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

Answer:

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

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

                          $I = \frac{1}{2} c \epsilon _O E_{rms}^2$

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

Making $E_{rms}$ the subject we have

                   $E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }$

Substituting values as given on the question

                $E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }$

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

The amplitude of the electric field is mathematically represented as

                  $A_E = \sqrt{2} * E_{rms}$

                         $= \sqrt{2} * 2.37*10^5$

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

The amplitude of the magnetic field is mathematically represented as

                       $A_M = \frac{A_E}{c}$

Substituting value

                      $A_M = \frac{3.35 *10^5}{3.0*10^8}$

                             $A_M= 1.12 *10^{-3} T$