Question

You are reading a book and decide to estimate the magnitude of the electric and magnetic fields of the light that is incident on the book from your desk lamp. You can treat the lamp as a point source that emits sinusoidal EM waves uniformly in all directions. The book is at 1 m from the 100 Watt light bulb. Assume that all the electric energy is transformed into visible light.

(Part a) Calculate I, the intensity of the light incident on your book?
(Part b) Find Eo and Bo, the amplitude of the electric and the magnetic fields of the EM waves emitted by the lamp.

Answer 1

a. 7.96 W/m² b. i. 0.205 V/m ii. 0.68 nT

Step-by-step explanation:

(Part a) Calculate I, the intensity of the light incident on your book?

Intensity, I = Power, P/Area,A

I = P/A where P = 100 W and A = 4πr² where r = distance of source from book = 1 m.

So, I = P/A

= 100 W/4π(1 m)²

= 25/π W/m²

= 7.96 W/m²

(Part b) Find Eo and Bo, the amplitude of the electric and the magnetic fields of the EM waves emitted by the lamp.

i. Eo the amplitude of the electric field

Intensity, I = E²/cμ₀ where E = r.m.s value of electric field, c = speed of light = 3 × 10⁸ m/s and μ₀ = permeability of free space = 4π × 10⁻⁷ H/m

Thus, E = √(I/cμ₀)

substituting the values of the variables into the equation, we have

E = √(I/cμ₀)

E = √(7.96 W/m²/[3 × 10⁸ m/s × 4π × 10⁻⁷ H/m])

E = √(7.96 W/m²/120π H/s)

E = √(0.0211 Ws/Hm²)

E = 0.145 V/m

Now E = E₀/√2 where E₀ = maximum value of electric field

So, E₀ = √2E

= √2 × 0.145 V/m

= 0.205 V/m

ii. Bo the amplitude of the magnetic field

Since c = E₀/B₀ where c = speed of light = 3 × 10⁸ m/s

B₀ = E₀/c

= 0.205 V/m ÷ 3 × 10⁸ m/s

= 0.068 × 10⁻⁸ T

= 0.68 × 10⁻⁹ T

= 0.68 nT