Question

The efficiency of a bulb of power \( 60 \) W is \( 16\% \). The peak value of the electric field produced by the electromagnetic radiation from the bulb at a distance of \( 2 \) m from the bulb is? \[ \left(\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2} \right) \]

• A. \( 24 \) V/m
• B. \( 16 \) V/m
• C. \( 9 \) V/m
• D. \( 12 \) V/m \

Hint:

To find the peak electric field of an electromagnetic wave, use the relation \( I = \frac{1}{2} \epsilon_0 c E_0^2 \), and solve for \( E_0 \).

Answer 1

The Correct Option is D

Step 1: Power radiated as electromagnetic waves 
The total power of the bulb is given as: \[ P = 60 W \] Since the efficiency of the bulb is \( 16\% \), the power converted into electromagnetic radiation is: \[ P_{\text{em}} = 0.16 \times 60 = 9.6 W \] 

Step 2: Intensity of the electromagnetic waves 
The intensity (\( I \)) of the electromagnetic radiation at a distance \( r \) from the source is given by: \[ I = \frac{P_{\text{em}}}{4 \pi r^2} \] Substituting the values: \[ I = \frac{9.6}{4 \pi (2)^2} \] \[ = \frac{9.6}{16 \pi} \] \[ = \frac{0.6}{\pi} \text{ W/m}^2 \] 

Step 3: Relation between intensity and peak electric field 
The intensity of an electromagnetic wave is related to the peak electric field \( E_0 \) by: \[ I = \frac{1}{2} \epsilon_0 c E_0^2 \] where: - \( \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{N}^{-1} \text{m}^{-2} \), - \( c = 3 \times 10^8 \) m/s. Rearranging for \( E_0 \): \[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \] Substituting values: \[ E_0 = \sqrt{\frac{2 \times \frac{0.6}{\pi}}{(8.85 \times 10^{-12}) (3 \times 10^8)}} \] Approximating: \[ E_0 \approx 12 \text{ V/m} \] Thus, the peak value of the electric field is 12 V/m.