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?

Options

A. `120"N C"^(-1)`
B. `134"N C"^(-1)`
C. `510"N C"^(-1)`
D. `153"N C"^(-1)`

Answer

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 an electromagnetic wave in vacuum. The relationship is given by:

\[ E = c \cdot B \]

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

### Step-by-Step Solution:

1. **Identify the given values:**
- The amplitude of the magnetic field \( B_0 = 510 \, \text{nT} \).
- Convert this to Tesla:
\[ B_0 = 510 \, \text{nT} = 510 \times 10^{-9} \, \text{T} \]

2. **Use the speed of light:**
- The speed of light in vacuum \( c = 3 \times 10^8 \, \text{m/s} \).

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

4. **Calculate the electric field amplitude:**
- Perform the multiplication:
\[ E = 3 \times 10^8 \times 510 \times 10^{-9} \]
- Simplifying this:
\[ E = 3 \times 510 \times 10^{-1} = 1530 \times 10^{-1} = 153 \, \text{N/C} \]

5. **Conclusion:**
- The amplitude of the electric field part of the wave is:
\[ E = 153 \, \text{N/C} \]

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