Question

A plane EM wave travelling along z direction is described by `E = E_0 sin (kz - ωt)hati` and `B = B_0 sin (kz - ωt)hatj`. show that 

1. The average energy density of the wave is given by `u_(av) = 1/4 ε_0E_0^2 + 1/4 B_0^2/mu_0`.
2. The time averaged intensity of the wave is given by `I_(av) = 1/2 cε_0 E_0^2`.

Answer 1

i. Total energy carried by electromagnetic wave is due to electric field vector and magnetic field vector. In electromagnetic wave, E and B vary from point to point and from moment to moment.

The energy density due to electric field `vecE` is `U_e = 1/2 ε_0E^2`

The energy density due to magnetic field B is `U_B = 1/(2mu_0) B^2`

Total average energy density of electromagnetic wave

`U = U_E + U_ = 1/2 ε_0E^2 + 1/(2mu_0) B^2`

`E = E_0 sin (kz - ωt)hati`

`B = B_0 sin (kz - ωt)hatj`

Average value of E² over cycle `E_0^2/2`

The average value of B² over a `B_0^2/2`

∴ `U_(av) = 1/2 ε_0 1/2 E_0^2 + 1/(2mu_0) * ((B_0^2))/2`

`u_(av) = 1/4 [ε_0E_0^2 + B_0^2/mu_0]`

ii. We know that E₀ ​= CB₀​ and C = `1/sqrt(mu_0ε_0)`

`1/4 B_0^2/mu_0 = 1/4 (E_0^2/C)/mu_0 = E_0^2/(4mu) mu_0ε_0 = 1/4 ε_0E_0^2`

U_B = U_E 

U_av = `1/4 ε_0E_0^2 + 1/4 B_0^2/(4 mu_0)`

∵ `E_0/B_0 = C` ⇒ `E_0^2/B_0^2 = C^2`

⇒ `E_0^2/B_0^2 = 1/(mu_0ε_0)` 

⇒ `B_0^2 = mu_0εE_0^2`

∴ U_av = `1/4 ε_0E_0^2 + (mu_0ε_0^2E_0^2)/(4 mu_0)`

= `1/4 ε_0E_0^2 + 1/4 ε_0 E_0^2`

U_av = `1/4 ε_0 E_0^2`

U_av = `1/4 ε_0 B_0^2/(mu_0ε_0) + B_0^2/(4mu_0)`  ......`[∵ E_0^2 = B_0^2/(mu_0ε_0)]`

= `B_0^2/(4mu_0) + B_0^2/(4mu_0)`

U_av = `B_0^2/(2mu_0)`

∴ U_E = U_B 

Time average intensity of wave

I_av = U_av . C = `1/2 (B_0^2C)/mu_0 = 1/2 ε_0E_0^2C`