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, we need to analyze the given equations for the electromagnetic wave and derive the relationships between the parameters involved: \( E_0 \), \( B_0 \), \( k \), and \( \omega \).

### Step-by-Step Solution:

1. **Identify the Given Equations**:
The electromagnetic wave is described by the equations:
\[ E = E_0 \sin(kx - \omega t) \]
\[ B = B_0 \sin(kx - \omega t) \]

2. **Understand the Relationship Between Electric and Magnetic Fields**:
In an electromagnetic wave, the ratio of the amplitudes of the electric field \( E_0 \) and the magnetic field \( B_0 \) is related to the speed of light \( c \):
\[ \frac{E_0}{B_0} = c \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

3. **Relate Wave Speed to Angular Frequency and Wave Number**:
The wave speed \( c \) can also be expressed in terms of the angular frequency \( \omega \) and the wave number \( k \):
\[ c = \frac{\omega}{k} \]

4. **Equate the Two Expressions for Wave Speed**:
Since both expressions represent the speed of light, we can set them equal to each other:
\[ \frac{E_0}{B_0} = \frac{\omega}{k} \]

5. **Cross Multiply to Derive a Relationship**:
Cross multiplying gives us:
\[ E_0 k = B_0 \omega \]

6. **Identify the Correct Option**:
Now we can analyze the options provided in the question:
- Option 1: \( E_0 k = B_0 \omega \) (Correct)
- Option 2: \( E_0 B_0 = \omega k \) (Incorrect)
- Option 3: \( E_0 \omega = B_0 k \) (Incorrect)
- Option 4: None of these (Incorrect)

Therefore, the correct answer is **Option 1**: \( E_0 k = B_0 \omega \).