Assume a bulb of efficiency `2.5%` as a point source. The peak values of electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m is respectively

Options

A. `2.5"V m"^(-1), 3.6xx10^(-8)T`
B. `4.2"V m"^(-1), 2.8xx10^(-8)T`
C. `4.08"V m"^(-1), 1.36xx10^(-8)T`
D. `3.6"V m"^(-1), 4.2xx10^(-8)T`

Explanation

To solve the problem of finding the peak values of electric and magnetic fields produced by a 100 W bulb with an efficiency of 2.5% at a distance of 3 m, we will follow these steps:

### Step 1: Calculate the Effective Power Output
Given the efficiency of the bulb is 2.5%, we first need to calculate the effective power output of the bulb.

\[ \text{Effective Power} = \text{Total Power} \times \text{Efficiency} = 100 \, \text{W} \times 0.025 = 2.5 \, \text{W} \]

### Step 2: Calculate the Intensity at a Distance of 3 m
Since the bulb is treated as a point source, the intensity \( I \) at a distance \( r \) can be calculated using the formula:

\[ I = \frac{P}{A} \]

where \( A \) is the surface area of a sphere given by \( A = 4 \pi r^2 \).

For \( r = 3 \, \text{m} \):

\[ A = 4 \pi (3)^2 = 36 \pi \, \text{m}^2 \]

Now, substituting the values:

\[ I = \frac{2.5 \, \text{W}}{36 \pi} \approx \frac{2.5}{113.097} \approx 0.022 \, \text{W/m}^2 \]

### Step 3: Calculate the Intensity due to Electric Field
Since the intensity is equally distributed between the electric and magnetic fields, we have:

\[ I_e = \frac{I}{2} = \frac{0.022}{2} = 0.011 \, \text{W/m}^2 \]

### Step 4: Calculate the Peak Electric Field \( E_0 \)
The intensity due to the electric field is given by:

\[ I_e = \frac{1}{2} \epsilon_0 c E_0^2 \]

Rearranging for \( E_0 \):

\[ E_0 = \sqrt{\frac{2 I_e}{\epsilon_0 c}} \]

Substituting the values where \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) and \( c = 3 \times 10^8 \, \text{m/s} \):

\[ E_0 = \sqrt{\frac{2 \times 0.011}{8.85 \times 10^{-12} \times 3 \times 10^8}} \]

Calculating the denominator:

\[ 8.85 \times 10^{-12} \times 3 \times 10^8 \approx 2.655 \times 10^{-3} \]

Now substituting back:

\[ E_0 = \sqrt{\frac{0.022}{2.655 \times 10^{-3}}} \approx \sqrt{8.29} \approx 2.88 \, \text{V/m} \]

### Step 5: Convert RMS to Peak Value
The peak value \( E_0 \) is related to the RMS value \( E_{rms} \) by:

\[ E_0 = \sqrt{2} \times E_{rms} \approx 1.41 \times 2.88 \approx 4.08 \, \text{V/m} \]

### Step 6: Calculate the Peak Magnetic Field \( B_0 \)
The peak magnetic field can be calculated using the relationship:

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

Substituting the values:

\[ B_0 = \frac{4.08}{3 \times 10^8} \approx 1.36 \times 10^{-8} \, \text{T} \]

### Final Results
- Peak Electric Field \( E_0 \approx 4.08 \, \text{V/m} \)
- Peak Magnetic Field \( B_0 \approx 1.36 \times 10^{-8} \, \text{T} \)

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