In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10^{10} Hz and amplitude 48 V m^{−1}.

**(a)** What is the wavelength of the wave?

**(b) **What is the amplitude of the oscillating magnetic field?

**(c) **Show that the average energy density of the **E **field equals the average energy density of the **B **field. [*c *= 3 × 10^{8} m s^{−1}.]

#### Solution

Frequency of the electromagnetic wave, *ν* = 2.0 × 10^{10} Hz

Electric field amplitude, *E*_{0} = 48 V m^{−1}

Speed of light, *c* = 3 × 10^{8} m/s

**(a)** Wavelength of a wave is given as:

`lambda=c/v`

`=(3xx10^8)/(2xx10^10)=0.015m`

**(b)** Magnetic field strength is given as:

`B_0=E_0/c`

`=48/(3xx10^8)=1.6xx10^-7 T`

**(c)** Energy density of the electric field is given as:

`U_E=1/2in_0E^2`

And, energy density of the magnetic field is given as:

`U_B=1/(2mu_0)B^2`

Where,

∈_{0} = Permittivity of free space

μ_{0} = Permeability of free space

We have the relation connecting *E* and *B* as:

*E* = *cB* … (1)

Where,

`c=1/(sqrt(in_0mu_0))` … (2)

Putting equation (2) in equation (1), we get

`E=1/sqrt(in_0mu_0)`

Squaring both sides, we get

`E=1/(in_0mu_0) B^2`

`in_0E^2=B^2/mu_0`