The average electric field of electromagnetic waves in certain region of free space is `9 xx 10^(–4) NC^(–1)`. Then the average magnetic field in the same region is of the order of

Options

A. `27 xx 10^(-4) T`
B. `3 xx 10^(-12) T`
C. `((1)/(3)) xx 10^(-12) T`
D. `3 xx 10^(12) T`

Explanation

To find the average magnetic field in the same region where the average electric field of electromagnetic waves is given, we can use the relationship between the electric field (E) and the magnetic field (B) in electromagnetic waves. The speed of light (c) is given by the equation:

\[ c = \frac{E}{B} \]

Where:
- \( E \) is the electric field strength,
- \( B \) is the magnetic field strength,
- \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Average electric field, \( E = 9 \times 10^{-4} \, \text{N/C} \)
- Speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Rearrange the formula to find B:**
\[ B = \frac{E}{c} \]

3. **Substitute the known values into the equation:**
\[ B = \frac{9 \times 10^{-4} \, \text{N/C}}{3 \times 10^8 \, \text{m/s}} \]

4. **Perform the calculation:**
- First, calculate the division:
\[ B = \frac{9}{3} \times \frac{10^{-4}}{10^8} = 3 \times 10^{-12} \, \text{T} \]

5. **Final answer:**
The average magnetic field \( B \) in the same region is \( 3 \times 10^{-12} \, \text{T} \).