In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10 H_z` and amplitude `48V_m^-1`
(a) What is the wavelength o f the wave?
(b) What is the amplitude of the oscillating magnetic field.
(c) Show that the average energy density of the field `E` equals the average energy density of the field `B.[c=3xx10^8 ms^-1]`.

Explanation

We are given that,
`E_(0) = 48 V//m, v = 2.0×x10^(10) Hz and c=3×x10^(8) V//m `
(a) Wavelength of the wave,
`lambda = c/v= (3xx10^(8)m//s)/(2.0xx10^(10)s^(-1))=1.5xx10^(-2)m`
(b) Amplitude of the oscillating magnetic filed,
`B_(0)=E_(0)/c=(48V//m)/(3xx10^(8)m//s)=1.6xx10^(-7)T`
(c) Total average energy density.
`u_("av")=1/2in_(0)E_(0)^(2)`
`=1/2(8.85xx10^(12))(48)^(2)J//M^(3)=1.0xx10^(-8)J//m^(3)`