A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magentic field is given by `vecB = 5 xx 10^(-8) hati T`. The corresponding electric field E is (speed of light `c = 3 xx 10^(8)ms^(-1)`

Options

A. `1.66xx 10^(-16) hat i V//M`
B. `-15 hati V//m`
C. `-1.66 xx 10^(-16) hati V//m`
D. `15hati V//m`

Answer

To find the corresponding electric field \( \vec{E} \) for the given magnetic field \( \vec{B} \) of an electromagnetic wave propagating in vacuum, we can use the relationship between the electric field and magnetic field in an electromagnetic wave.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Frequency of the wave, \( f = 25 \, \text{GHz} = 25 \times 10^9 \, \text{Hz} \)
- Magnetic field, \( \vec{B} = 5 \times 10^{-8} \, \hat{i} \, \text{T} \)
- Speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Determine the Direction of Propagation:**
- The wave is propagating along the z-direction. Therefore, the wave vector \( \vec{k} \) points in the \( \hat{k} \) direction.

3. **Use the Relationship Between Electric Field and Magnetic Field:**
- The relationship between the electric field \( \vec{E} \) and the magnetic field \( \vec{B} \) in an electromagnetic wave is given by:
\[ \vec{E} = c \cdot \vec{B} \times \hat{k} \]
- Here, \( \hat{k} \) is the unit vector in the direction of propagation (z-direction).

4. **Calculate the Cross Product:**
- Given \( \vec{B} = 5 \times 10^{-8} \, \hat{i} \):
\[ \vec{B} \times \hat{k} = (5 \times 10^{-8} \, \hat{i}) \times \hat{k} \]
- Using the right-hand rule, \( \hat{i} \times \hat{k} = -\hat{j} \):
\[ \vec{B} \times \hat{k} = 5 \times 10^{-8} (-\hat{j}) = -5 \times 10^{-8} \hat{j} \]

5. **Calculate the Electric Field:**
- Now, substitute into the equation for \( \vec{E} \):
\[ \vec{E} = c \cdot \vec{B} \times \hat{k} = 3 \times 10^8 \, \text{m/s} \cdot (-5 \times 10^{-8} \hat{j}) \]
- This simplifies to:
\[ \vec{E} = -15 \, \hat{j} \, \text{V/m} \]

6. **Final Result:**
- Therefore, the corresponding electric field \( \vec{E} \) is:
\[ \vec{E} = 15 \, \hat{j} \, \text{V/m} \]