A plane electromagnetic wave of frequency `40 MHz` travels in free space in the `x` -direction. At some point and at some instant, the electric field `vec(E)` has its maximum value of 750 N/C in `y` -direction. The wavelength (in m) of the wave is

Answer

To find the wavelength of a plane electromagnetic wave, we can use the relationship between the speed of light, frequency, and wavelength. The formula is given by:

\[ c = f \cdot \lambda \]

Where:
- \( c \) is the speed of light in a vacuum (approximately \( 3 \times 10^8 \) m/s),
- \( f \) is the frequency of the wave,
- \( \lambda \) is the wavelength.

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Frequency \( f = 40 \text{ MHz} = 40 \times 10^6 \text{ Hz} \)
- Speed of light \( c = 3 \times 10^8 \text{ m/s} \)

**Step 2: Rearrange the formula to solve for wavelength \( \lambda \).**
\[ \lambda = \frac{c}{f} \]

**Step 3: Substitute the values into the equation.**
\[ \lambda = \frac{3 \times 10^8 \text{ m/s}}{40 \times 10^6 \text{ Hz}} \]

**Step 4: Perform the calculation.**
\[ \lambda = \frac{3 \times 10^8}{40 \times 10^6} = \frac{3}{40} \times 10^{8-6} = \frac{3}{40} \times 10^2 \]
\[ \lambda = \frac{3}{40} \times 100 = \frac{300}{40} = 7.5 \text{ m} \]

**Step 5: Finalize the answer.**
\[ \lambda = 7.5 \text{ m} \]

### Final Answer:
The wavelength of the wave is \( 7.5 \text{ m} \).