Question

The time-average power carried by a UPEMW propagating in vacuum is 0.05 W/m2. i) What is the amplitude value of the electric field and the amplitude value of the magnetic field in the wave

Answer 1

The correct solution is "11.51 mA".

Step-by-step explanation:

Given:

Time average power,

`P_(avg)=0.05 \ W/m^2`

n = 377

As we now,

⇒
`P_(avg)=(E_0^2)/(n)`

or,

⇒
`E_0^2=0.05* 377`

⇒
`=4.341 \ V`

hence,

⇒
`H_0=(E_0)/(n)`

By putting the values, we get

`=(4.341)/(377)`

`=11.51 \ mA`