We can reasonably model a 75 W incandescent light bulb as a sphere 6.0 cm in diameter. Typically, only about 5% of the energy goes to visible light; the rest goes largely to non-visible light infrared radiation.
(a) What is the visible light intensity (in W/m^2) at the surface of the bulb.
(b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?
I am really lost on how to solve this problem. The only thing I know for sure that I need to do is: (75)(.05)=3.75 W for the visible light. Can you please show your work?
find the surface area of the sphere (4 pi r^2) where r = .03 meters
then you have .05*75/ surface area
Your text has the relationship between intensity and E and B amplitudes.
Sure! Let's break down the problem step-by-step.
(a) To find the visible light intensity at the surface of the bulb, we need to calculate the power per unit area.
Given:
Total power of the bulb = 75 W
Percentage of energy going to visible light = 5% = 0.05
Visible light power = Total power * Percentage of energy going to visible light
Visible light power = 75 W * 0.05 = 3.75 W
The intensity is defined as the power per unit area, so we need to divide the visible light power by the surface area of the sphere.
Surface area of a sphere = 4πr^2, where r is the radius of the sphere.
Given:
Diameter of the sphere = 6.0 cm
Radius of the sphere = Diameter / 2 = 6.0 cm / 2 = 3.0 cm = 0.03 m
Surface area of the sphere = 4π(0.03 m)^2 = 0.036π m^2
Visible light intensity = Visible light power / Surface area of the sphere
Visible light intensity = 3.75 W / 0.036π m^2 ≈ 33.01 W/m^2
So, the visible light intensity at the surface of the bulb is approximately 33.01 W/m^2.
(b) To find the amplitudes of the electric and magnetic fields, we need to use the relationship between intensity and amplitude.
The intensity of an electromagnetic wave is given by the formula:
Intensity = (ε_0 * c * E_0^2) / 2,
where ε_0 is the vacuum permittivity, c is the speed of light, and E_0 is the amplitude of the electric field.
Given that the intensity is 33.01 W/m^2 and using the above formula, we can solve for E_0.
Rearranging the formula, we get:
E_0^2 = (2 * Intensity) / (ε_0 * c)
Now, we can substitute the respective values:
ε_0 (vacuum permittivity) ≈ 8.854 x 10^(-12) F/m
c (speed of light) ≈ 3.0 x 10^8 m/s
Intensity = 33.01 W/m^2
E_0^2 = (2 * 33.01 W/m^2) / (8.854 x 10^(-12) F/m * 3.0 x 10^8 m/s)
E_0^2 ≈ 2.367 x 10^20 V^2/m^2
Finally, we take the square root to find the amplitude of the electric field:
E_0 ≈ √(2.367 x 10^20 V^2/m^2) ≈ 4.866 x 10^9 V/m
The amplitude of the electric field at the surface of the bulb is approximately 4.866 x 10^9 V/m.
Since light is an electromagnetic wave, the magnetic field amplitude can be found using the formula:
E_0 / B_0 = c,
where B_0 is the amplitude of the magnetic field. Rearranging the formula, we get:
B_0 = E_0 / c
Substituting the values:
B_0 = (4.866 x 10^9 V/m) / (3.0 x 10^8 m/s)
B_0 = 16.22 T
So, the amplitude of the magnetic field at the surface of the bulb is approximately 16.22 T.
Sure! Let's break down the problem into smaller steps to solve it.
Step 1: Calculate the total power radiated by the bulb.
We know that the bulb emits 75 W of power, and only 5% of that power is in visible light. So, we can calculate the total power radiated by the bulb as follows:
Visible light power = 75 W * 0.05 = 3.75 W
Step 2: Calculate the surface area of the bulb.
The bulb is modeled as a sphere 6.0 cm in diameter. To calculate the surface area of a sphere, we can use the formula:
Surface area of a sphere = 4πr^2
where r is the radius of the sphere. In this case, the radius is half the diameter, so r = 6.0 cm / 2 = 3.0 cm.
Convert the radius to meters: 3.0 cm = 0.03 m.
Now, we can calculate the surface area:
Surface area of the bulb = 4π(0.03 m)^2
Step 3: Calculate the visible light intensity.
The visible light intensity (I) is defined as the power per unit area. We can calculate it by dividing the visible light power by the surface area of the bulb:
Visible light intensity = Visible light power / Surface area of the bulb
Substituting in the values, we have:
Visible light intensity = 3.75 W / (4π(0.03 m)^2)
Now, we can evaluate this expression to get the visible light intensity in W/m^2.
Step 4: Calculate the amplitudes of the electric and magnetic fields.
The amplitude of an electromagnetic wave is related to its intensity. Specifically, the intensity is proportional to the square of the amplitude.
So, the amplitude (A) can be calculated using the equation:
Amplitude = √(Intensity / (ε0 * c))
where ε0 is the permittivity of free space (8.854 x 10^-12 F/m) and c is the speed of light (3 x 10^8 m/s).
For the electric field amplitude, substitute the calculated intensity into the formula:
Electric field amplitude = √(Visible light intensity / (ε0 * c))
For the magnetic field amplitude, the relation is given by:
Magnetic field amplitude = Electric field amplitude / c
Evaluate these expressions to find the respective amplitudes of the electric and magnetic fields at the surface of the bulb.
By following these steps, you should be able to calculate both the visible light intensity and the amplitudes of the electric and magnetic fields.