The amplitude of the magnetic field part of harmonic electromagnetic wave in vacuum is `B_(0) = 510 nt`. What is the amplitude of the electric field part of the wave ?

Answer

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

\[ E₀ = c \cdot B₀ \]

where:
- \( E₀ \) is the amplitude of the electric field,
- \( B₀ \) is the amplitude of the magnetic field,
- \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s.

### Step-by-Step Solution:

1. **Identify the given values**:
- Amplitude of the magnetic field, \( B₀ = 510 \, \text{nT} = 510 \times 10^{-9} \, \text{T} \)
- Speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Use the relationship to find \( E₀ \)**:
\[ E₀ = c \cdot B₀ \]

3. **Substitute the values into the equation**:
\[ E₀ = (3 \times 10^8 \, \text{m/s}) \cdot (510 \times 10^{-9} \, \text{T}) \]

4. **Calculate \( E₀ \)**:
\[ E₀ = 3 \times 510 \times 10^{-1} \, \text{N/C} \]
\[ E₀ = 1530 \times 10^{-1} \, \text{N/C} \]
\[ E₀ = 153 \, \text{N/C} \]

5. **Final Answer**:
The amplitude of the electric field part of the wave is \( E₀ = 153 \, \text{N/C} \).