Askiitians Tutor Team
Last Activity: 5 Months ago
To determine the average Poynting vector 1.5 meters away from a 100-watt bulb that emits radiation equally in all directions, we need to understand a few concepts related to electromagnetic radiation and the Poynting vector itself.
Understanding the Poynting Vector
The Poynting vector, represented as **S**, describes the directional energy flux (the rate of energy transfer per unit area) of an electromagnetic field. It is calculated using the formula:
S = E × H
where **E** is the electric field strength and **H** is the magnetic field strength. However, for our purposes, we can simplify this by considering the intensity of the radiation emitted by the bulb.
Calculating the Intensity
Since the bulb emits radiation uniformly in all directions, we can treat it as a point source. The intensity **I** of the radiation at a distance **r** from the source can be calculated using the formula:
I = P / A
where **P** is the power of the bulb (100 watts) and **A** is the surface area of a sphere with radius **r** (1.5 meters in this case).
Surface Area of a Sphere
The surface area **A** of a sphere is given by:
A = 4πr²
Substituting **r = 1.5 m**:
A = 4π(1.5)² = 4π(2.25) = 9π ≈ 28.27 m²
Finding the Intensity
Now, we can calculate the intensity:
I = P / A = 100 W / 28.27 m² ≈ 3.53 W/m²
Relating Intensity to the Poynting Vector
The average Poynting vector can be directly related to the intensity of the radiation. In free space, the average magnitude of the Poynting vector is equal to the intensity:
|S| = I
Thus, the average Poynting vector at a distance of 1.5 meters from the bulb is:
|S| ≈ 3.53 W/m²
Summary of Findings
In summary, the average Poynting vector 1.5 meters away from a 100-watt bulb, which emits radiation uniformly in all directions, is approximately 3.53 W/m². This value represents the average energy flux per unit area at that distance, indicating how much energy is passing through a square meter of space around the bulb.