Question

Only about 10% of the electromagnetic energy from an incandescent lightbulb is visible light. The bulb radiates most of its energy in the infrared part of the electromagnetic spectrum.1. If you place a 100-W lightbulb 2.2 m away from you, what is the intensity of the infrared radiation at your location? Assume that bulb radiates only visible light and infrared radiation.
2. What is the average infrared energy density?
3. What is the approximate magnitude of the infrared electric field?
4. What is the approximate magnitude of the infrared magnetic field?

Answer 1

Answer:

A) Intensity = 1.48 W/m²

B) Average infrared energy density = 4.93 x 10^(-9) J/m³

C) Magnitude of the infrared electric field = 31.2 N/C

D) Magnitude of the infrared magnetic field = 1.04 x 10^(-7) T

Explanation:

-We are given that;

Intensity of infrared radiation is 2.2m away from the light bulb.

-Power of light bulb = 100W

-Since 10% Electromagnetic energy from an incandescent lightbulb is visible light, thus 90% of electromagnetic energy emitted by light bulb is infrared.

A) The power of the infrared radiation would be = 90% of the total electromagnetic power emitted by the light bulb.

Thus,

power (P) = 90% x 100 = 90 W

The formula for intensity of radiation for an area (A) is given as;

I = P/A

Where P is power and A is area.

Area = 4πR²

R = 2.2m

Thus, A = 4π x 2.2² = 60.82 m²

Thus, I = 90/60.82 = 1.48 W/m²

B) The equation for the average energy density is given as;

I = u'c

Where I is intensity and u' is average energy density and c is the speed of electromagnetic wave in air which is 3 x 10^(8) m/s

Thus, making u' the subject,

u' = I/c

u' = 1.48/3 x 10^(8)

= 4.93 x 10^(-9) J/m³

C) The equation that defines the relationship between energy density and maximum magnitude of the magnetic field is given as;

u' = B(max)²/2μo where μo is permeability of free space and given as 1.257 x 10^(-6) T.m/A

Thus, B(max) = √(2μo•u')

B(max) = √(2 x 1.257 x 10^(-6) x 4.93 x 10^(-9)) = 1.04 x 10^(-7) T

Maximum value of magnitude of electric field is given as;

Emax = cBmax

Thus, Emax = 3 x 10^(8) x 1.04 x 10^(-7) = 31.2 N/C

Answer 2

Answer:

1. 1.48 W/$m^{2}$

2. 4.93 x $10^{-9} J/m^{3}$

3. 33.3 V/m

4. 1.11 x $10^{-7} T$

Explanation:

The step by step explanation to the question is contained in the attached images;