An observer is 1.8 m from an isotropic point light source whose power is 250W. Calculate the rms value of the electric and magnetic fields due to the source at the position of the observer.

Explanation

To solve the problem of finding the RMS values of the electric and magnetic fields due to an isotropic point light source, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Power of the light source, \( P = 250 \, \text{W} \)
- Distance from the source to the observer, \( R = 1.8 \, \text{m} \)

2. **Calculate the Area of the Sphere:**
The area \( A \) over which the power is distributed is given by the surface area of a sphere:
\[ A = 4 \pi R^2 \]
Substituting the value of \( R \):
\[ A = 4 \pi (1.8)^2 \approx 4 \pi (3.24) \approx 40.6 \, \text{m}^2 \]

3. **Calculate the Intensity \( I \):**
The intensity \( I \) is defined as power per unit area:
\[ I = \frac{P}{A} \]
Substituting the values:
\[ I = \frac{250}{40.6} \approx 6.17 \, \text{W/m}^2 \]

4. **Relate Intensity to Electric Field:**
The intensity can also be expressed in terms of the RMS electric field \( E_{\text{RMS}} \):
\[ I = \frac{1}{2} \epsilon_0 c E_{\text{RMS}}^2 \]
Rearranging for \( E_{\text{RMS}} \):
\[ E_{\text{RMS}}^2 = \frac{2I}{\epsilon_0 c} \]
Thus,
\[ E_{\text{RMS}} = \sqrt{\frac{2I}{\epsilon_0 c}} \]

5. **Substituting Constants:**
- The permittivity of free space, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)
- The speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

Now substituting the values:
\[ E_{\text{RMS}} = \sqrt{\frac{2 \times 6.17}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]

6. **Calculating \( E_{\text{RMS}} \):**
First, calculate the denominator:
\[ \epsilon_0 c = 8.85 \times 10^{-12} \times 3 \times 10^8 \approx 2.655 \times 10^{-3} \]
Now substituting back:
\[ E_{\text{RMS}} = \sqrt{\frac{12.34}{2.655 \times 10^{-3}}} \approx \sqrt{4651.8} \approx 68.2 \, \text{V/m} \]

7. **Calculate the Magnetic Field \( B_{\text{RMS}} \):**
The relationship between the electric field and the magnetic field in electromagnetic waves is given by:
\[ B_{\text{RMS}} = \frac{E_{\text{RMS}}}{c} \]
Substituting the values:
\[ B_{\text{RMS}} = \frac{68.2}{3 \times 10^8} \approx 2.27 \times 10^{-7} \, \text{T} \]

### Final Results:
- RMS Electric Field, \( E_{\text{RMS}} \approx 68.2 \, \text{V/m} \)
- RMS Magnetic Field, \( B_{\text{RMS}} \approx 2.27 \times 10^{-7} \, \text{T} \)