The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is `B_0=510 nT`. What is the amplitude of the electric field part of the wave?

Explanation

To find the amplitude of the electric field part of a harmonic electromagnetic wave given the amplitude of the magnetic field, we can use the relationship between the electric field (E) and the magnetic field (B) in electromagnetic waves. The relationship is given by:

\[ E_0 = c \cdot B_0 \]

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The amplitude of the magnetic field \( B_0 = 510 \, \text{nT} \).
- Convert \( B_0 \) from nanotesla to tesla:
\[ B_0 = 510 \, \text{nT} = 510 \times 10^{-9} \, \text{T} \]

2. **Use the Relationship Between E and B:**
- The relationship is:
\[ E_0 = c \cdot B_0 \]

3. **Substitute the Values:**
- Substitute \( c = 3 \times 10^8 \, \text{m/s} \) and \( B_0 = 510 \times 10^{-9} \, \text{T} \):
\[ E_0 = (3 \times 10^8 \, \text{m/s}) \cdot (510 \times 10^{-9} \, \text{T}) \]

4. **Calculate the Electric Field Amplitude:**
- Perform the multiplication:
\[ E_0 = 3 \times 510 \times 10^{-1} \, \text{V/m} \]
- Calculate \( 3 \times 510 = 1530 \):
\[ E_0 = 1530 \times 10^{-1} \, \text{V/m} = 153 \, \text{V/m} \]

5. **Final Result:**
- The amplitude of the electric field part of the wave is:
\[ E_0 = 153 \, \text{V/m} \]

### Summary:
The amplitude of the electric field part of the wave is \( 153 \, \text{V/m} \).