Question

A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is in the x direction and the magnetic field is in the -y direction.(a) What is the direction of propagation of the wave?(b) If the intensity of the wave at point P is 7.43kW/cm2, what is the electric and magnetic field amplitude at that point? (c = 3.00 ×108 m/s, μ0 = 4π× 10-7 T · m/A,ε0 = 8.85 × 10-12 C2/N · m2)

Answer 1

a

The direction of the wave propagation is the negative z -axis

b

The amplitude of electric and magnetic field are
`A_E= 3.35*10^5 V/m` ,

`A_M= 1.12 *10^(-3) T` respectively

Step-by-step explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive x-axis i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative z -axis

The Intensity of the wave is mathematically represented as

`I = (1)/(2) c \epsilon _O E_(rms)^2`

Given that
`I = 7.43 (kW)/(cm^2) = 7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 (W)/(m^2)`

Making
`E_(rms)` the subject we have

`E_(rms) = \sqrt{(I)/(0.5*c*\epsilon_o) }`

Substituting values as given on the question

`E_(rms) = \sqrt{(7.43 *10^7[(W)/(m^2) ])/(0.5 * 3.08*10^8 *8.85*10^(-12)) }`

`= 2.37*10^5 \ V/m`

The amplitude of the electric field is mathematically represented as

`A_E = √(2) * E_(rms)`

`= √(2) * 2.37*10^5`

`A_E= 3.35*10^5 V/m`

The amplitude of the magnetic field is mathematically represented as

`A_M = (A_E)/(c)`

Substituting value

`A_M = (3.35 *10^5)/(3.0*10^8)`

`A_M= 1.12 *10^(-3) T`