The intensity of the light from a bulb incident on a surface is 0 . 22 W m - 2 . The amplitude of the magnetic field in this light-wave is _____ × 10 - 9 T (Given : Permittivity of vacuum ϵ 0 = 8 . 85 × 10 - 12 C 2 N - 1 m - 2 , speed of light in vacuum c = 3 × 10 8 m s - 1 )

Step-by-step solution

The intensity is defined as energy per unit time per unit area. Therefore, I = d E A d t . Energy density is energy per unit volume. Therefore, U d = d E A d x . Dividing I U d = d x d t = c ⇒ U d = I c = 0 . 22 3 × 10 8 = 22 3 × 10 - 10   J   m - 3 Magnetic energy density will be half of total energy density, Therefore, U b = 11 3 × 10 - 10   J   m - 3 . Now energy density in magnetic field is given by, U b = B 2 2 μ 0 . Therefore, B rms 2 2 μ 0 = 11 3 × 10 -