Suppose that the electric field amplitude of an electromagnetic wave us `E_0=120 N// C` and that its frequency is `50.0 MHz`.
(a) Determine` B_0, omega, k` and `lambda`,
(b) Find expressions for `E` and `B`.

Explanation

Frequency of the electromangeitc wave `v=2.0xx10^(10)Hz`
Electric field amplitude, `E_(0)=4S Vm^(-1)`
Speedof light c=`3xx10^(8)m//s`
(a) Wavelength of a wave is given as:
`gamma=(c)/(v)`
`=(3xx10^(8))/(2xx10^(10))=0.015m`
(b) Magnetic field strength is given as:
`B_(@)=(E_(@))/(c)`
`=(48)/(3xx10^(8))=1.6xx10^(-7)T`
(c) Energy density of the electric field is given as:
`U_(f)=(1)/(2)epsilon_(0)E^(2)`
And, energy density of the magnetic fields is given as:
`U_(B)=(1)/(2mu_(@))B^(2)`
whre,
`epsilon_(0)=` permittivity of free space
`mu_(0)=` permeability of free space
we have the relation connecting E and B as:
`E=cB`.......(i)
where
`c=(1)/sqrt(epsilon_(0)mu_(0))`...........(2)
Putting equation (2) in equation (1), we get
`E=(1)/sqrt(epsilon_(0)mu_(0))B`
Squaring both sides, we get
`E^(2)=(1)/(epsilon_(0)mu_(0))B^(2)`
`epsilon_(0)E^(2)=(B^(2))/(mu_(0))`
`(1)/(2)epsiolon_(0)E^(2)=(1)/(2)(B^(2))/(mu_(0))`
`rArrU_(E)=U_(B)`