Find the value of Electric field amplitude if the Magnetic field's amplitude is `6 xx 10^8` T?

Options

A. `2 Vm^(-1)`
B. `74 xx 10^(3) Vm^(-1)`
C. `7 xx 10^(3) Vm^(-1)`
D. `17 xx 10^(3) Vm^(-1)`

Explanation

To find the value of the electric field amplitude (E₀) given the magnetic field amplitude (B₀), we can use the relationship between the electric field and magnetic field in an electromagnetic wave. The relationship is given by:

\[ E_0 = c \cdot B_0 \]

where:
- \( E_0 \) is the electric field amplitude,
- \( B_0 \) is the magnetic field amplitude,
- \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s.

### Step-by-Step Solution:

1. **Identify the given values**:
- Magnetic field amplitude \( B_0 = 6 \times 10^{-8} \) T.
- Speed of light \( c = 3 \times 10^8 \) m/s.

2. **Use the relationship to find E₀**:
\[ E_0 = c \cdot B_0 \]

3. **Substitute the values into the equation**:
\[ E_0 = (3 \times 10^8 \, \text{m/s}) \cdot (6 \times 10^{-8} \, \text{T}) \]

4. **Perform the multiplication**:
\[ E_0 = 3 \times 6 \times 10^{8 - 8} = 18 \times 10^{0} = 18 \, \text{V/m} \]

5. **Final Result**:
The electric field amplitude \( E_0 \) is \( 18 \, \text{V/m} \).