A radiation is emitted by 100 watt bulb and it generates an electric field and magnetic field. 2.4% of the power is converted into radiation. Calculate the peak value of electric field (V/m) at a distance 2m from the bulb? (Take $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} / C^{2}$ and speed of light = $3 \times 10^{8} m / s$ )

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 6.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} Intensity, \frac{Power}{Area} = \frac{1}{2} ε_{0} E_{0}^{2} c \\ \frac{2.4}{100} \times \frac{100}{4 π {(2)}^{2}} = \frac{1}{2} ε_{0} E_{0}^{2} c \\ \Rightarrow E = 6 V / m \end{array}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!