In an electromagnetic wave in free space the root mean square value of the electric field is `E_(rms)=6 V//m`. The peak value of the magnetic field is

Options

A. `2.83xx10^(-8)T`
B. `0.70xx10^(-8)T`
C. `4.23xx10^(-8)T`
D. `1.41xx10^(-8)T`

Explanation

To find the peak value of the magnetic field \( B_0 \) in an electromagnetic wave given the root mean square (RMS) value of the electric field \( E_{\text{rms}} = 6 \, \text{V/m} \), we can follow these steps:

### Step 1: Relate \( E_{\text{rms}} \) to the peak electric field \( E_0 \)

The relationship between the RMS value of the electric field \( E_{\text{rms}} \) and the peak value \( E_0 \) is given by:

\[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} \]

### Step 2: Calculate the peak electric field \( E_0 \)

We can rearrange the equation to solve for \( E_0 \):

\[ E_0 = E_{\text{rms}} \times \sqrt{2} \]

Substituting the given value:

\[ E_0 = 6 \, \text{V/m} \times \sqrt{2} = 6\sqrt{2} \, \text{V/m} \]

### Step 3: Use the relationship between electric and magnetic fields

In an electromagnetic wave, the relationship between the peak electric field \( E_0 \) and the peak magnetic field \( B_0 \) is given by:

\[ \frac{E_0}{B_0} = c \]

where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step 4: Solve for the peak magnetic field \( B_0 \)

Rearranging the equation gives us:

\[ B_0 = \frac{E_0}{c} \]

Substituting the value of \( E_0 \) and \( c \):

\[ B_0 = \frac{6\sqrt{2} \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]

### Step 5: Calculate \( B_0 \)

Calculating the value:

\[ B_0 = \frac{6\sqrt{2}}{3 \times 10^8} \, \text{T} \]

\[ B_0 = \frac{2\sqrt{2}}{10^8} \, \text{T} \]

### Final Answer

Thus, the peak value of the magnetic field \( B_0 \) is:

\[ B_0 = \frac{2\sqrt{2}}{10^8} \, \text{T} \]