Question : A source of light of power 100 W emits light in all directions. Calculate electric and magnetic field at a distance of 20 cm from the source.
Doubt by Aishwarya
Solution :
Power (P) = 100 W
r = 20 cm = 0.20 m
r = 20 cm = 0.20 m
The source is emitting the in all the directions uniformly so the surface must be a sphere of radius of 0.20 m
We know,
Intensity = Power / Area
I = P/πr²
I = 100/4π(0.2)²
I = 100/0.16π
I = 625/π W/m²
We know,
Intensity = Power / Area
I = P/πr²
I = 100/4π(0.2)²
I = 100/0.16π
I = 625/π W/m²
We know, In an EMW, the half of the Intensity is provided by Electric Field and the other half is provided by Magnetic Field
So, the Intensity of Electric field will be
IE = I/2 = [625/π]/2
IE = I/2 = 625/2π W/m²
So, the Intensity of Electric field will be
IE = I/2 = [625/π]/2
IE = I/2 = 625/2π W/m²
Also,
Intensity = Energy Density × Velocity of Wave
Electric Field Intensity = Energy Density of Electric Field × Velocity of EMW
IE = ½ [ε₀Erms² × c]
2IE/cε₀ = Erms²
Erms² = 2IE/cε₀
Erms² = [2×625]/[2×3.14×3×108×8.85×10-12]
Erms² = [1250/166.734]×104
Erms² = 7.49697×104
Erms = 2.738×102 V/m
E₀/√2 = 2.738×102
E₀ = √2 × 2.738×102
E₀ = 1.414 × 2.738×102
E₀ = 3.871×102 V/m
We know
E₀/B₀ = c
B₀ = E₀/c
B₀ = E₀/c
B₀ = [3.871×102]/[3×108]
B₀ = [3.871×102]/[3×108]
B₀ = 1.29×10-6
B₀ = 1.29×10-6 T
Intensity = Energy Density × Velocity of Wave
Electric Field Intensity = Energy Density of Electric Field × Velocity of EMW
IE = ½ [ε₀Erms² × c]
2IE/cε₀ = Erms²
Erms² = 2IE/cε₀
Erms² = [2×625]/[2×3.14×3×108×8.85×10-12]
Erms² = [1250/166.734]×104
Erms² = 7.49697×104
Erms = 2.738×102 V/m
E₀/√2 = 2.738×102
E₀ = √2 × 2.738×102
E₀ = 1.414 × 2.738×102
E₀ = 3.871×102 V/m
We know
E₀/B₀ = c
B₀ = E₀/c
B₀ = E₀/c
B₀ = [3.871×102]/[3×108]
B₀ = [3.871×102]/[3×108]
B₀ = 1.29×10-6
B₀ = 1.29×10-6 T
