Question

The intensity of the light from a bulb incident on a surface is 0.22 W/m². The amplitude of the magnetic field in this light-wave is ___ × 10^–9 T.

(Given : Permittivity of vacuum ε₀ = 8.85 × 10^–12C²N^–1–m^–2, speed of light in vacuum c = 3 × 10⁸ ms^–1)

Answer 1

Correct Answer: 43

To find the amplitude of the magnetic field (B₀) in the light wave, we will use the formula that relates the intensity (I) of an electromagnetic wave to its electric field amplitude (E₀) and magnetic field amplitude (B₀). The formula for intensity is:

I = ½ cε₀E₀²

Given: I = 0.22 W/m², ε₀ = 8.85 × 10^–12C²N^–1m^–2, c = 3 × 10⁸ m/s.

Rearranging the formula for E₀:

E₀ = √((2I)/(cε₀))

Substitute values to calculate E₀:

E₀ = √((2 × 0.22)/(3 × 10⁸ × 8.85 × 10^–12))

E₀ = √((0.44)/(2.655 × 10^–3))

E₀ = √(165.84 × 10³)

E₀ = 12.88 × 10³ V/m

The relation between E₀ and B₀ is given by:

B₀ = E₀/c

Substitute to find B₀:

B₀ = 12.88 × 10³/(3 × 10⁸)

B₀ = 4.2933 × 10^–5 T

Express B₀ in terms of 10^–9 T:

B₀ = 42.933 × 10^–9 T

The amplitude of the magnetic field is 43 × 10^–9 T, which fits the provided range of 43, 43.