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

Explanation

To find the peak value of the magnetic field (B₀) in an electromagnetic wave given the root mean square (RMS) value of the electric field (Eₘₛ), we can follow these steps:

### Step 1: Understand the relationship between RMS and peak values
The RMS value of the electric field (Eₘₛ) is related to the peak value of the electric field (E₀) by the formula:
\[ Eₘₛ = \frac{E₀}{\sqrt{2}} \]
Given that \(Eₘₛ = 6 \, \text{V/m}\), we can rearrange this formula to find \(E₀\):
\[ E₀ = Eₘₛ \times \sqrt{2} \]

### Step 2: Calculate the peak value of the electric field
Substituting the given value of \(Eₘₛ\):
\[ E₀ = 6 \, \text{V/m} \times \sqrt{2} \]
Calculating this gives:
\[ E₀ = 6 \times 1.414 \approx 8.49 \, \text{V/m} \]

### Step 3: Use the relationship between electric and magnetic fields
In an electromagnetic wave, the peak values of the electric field (E₀) and magnetic field (B₀) are related by the speed of light (c):
\[ B₀ = \frac{E₀}{c} \]
where \(c \approx 3 \times 10^8 \, \text{m/s}\).

### Step 4: Substitute the values to find B₀
Now substituting \(E₀\) and \(c\) into the equation:
\[ B₀ = \frac{8.49 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \]
Calculating this gives:
\[ B₀ \approx 2.83 \times 10^{-8} \, \text{T} \]

### Final Answer
Thus, the peak value of the magnetic field is approximately:
\[ B₀ \approx 2.83 \times 10^{-8} \, \text{T} \]