# A half wave antenna is used to produce an electromagnetic wave which has a maximum electric field 600 V/m.a. Determine the magnetic field of this wave. b. Determine the total average energy density...

A half wave antenna is used to produce an electromagnetic wave which has a maximum electric field 600 V/m.

a. Determine the magnetic field of this wave.

b. Determine the total average energy density of this wave.

c. Determine the length of each of the two wires used in the antenna if the frequency of the wave produced is 3.2 GHz.

### 1 Answer | Add Yours

a)

Suppose you have an electromagnetic wave with the Eand Bvectors like in the figure below. Than the direction of wave propagation is along the z axis (see the figure).We can write

`E_x = E_max*sin[2*pi*(z/lambda -t/T)]`

`B_y =B_max*sin[2*pi*(z/lambda-t/T)]`

The Maxwell equation relating the vectors **E** and **B** is

`grad xx E = -(delB)/(delt)`

which for the case of the wave that has only `E_x` and `B_y` components is written as

`(delE_x)/(delz) =-(delB_y)/(delt)` or replacing with the values above

`(2*pi)/lambda*E_max =(2*pi)/T*B_max`

`E_max = lambda/T *B_max = c*B_max`

`B_max = 600/(3*10^8) = 2*10^-6 T`

b)

The average energy density is

`w = (epsilon_0*E_max^2)/2 = B_max^2/(2*mu_0)`

`w = 8.854*10^-12*600^2/2 =1.59*10^-6 J/m^3`

c)

The antenna is half wave, this means that the length of each wire is

`L =lambda/2 =(1/2)*(c/nu) =0.5*3*10^8/3.2*10^9 =`

`=4.6875*10^-2 m =4.6875 cm`

**Sources:**