Question

A point source of electromagnetic radiation has an average output power of $800 \mathrm{~W}$. The maximum value of electric field at a distance of $4.0 \mathrm{~m}$ from the source is (A) $64.7 \mathrm{Vm}^{-1}$ (B) $57.8 \mathrm{Vm}^{-1}$ (C) $56.72 \mathrm{Vm}^{-1}$ (D) $54.77 \mathrm{Vm}^{-1}$

Name: Problem 2171, Chapter 15: JEE Physics
Uploaded: 2021-10-02T10:46:36-08:00
Duration: 3 min 7 s
Channel: Ankur Khosla
Description: A point source of electromagnetic radiation has an average output power of 800 W. The maximum value of electric field at a distance of 4.0 m from the source is (A) 64.7 Vm^-1 (B) 57.8 Vm^-1 (C) 56.72 Vm^-1 (D) 54.77 Vm^-1

   A point source of electromagnetic radiation has an average output power of $800 \mathrm{~W}$. The maximum value of electric field at a distance of $4.0 \mathrm{~m}$ from the source is
(A) $64.7 \mathrm{Vm}^{-1}$
(B) $57.8 \mathrm{Vm}^{-1}$
(C) $56.72 \mathrm{Vm}^{-1}$
(D) $54.77 \mathrm{Vm}^{-1}$

JEE Physics

StemEZ 1st Edition

Chapter 15, Problem 2171

Instant Answer

Solved by Expert Ankur Khosla

10/02/2021

Step 1

0 m from the source. Show more…

Show all steps

Thanks for your feedback!

A point source of electromagnetic radiation has an average output power of $800 \mathrm{~W}$. The maximum value of electric field at a distance of $4.0 \mathrm{~m}$ from the source is (A) $64.7 \mathrm{Vm}^{-1}$ (B) $57.8 \mathrm{Vm}^{-1}$ (C) $56.72 \mathrm{Vm}^{-1}$ (D) $54.77 \mathrm{Vm}^{-1}$

Play audio

Feedback

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Intensity of Electromagnetic Waves

Intensity is defined as the power per unit area transmitted by an electromagnetic wave. For a point source emitting isotropically, it is calculated by dividing the total power by the surface area of the sphere over which the power is spread.

Electric Field Amplitude and the Poynting Vector

The Poynting vector relates the electromagnetic energy flow to the electric and magnetic fields. For a propagating wave in free space, the average intensity is related to the peak electric field by the equation I = (1/2)c??E?². This relation is used to compute the maximum electric field amplitude from the measured or given intensity.

Inverse Square Law

This law states that the intensity of radiation from a point source decreases as the square of the distance from the source, because the energy spreads out uniformly over the surface area of a sphere (which increases as 4?r²). This is key in determining how power is distributed in space.

Please give Ace some feedback