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 a plane electromagnetic wave propagating in vacuum, we can use the relationship between the electric field, magnetic field, and the direction of propagation.

### Step-by-step Solution:

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

2. **Determine the Wavelength**:
The wavelength \( \lambda \) of the electromagnetic wave can be calculated using the formula:
\[ \lambda = \frac{c}{f} \]
Substituting the values:
\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{25 \times 10^9 \, \text{Hz}} = \frac{3}{25} \times 10^{-1} \, \text{m} = 0.012 \, \text{m} = 1.2 \, \text{cm} \]

3. **Use the Relationship Between \( \vec{E} \) and \( \vec{B} \)**:
In an electromagnetic wave, the electric field \( \vec{E} \), magnetic field \( \vec{B} \), and the direction of propagation are related by:
\[ \vec{E} = c \vec{B} \times \hat{k} \]
where \( \hat{k} \) is the unit vector in the direction of propagation (which is along the z-axis, \( \hat{k} = \hat{z} \)).

4. **Calculate \( \vec{E} \)**:
Given \( \vec{B} = 5 \times 10^{-8} \, \hat{j} \), we can compute \( \vec{E} \):
\[ \vec{E} = c \vec{B} \times \hat{k} = (3 \times 10^8) (5 \times 10^{-8} \, \hat{j}) \times \hat{k} \]
Using the right-hand rule for the cross product:
\[ \hat{j} \times \hat{k} = \hat{i} \]
Thus,
\[ \vec{E} = 3 \times 10^8 \times 5 \times 10^{-8} \hat{i} = 15 \hat{i} \, \text{V/m} \]

5. **Final Result**:
The corresponding electric field \( \vec{E} \) is:
\[ \vec{E} = 15 \hat{i} \, \text{V/m} \]