Question

We can reasonably model a 100-W incandescent lightbulb as a sphere 6.0cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation.A) What is the visible light intensity at the surface of the bulb?B) What is the amplitude of the electric field at this surface, for a sinusoidal wave with this intensity?C) What is the amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity?

Answer 1

Answer:

a)  I = 5.65 10⁻² W / m², b)  E = 0.238 N / C, c)   B = 7.93 10⁻¹⁰ T

Explanation:

a) Let's find the amount of power in the region of visible light

     P₁ = 0.05 P

     P₁ = 0.05 100

     P₁ = 5 W

Power is distributed on the surface of a sphere

         P₁ = I / A

         I = P₁ A

   

The area of ​​the sphere

         A = 4π r²

         I = P₁ 4π (d / 2)²

Let's calculate

        I = 5 4π (0.06 / 2)²

        I = 5.65 10⁻² W / m²

b) The intensity is the square of the electric field

          I = E .E = E²

         E = √ I

         E = √ 5.65 10⁻²

         E = 0.238 N / C

c) The electric and magnetic field of an electromagnetic wave are related

                  .E = cB

                  B = E / c

                   B = 0.238 / 3 10⁸

                  B = 0.0793 10⁻⁸ T

                  B = 7.93 10⁻¹⁰ T