_(av) = u_(av) c = (1)/(2) c epsilon_(0) E_(0)^(2)`