Question

For the electromagnetic field described by the equations below, show (a) under what conditions of ω and k does the field satisfy Maxwell’s equations? And (b), suppose that ω = 1010 s−1 and E0 = 10 kV/m. What is the wavelength? What is the energy density in joules per cubic meter, averaged over a large region? From this calculate the power density, the energy flow in joules per square meter per second.

Answer 1

Answer:

a) w / k = c , b)    λ  = 1,887 10⁻¹ m ,   $u_{E}$ =  4.42 10⁻⁴ J / m³

Explanation:

a) Maxwell's equations when solved for an electromagnetic wave result in

              E = E₀ cos (kx - wt)

             B = B₀ cos (kx -wt)

Where k is the vector ce wave and w the angular velocity

            k = 2π / λ

            w = 2π f

Let's divide the two equations

           w / k = f λ  

          w / k = c

Therefore, for w and k to be a solution to Maxwell's equations, their relationship must be equal to the speed of light.

b) If w = 10¹⁰ s⁻¹

      w = 2π f

       f = w / 2π

       f = 10¹⁰ / 2π

       f = 1.59 10⁹ Hz

The speed of light is

       c = λ  f

       λ  = c / f

       λ  = 3 10⁸ / 1.59 10⁹

       λ  = 1,887 10⁻¹ m

Energy density is

      $u_{E}$ = ½ ε₀ E₀²

      $u_{E}$ = ½ 8.85 10⁻¹² (10 10³)²

      $u_{E}$ =  4.42 10⁻⁴ J / m³

Power is energy per unit of time

          P =  $u_{E}$ / t

We calculate for every second

         P = 4.42 10⁻⁴ W / m³

The flow or intensity of energy is

           I = S = c u

           I = 3. 108 4.42 10⁻⁴

           I = 1.33 10⁵ W / m2