Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric filed at a distance 4.0 m from the source is

Options:

64.7 Vm−1

57.8 Vm−1

56.72 Vm−1

5477 Vm−1

Correct Answer:

5477 Vm−1

Explanation:

Intensity of electromagnetic wave is $ I=\frac{P_{av}}{4\pi × r^2}=\frac{E^2_0}{2\mu_0c}$

$ or \, E_0= \sqrt{\frac{\mu_0cP_{av}}{2\pi r^2}}$

$=\sqrt{\frac{(4\pi ×10^{-7})×(3×10^8)×800 }{2\pi ×(4)^2}}$

$ = 54.77 Vm^{-1}$