A point source of electromagnetic radiation has an average power output of 800W. The mximum value of electric field at a distance 3.5 m from the source will be 62.6 V/m, the maximum value of magnetic field will be -

Options

A. `2.09 xx 10^(-5)` T
B. `2.09 xx 10^(-6)` T
C. `2.09 xx 10^(-7)` T
D. `2.09 xx 10^(-8)` T

Explanation

To find the maximum value of the magnetic field (B₀) at a distance of 3.5 m from a point source of electromagnetic radiation with a given average power output and maximum electric field, we can use the relationship between the electric field (E₀) and the magnetic field (B₀) in electromagnetic waves.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Average power output (P) = 800 W
- Maximum electric field (E₀) = 62.6 V/m
- Speed of light (c) = 3 × 10^8 m/s

2. **Use the Relationship Between E₀ and B₀:**
The relationship between the maximum electric field (E₀) and the maximum magnetic field (B₀) in an electromagnetic wave is given by:
\[ E₀ = c \cdot B₀ \]
Rearranging this equation to solve for B₀ gives:
\[ B₀ = \frac{E₀}{c} \]

3. **Substitute the Known Values:**
Substitute the known values of E₀ and c into the equation:
\[ B₀ = \frac{62.6 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

4. **Calculate B₀:**
Performing the calculation:
\[ B₀ = \frac{62.6}{3 \times 10^8} \approx 2.08667 \times 10^{-7} \, \text{T} \]

5. **Round to Appropriate Significant Figures:**
Rounding the answer to three significant figures gives:
\[ B₀ \approx 2.09 \times 10^{-7} \, \text{T} \]

### Final Answer:
The maximum value of the magnetic field (B₀) at a distance of 3.5 m from the source is approximately **2.09 × 10⁻⁷ T**.