A beam has intensity `2.5xx10^(14)Wm^(-2)`. The ratio of electric and magnetic fields in the beam is

Options

A. `2.98xx10^(8)NC^(-1)T^(-1)`
B. `8.32xx10^(5)NCT^(-1)`
C. `6.22xx10^(7)NC^(-1)T^(-1)`
D. `2.88xx10^(6)NCT^(-1)`

Answer

To find the ratio of the electric field (E) to the magnetic field (B) in an electromagnetic wave given the intensity (I) of the beam, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between intensity, electric field, and magnetic field:**
The intensity (I) of an electromagnetic wave is given by the formula:
\[ I = \frac{1}{2} \epsilon_0 c E^2 \]
where:
- \( I \) is the intensity,
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \, \text{m/s} \)),
- \( E \) is the electric field strength.

2. **Rearranging the formula to find E:**
From the intensity formula, we can rearrange to find \( E \):
\[ E = \sqrt{\frac{2I}{\epsilon_0 c}} \]

3. **Substituting the known values:**
Given \( I = 2.5 \times 10^{14} \, \text{W/m}^2 \):
\[ E = \sqrt{\frac{2 \times (2.5 \times 10^{14})}{(8.85 \times 10^{-12})(3 \times 10^8)}} \]

4. **Calculating the denominator:**
Calculate \( \epsilon_0 c \):
\[ \epsilon_0 c = (8.85 \times 10^{-12}) \times (3 \times 10^8) = 2.655 \times 10^{-3} \]

5. **Calculating E:**
Substitute back into the equation for \( E \):
\[ E = \sqrt{\frac{5 \times 10^{14}}{2.655 \times 10^{-3}}} \]
\[ E = \sqrt{1.885 \times 10^{17}} \approx 4.34 \times 10^8 \, \text{V/m} \]

6. **Finding the magnetic field (B):**
The relationship between the electric field and the magnetic field in an electromagnetic wave is given by:
\[ B = \frac{E}{c} \]
Substituting the values:
\[ B = \frac{4.34 \times 10^8}{3 \times 10^8} \approx 1.45 \, \text{T} \]

7. **Finding the ratio of E to B:**
The ratio of the electric field to the magnetic field is:
\[ \frac{E}{B} = \frac{4.34 \times 10^8}{1.45} \approx 3 \times 10^8 \, \text{(units: V/m per T)} \]

### Final Answer:
The ratio of the electric field to the magnetic field in the beam is approximately:
\[ \frac{E}{B} \approx 3 \times 10^8 \, \text{(V/m per T)} \]