An electromagnetic wave going through vacuum is described by
` E= E_0 sin(kx- omega t), B=B_0sin(kx-omega t)`.
Then

Options

A. `(E_0)k =(B_0)omega`
B. `(E_0) (B_0)=omega k`
C. `(E_0)omega = B_0 k `
D. none of these .

Answer

To solve the problem regarding the electromagnetic wave described by the equations \( E = E_0 \sin(kx - \omega t) \) and \( B = B_0 \sin(kx - \omega t) \), we need to analyze the relationships between the electric field \( E_0 \), magnetic field \( B_0 \), wave number \( k \), and angular frequency \( \omega \).

### Step-by-Step Solution:

1. **Understand the relationship between \( E_0 \), \( B_0 \), \( k \), and \( \omega \):**
The speed of light \( c \) in vacuum is given by the relationship:
\[ c = \frac{E_0}{B_0} \]
We also know that the wave speed can be expressed in terms of wave number \( k \) and angular frequency \( \omega \):
\[ c = \frac{\omega}{k} \]

2. **Set the two expressions for \( c \) equal to each other:**
Since both expressions represent the speed of light, we can equate them:
\[ \frac{E_0}{B_0} = \frac{\omega}{k} \]

3. **Cross-multiply to find a relationship between \( E_0 \), \( B_0 \), \( k \), and \( \omega \):**
Rearranging the equation gives:
\[ E_0 k = B_0 \omega \]
This shows that option 1, \( E_0 k = B_0 \omega \), is correct.

4. **Evaluate the other options:**
- **Option 2:** \( E_0 B_0 = \omega k \) - This does not hold true based on our derived relationship.
- **Option 3:** \( E_0 \omega = B_0 k \) - This is also incorrect as it does not match the derived relationship.
- **Option 4:** None of these - This is incorrect since we have found a valid relationship.

5. **Conclusion:**
The only correct expression from the options provided is:
\[ E_0 k = B_0 \omega \]

### Final Answer:
The correct option is **Option 1: \( E_0 k = B_0 \omega \)**.